Explanation:
Okay, so I need to find the value of $ a_1 $ given that the sequence is strictly increasing, and for every positive integer $ k $, the subsequence $ a_{2k-1}, a_{2k}, a_{2k+1} $ is geometric, and the subsequence $ a_{2k}, a_{2k+1}, a_{2k+2} $ is arithmetic. Also, $ a_{13} = 2016 $.

First, let me make sure I understand the problem correctly. The sequence is strictly increasing, so each term is larger than the previous one. Then, for any $ k $, if I take the terms at positions $ 2k-1 $, $ 2k $, $ 2k+1 $, they form a geometric sequence. Similarly, the terms at positions $ 2k $, $ 2k+1 $, $ 2k+2 $ form an arithmetic sequence.

So, for example, when $ k = 1 $, the first three terms $ a_1, a_2, a_3 $ form a geometric sequence. Then, the terms $ a_2, a_3, a_4 $ form an arithmetic sequence. Similarly, for $ k = 2 $, the terms $ a_3, a_4, a_5 $ form a geometric sequence, and $ a_4, a_5, a_6 $ form an arithmetic sequence, and so on.

Since the sequence is strictly increasing, all the terms are positive integers and each term is larger than the previous.

Let me try to write down the relations for the first few terms and see if I can find a pattern or recurrence relation.

Starting with $ k = 1 $:

1. $ a_1, a_2, a_3 $ is geometric. So, $ a_2^2 = a_1 \cdot a_3 $.
2. $ a_2, a_3, a_4 $ is arithmetic. So, $ 2a_3 = a_2 + a_4 $. Therefore, $ a_4 = 2a_3 - a_2 $.

For $ k = 2 $:

1. $ a_3, a_4, a_5 $ is geometric. So, $ a_4^2 = a_3 \cdot a_5 $. Therefore, $ a_5 = \frac{a_4^2}{a_3} $.
2. $ a_4, a_5, a_6 $ is arithmetic. So, $ 2a_5 = a_4 + a_6 $. Therefore, $ a_6 = 2a_5 - a_4 $.

For $ k = 3 $:

1. $ a_5, a_6, a_7 $ is geometric. So, $ a_6^2 = a_5 \cdot a_7 $. Therefore, $ a_7 = \frac{a_6^2}{a_5} $.
2. $ a_6, a_7, a_8 $ is arithmetic. So, $ 2a_7 = a_6 + a_8 $. Therefore, $ a_8 = 2a_7 - a_6 $.

And so on. It seems like there's a pattern here. Each odd-indexed term (starting from $ a_1 $) is part of a geometric sequence with the next two terms, and then the following term is determined by the arithmetic sequence.

But since the sequence is strictly increasing, all these terms must satisfy $ a_1 < a_2 < a_3 < a_4 < \dots $. Also, since they are integers, the geometric sequences must have integer ratios, and the arithmetic sequences must have integer differences.

But let me think about how to express the terms in terms of previous terms.

Let me try to express $ a_3 $, $ a_4 $, $ a_5 $, etc., in terms of $ a_1 $ and $ a_2 $. Let me define variables for each term. Let me denote $ a_1 = x $, $ a_2 = y $. Then, since $ a_1, a_2, a_3 $ is geometric, $ a_3 = \frac{a_2^2}{a_1} = \frac{y^2}{x} $. But since $ a_3 $ must be an integer, $ x $ divides $ y^2 $.

Similarly, from the arithmetic sequence $ a_2, a_3, a_4 $, $ a_4 = 2a_3 - a_2 = 2 \cdot \frac{y^2}{x} - y $.

Then, moving to the next geometric sequence $ a_3, a_4, a_5 $. Since this is geometric, $ a_4^2 = a_3 \cdot a_5 $, so $ a_5 = \frac{a_4^2}{a_3} $. Let me compute that:

$ a_5 = \frac{(2 \cdot \frac{y^2}{x} - y)^2}{\frac{y^2}{x}} $

Let me simplify that:

First, compute numerator:

$ (2 \cdot \frac{y^2}{x} - y)^2 = \left( \frac{2y^2 - xy}{x} \right)^2 = \frac{(2y^2 - xy)^2}{x^2} $

Then divide by $ \frac{y^2}{x} $:

$ \frac{(2y^2 - xy)^2}{x^2} \div \frac{y^2}{x} = \frac{(2y^2 - xy)^2}{x^2} \cdot \frac{x}{y^2} = \frac{(2y^2 - xy)^2}{x y^2} $

Simplify numerator:

Factor out terms:

Let me factor $ 2y^2 - xy = y(2y - x) $

Therefore:

$ \frac{(y(2y - x))^2}{x y^2} = \frac{y^2 (2y - x)^2}{x y^2} = \frac{(2y - x)^2}{x} $

So $ a_5 = \frac{(2y - x)^2}{x} $

Since $ a_5 $ must be an integer, $ x $ divides $ (2y - x)^2 $.

Similarly, moving on, we can compute $ a_6 $ from the arithmetic sequence $ a_4, a_5, a_6 $:

$ a_6 = 2a_5 - a_4 = 2 \cdot \frac{(2y - x)^2}{x} - \left( 2 \cdot \frac{y^2}{x} - y \right) $

Let me compute this:

First term: $ 2 \cdot \frac{(2y - x)^2}{x} $

Second term: $ 2 \cdot \frac{y^2}{x} - y = \frac{2y^2 - xy}{x} $

Therefore:

$ a_6 = \frac{2(2y - x)^2}{x} - \frac{2y^2 - xy}{x} = \frac{2(2y - x)^2 - (2y^2 - xy)}{x} $

Let me expand $ 2(2y - x)^2 $:

First compute $ (2y - x)^2 = 4y^2 - 4xy + x^2 $

Multiply by 2: 8y² - 8xy + 2x²

Subtract $ (2y² - xy) $:

So total numerator:

$ 8y² - 8xy + 2x² - 2y² + xy = (8y² - 2y²) + (-8xy + xy) + 2x² = 6y² -7xy + 2x² $

Therefore, $ a_6 = \frac{6y² -7xy + 2x²}{x} = \frac{6y²}{x} -7y + 2x $

Hmm. So $ a_6 = \frac{6y^2 -7xy + 2x^2}{x} $. Let me check that again:

Wait, the numerator was 6y² -7xy +2x². So, dividing by x:

That's (6y²)/x - 7y + 2x. Correct.

Since $ a_6 $ must be an integer, so x divides 6y² -7xy +2x². Which is the same as x divides 6y² -7xy +2x².

But maybe we can factor this expression?

Let me try to factor 6y² -7xy +2x². Let me think of it as quadratic in y:

6y² -7x y +2x².

Let me try to factor:

Looking for two numbers a and b such that a*b = 6*2x² = 12x² and a + b = -7x. Wait, but maybe not. Alternatively, factor as (Ay + Bx)(Cy + Dx). Let me try:

Multiply out (Ay + Bx)(Cy + Dx) = AC y² + (AD + BC)xy + BD x². So we need AC = 6, BD = 2, and AD + BC = -7.

Possible A and C: factors of 6: 6 and 1, 3 and 2. Similarly, BD = 2: possible B and D: 2 and 1, 1 and 2, or negatives. Since coefficients are integers, and since the sequence is increasing, maybe positive coefficients?

Wait, but since the sequence is strictly increasing, and a1 < a2 < a3 < ..., so x < y < a3 < a4 < ... So x and y are positive integers, with y > x.

Let me try factors. Let me assume A and C positive, B and D positive.

Let me try AC = 6. Let's take A=3, C=2. Then BD = 2. Let me try B=2, D=1. Then AD + BC = 3*1 + 2*2 = 3 + 4 = 7. But we need AD + BC = -7. So if I take negative signs. Wait, but if we have BD=2, and if B and D are negative, but since y and x are positive, maybe the coefficients can be negative? But the overall expression is 6y² -7xy + 2x², which is positive? Not sure. Wait, if x and y are positive, then 6y² -7xy +2x² might be positive or negative? Well, since the sequence is increasing, so for example, a_6 must be larger than a_5, which must be larger than a_4, etc. So perhaps the expression 6y² -7xy +2x² is positive?

But regardless, let me check. Suppose that 6y² -7xy +2x² factors as (3y - 2x)(2y - x). Let me check:

(3y - 2x)(2y - x) = 6y² -3xy -4xy + 2x² = 6y² -7xy +2x². Yes! Perfect.

So 6y² -7xy +2x² = (3y -2x)(2y -x). Therefore, $ a_6 = \frac{(3y - 2x)(2y -x)}{x} $.

Therefore, $ a_6 = \frac{(3y -2x)(2y -x)}{x} $.

Since $ a_6 $ must be an integer, $ x $ divides $ (3y -2x)(2y -x) $.

But since earlier, we had that $ x $ divides $ y^2 $ from the geometric sequence $ a_1, a_2, a_3 $. Let me recall that $ a_3 = y^2 / x $, so x divides y². Let me denote that as x | y². So maybe we can write y = x * m, where m is a rational number? But since y and x are integers, m must be rational. However, since x divides y², then if we write m as a reduced fraction, say m = p/q where p and q are coprime integers, then x must be divisible by q².

Alternatively, maybe it's better to suppose that x divides y², so let me write y = kx, where k is a rational number. Wait, but since y is an integer, and x is an integer, then if x divides y², then y² = x * n for some integer n.

Alternatively, maybe we can let d = gcd(x, y), and write x = d * m, y = d * n, where m and n are coprime positive integers. Then since x divides y² = d² n², so x divides d² n². Since x = d m, then d m divides d² n², which implies m divides d n². Since m and n are coprime, m divides d. Let me denote d = m * t, so that x = d m = m^2 t, and y = d n = m t n. Therefore, since m and n are coprime, and t is a positive integer.

So, substituting back, we can write:

x = m² t

y = m t n

With m, n, t positive integers, gcd(m, n) = 1.

This substitution might help in simplifying expressions. Let me try this substitution.

So, let me define variables:

Let $ x = m^2 t $

$ y = m t n $

Where $ \gcd(m, n) = 1 $, and $ m, t, n $ are positive integers.

Now, since $ a_3 = y² / x = (m² t² n²) / (m² t) ) = t n² $. So $ a_3 = t n² $.

Similarly, $ a_4 = 2 a_3 - y = 2 t n² - m t n = t n (2n - m) $.

Then $ a_5 = (2y -x)^2 / x = (2 m t n - m² t)^2 / (m² t ) = [ m t (2n - m) ]² / (m² t ) = m² t² (2n - m)^2 / (m² t ) = t (2n - m)^2 $.

So $ a_5 = t (2n - m)^2 $.

Then $ a_6 = (3y - 2x)(2y -x)/x $. Let me compute that:

First compute 3y - 2x = 3 m t n - 2 m² t = m t (3n - 2m)

Similarly, 2y - x = 2 m t n - m² t = m t (2n - m)

Therefore, (3y - 2x)(2y -x) = m t (3n - 2m) * m t (2n - m) = m² t² (3n - 2m)(2n - m)

Then divide by x = m² t:

So $ a_6 = [ m² t² (3n - 2m)(2n - m) ] / [ m² t ] = t (3n - 2m)(2n - m) $

Therefore, $ a_6 = t (3n - 2m)(2n - m) $

Similarly, let me check whether these expressions make sense. Since the sequence is strictly increasing, we can derive some inequalities.

First, since the sequence is strictly increasing:

$ a_1 < a_2 < a_3 < a_4 < a_5 < a_6 < ... $

So, $ x = m² t < y = m t n $. Therefore, m² t < m t n ⇒ m < n. Since m and n are positive integers, m < n.

Also, $ a_3 = t n² > y = m t n $ ⇒ t n² > m t n ⇒ n > m. Which is consistent with m < n.

Then $ a_4 = t n (2n - m) > a_3 = t n² $. Therefore:

$ t n (2n - m) > t n² ⇒ 2n - m > n ⇒ 2n - m > n ⇒ n > m $, which is again true.

Next, $ a_5 = t (2n - m)^2 $. Compare with $ a_4 = t n (2n - m) $. So:

$ t (2n - m)^2 > t n (2n - m) ⇒ (2n - m)^2 > n (2n - m) $, since 2n - m is positive? Let me check. Since the sequence is increasing, and a_4 is t n (2n - m), which must be greater than a_3 = t n². So 2n - m > n ⇒ n > m as before. Therefore 2n - m is positive. Thus, we can divide both sides by (2n - m):

$ (2n - m) > n ⇒ 2n - m > n ⇒ n > m $, which is true.

So that's okay.

Similarly, $ a_6 = t(3n - 2m)(2n - m) $. Compare with a_5 = t (2n - m)^2. So:

a_6 > a_5 ⇒ t(3n - 2m)(2n - m) > t (2n - m)^2 ⇒ (3n - 2m) > 2n - m ⇒ 3n - 2m > 2n - m ⇒ n > m. Which is true.

So all the inequalities so far hold as long as m < n.

So, with that substitution, we can perhaps find a pattern. Let me see if I can find expressions for subsequent terms.

But maybe instead of going further, I need to find a general formula for the terms. Let me see.

Alternatively, maybe there's a recursive relation here. Let me check. Let me note that each time, the odd terms are determined by the previous terms, and the even terms as well.

Alternatively, since the sequence alternates between geometric and arithmetic sequences, perhaps there's a recurrence relation.

But let me see. Let me try to compute $ a_7 $, which is part of the geometric sequence $ a_5, a_6, a_7 $. So:

$ a_6^2 = a_5 \cdot a_7 ⇒ a_7 = a_6² / a_5 $.

From above, $ a_5 = t (2n - m)^2 $, $ a_6 = t (3n - 2m)(2n - m) $.

Therefore, $ a_7 = [ t^2 (3n - 2m)^2 (2n - m)^2 ] / [ t (2n - m)^2 ] = t (3n - 2m)^2 $.

So $ a_7 = t (3n - 2m)^2 $.

Similarly, then $ a_8 $ is determined by the arithmetic sequence $ a_6, a_7, a_8 $. So:

$ 2a_7 = a_6 + a_8 ⇒ a_8 = 2a_7 - a_6 = 2 t (3n - 2m)^2 - t (3n - 2m)(2n - m) $.

Factor out t (3n - 2m):

$ a_8 = t (3n - 2m) [ 2(3n - 2m) - (2n - m) ] $

Compute inside the brackets:

$ 2(3n - 2m) - (2n - m) = 6n - 4m - 2n + m = 4n - 3m $

Thus, $ a_8 = t (3n - 2m)(4n - 3m) $

Similarly, moving on to the next geometric sequence $ a_7, a_8, a_9 $:

$ a_8^2 = a_7 \cdot a_9 ⇒ a_9 = a_8² / a_7 $

Let me compute that:

$ a_8 = t (3n - 2m)(4n - 3m) $

$ a_7 = t (3n - 2m)^2 $

Therefore:

$ a_9 = [ t^2 (3n - 2m)^2 (4n - 3m)^2 ] / [ t (3n - 2m)^2 ] = t (4n - 3m)^2 $

Similarly, moving on to the arithmetic sequence $ a_8, a_9, a_{10} $:

$ a_{10} = 2a_9 - a_8 = 2 t (4n - 3m)^2 - t (3n - 2m)(4n - 3m) $

Factor out t (4n - 3m):

$ a_{10} = t (4n - 3m) [ 2(4n - 3m) - (3n - 2m) ] $

Compute inside the brackets:

$ 2(4n - 3m) - (3n - 2m) = 8n - 6m - 3n + 2m = 5n - 4m $

Thus, $ a_{10} = t (4n - 3m)(5n - 4m) $

Hmm, I see a pattern here. It seems like for the odd terms, $ a_{2k+1} $, they are of the form $ t \times (\text{some linear expression in } n \text{ and } m)^2 $, and the even terms $ a_{2k} $ are of the form $ t \times (\text{some linear expression in } n \text{ and } m) \times (\text{another linear expression}) $.

Looking at the expressions for $ a_3, a_5, a_7, a_9 $:

- $ a_3 = t n^2 $
- $ a_5 = t (2n - m)^2 $
- $ a_7 = t (3n - 2m)^2 $
- $ a_9 = t (4n - 3m)^2 $

Similarly, even terms:

- $ a_4 = t n(2n - m) $
- $ a_6 = t(3n - 2m)(2n - m) $
- $ a_8 = t(3n - 2m)(4n - 3m) $
- $ a_{10} = t(4n - 3m)(5n - 4m) $

So, it seems like the odd-indexed terms follow a pattern where each term is $ t \times (kn - (k-1)m)^2 $, where k increases by 1 each time. Starting with k=1: $ a_3 = t(n)^2 $, then k=2: $ t(2n - m)^2 $, then k=3: $ t(3n - 2m)^2 $, etc.

Similarly, for even-indexed terms, they are products of consecutive expressions. For example, $ a_4 = t \times n \times (2n - m) $, $ a_6 = t \times (2n - m) \times (3n - 2m) $, $ a_8 = t \times (3n - 2m) \times (4n - 3m) $, $ a_{10} = t \times (4n - 3m) \times (5n - 4m) $, and so on.

This suggests that for the odd term $ a_{2k+1} $, it's $ t \times (kn - (k - 1)m)^2 $, and for the even term $ a_{2k} $, it's $ t \times (kn - (k - 1)m)( (k + 1)n - km ) $.

Let me check for $ k = 1 $:

- $ a_3 $, which is $ a_{2*1 + 1} $, so k = 1: $ t*(1n - 0m)^2 = t n² $. Correct.
- $ a_4 $, which is $ a_{2*1} $: t*(1n - 0m)*(2n - 1m) = t*n*(2n - m). Correct.

Similarly for $ k = 2 $:

- $ a_5 $: t*(2n - 1m)^2. Correct.
- $ a_6 $: t*(2n - m)*(3n - 2m). Correct.

So this seems like a general formula. Therefore, we can express $ a_{2k} $ as $ t \times (kn - (k - 1)m) \times ((k + 1)n - km) $, and $ a_{2k + 1} $ as $ t \times (kn - (k - 1)m)^2 $.

Therefore, if I can find expressions for $ a_{13} $, which is an odd-indexed term. Since 13 is odd, let me see. Let me figure out which term $ a_{13} $ corresponds to.

The odd indices are 1, 3, 5, 7, 9, 11, 13,... So $ a_{13} $ is the 7th odd term. Let me check:

For the formula $ a_{2k + 1} = t [k n - (k - 1)m]^2 $. Wait, when k=1, we get a_3. So if $ a_{2k + 1} $, then for k=1: a_3, k=2: a_5, k=3: a_7, k=4: a_9, k=5: a_{11}, k=6: a_{13}. Therefore, for $ a_{13} $, k = 6.

Therefore, $ a_{13} = t [6n - 5m]^2 $.

Given that $ a_{13} = 2016 $, so:

$ t (6n - 5m)^2 = 2016 $.

So, I need to find integers m, n, t with m < n, and $ \gcd(m, n) = 1 $, such that $ t (6n -5m)^2 = 2016 $. Also, since the sequence is strictly increasing, all the terms must be positive integers, and we have previous relations.

Additionally, since $ a_1 = x = m² t $, so once I find m, n, t, I can compute $ a_1 $.

Therefore, my task reduces to solving $ t (6n -5m)^2 = 2016 $, with the constraints that m, n, t are positive integers, m < n, $ \gcd(m, n) = 1 $, and also ensuring that all previous terms are integers and strictly increasing.

But first, let's factorize 2016 to see possible values for t and $ (6n -5m)^2 $.

2016 factors as:

Let me compute 2016:

Divide by 2: 2016 / 2 = 1008

/2 = 504

/2 = 252

/2 = 126

/2 = 63

So 2^5 * 63.

63 is 7*9 = 7*3².

So overall, prime factorization of 2016 is 2^5 * 3² * 7^1.

Therefore, 2016 = 2^5 * 3² * 7.

Now, since $ t \times (6n - 5m)^2 = 2016 $, so (6n -5m)^2 must be a divisor of 2016. Let me denote $ k = 6n -5m $, then $ k^2 $ divides 2016, so $ k^2 $ must be a square divisor of 2016.

Therefore, first, I need to find all square divisors of 2016.

The square divisors of 2016 are of the form (product of primes in exponents even numbers less than or equal to their exponents in 2016).

Given the prime factorization 2^5 * 3² * 7^1.

For square divisors, exponents for each prime must be even numbers:

- For prime 2: exponent can be 0, 2, 4 (since original exponent is 5, max even exponent less than or equal to 5 is 4)
- For prime 3: exponent can be 0, 2 (original exponent is 2)
- For prime 7: exponent must be 0 (since original exponent is 1, which is odd, so maximum even exponent is 0)

Therefore, square divisors of 2016 are:

- 2^0 * 3^0 * 7^0 = 1
- 2^2 * 3^0 * 7^0 = 4
- 2^4 * 3^0 * 7^0 = 16
- 2^0 * 3^2 * 7^0 = 9
- 2^2 * 3^2 * 7^0 = 4*9 = 36
- 2^4 * 3^2 * 7^0 = 16*9 = 144

Therefore, the square divisors of 2016 are 1, 4, 9, 16, 36, 144. Therefore, $ k^2 $ must be one of these, so k is 1, 2, 3, 4, 6, 12.

Therefore, possible values of k are 1,2,3,4,6,12. Therefore:

Case 1: $ k = 1 $, then $ (6n -5m)^2 = 1 $, so $ 6n -5m = 1 $, and t = 2016 / 1 = 2016.

Case 2: $ k = 2 $, then $ (6n -5m)^2 = 4 $, so $ 6n -5m = 2 $, t = 2016 / 4 = 504.

Case 3: $ k = 3 $, $ (6n -5m)^2 = 9 $, so $ 6n -5m = 3 $, t = 2016 / 9 = 224.

Case 4: $ k = 4 $, $ (6n -5m)^2 = 16 $, so $ 6n -5m = 4 $, t = 2016 / 16 = 126.

Case 5: $ k = 6 $, $ (6n -5m)^2 = 36 $, so $ 6n -5m = 6 $, t = 2016 / 36 = 56.

Case 6: $ k = 12 $, $ (6n -5m)^2 = 144 $, so $ 6n -5m = 12 $, t = 2016 / 144 = 14.

Therefore, these are the possible cases. Now, for each case, I need to check if there exist positive integers m, n with m < n, gcd(m,n) = 1, and also ensuring that the previous terms are strictly increasing and satisfy the required properties.

Let me analyze each case one by one.

Case 1: k = 1, so 6n -5m = 1, t = 2016

So, equation: 6n -5m = 1 ⇒ 6n = 5m +1 ⇒ n = (5m +1)/6. Since n must be an integer, 5m +1 must be divisible by 6. Let me solve for m:

5m +1 ≡ 0 mod 6 ⇒ 5m ≡ -1 mod 6 ⇒ 5m ≡ 5 mod 6 ⇒ m ≡ 1 mod 6 (since 5 is invertible modulo 6, inverse of 5 mod 6 is 5, because 5*5=25≡1 mod6). Therefore, m = 6t +1 for some integer t ≥ 0. Since m must be a positive integer, t ≥ 0. Let me check for t=0: m=1, n=(5*1 +1)/6 = 6/6 =1. But m=1, n=1. However, m < n is required. But here m = n = 1, which violates m < n. Therefore, t=0 is invalid. Next, t=1: m=7, n=(5*7 +1)/6 = (35 +1)/6=36/6=6. So m=7, n=6. But m=7, n=6 ⇒ m > n, which is invalid. Wait, that can't be. Wait, m = 6t +1, for t=1: m=6*1 +1=7. Then n=(5*7 +1)/6=36/6=6. So m=7, n=6. But m > n, which violates m < n. Therefore, invalid. Next t=2: m=6*2 +1=13, then n=(5*13 +1)/6=(65 +1)/6=66/6=11. So m=13, n=11 ⇒ m > n. Same problem. Similarly, t=3: m=19, n=(5*19 +1)/6=(95 +1)/6=96/6=16, m=19 > n=16. Not good. Wait, so this seems like for all t ≥0, m = 6t +1 and n=(5m +1)/6. Let me check if n can be greater than m? Let me write the equation 6n -5m =1. So, rearranged as 6n =5m +1 ⇒ n = (5m +1)/6. For n > m, we have:

(5m +1)/6 > m ⇒ 5m +1 >6m ⇒ 1 > m ⇒ m < 1. But m is a positive integer, so m ≥1. Therefore, m <1 is impossible. Therefore, n cannot be greater than m. Therefore, in this case, there are no solutions where m < n. Therefore, Case 1 is impossible.

Case 2: k=2, so 6n -5m =2, t=504.

Equation: 6n -5m =2 ⇒ 6n =5m +2 ⇒ n=(5m +2)/6. For n to be integer, 5m +2 must be divisible by 6. Let me solve:

5m +2 ≡0 mod6 ⇒5m ≡-2 ≡4 mod6 ⇒Multiply both sides by inverse of 5 mod6, which is 5 again:

m ≡4*5 ≡20 ≡2 mod6 ⇒ m ≡2 mod6. So m=6t +2, t≥0.

Let me compute n for m=6t +2:

n=(5*(6t +2) +2)/6=(30t +10 +2)/6=(30t +12)/6=5t +2.

Therefore, n=5t +2, m=6t +2. Now, m <n ?

Check m < n ⇒ 6t +2 <5t +2 ⇒ t <0. But t ≥0. Therefore, no solution. Therefore, in this case, m ≥n for all t ≥0. Therefore, no solution here. Case 2 is invalid.

Case 3: k=3, so 6n -5m =3, t=224.

Equation: 6n -5m =3 ⇒ 6n =5m +3 ⇒n=(5m +3)/6. So 5m +3 must be divisible by 6.

Let me solve 5m +3 ≡0 mod6 ⇒5m ≡ -3 ≡3 mod6 ⇒Multiply both sides by inverse of 5 mod6 (which is 5):

m ≡3*5=15≡3 mod6 ⇒ m≡3 mod6. Therefore, m=6t +3 for t ≥0.

Compute n=(5*(6t +3)+3)/6=(30t +15 +3)/6=(30t +18)/6=5t +3.

So n=5t +3, m=6t +3. Then check m <n:

6t +3 <5t +3 ⇒ t <0. Not possible. Therefore, again, no solution here. Case 3 invalid.

Case 4: k=4, 6n -5m=4, t=126.

Equation: 6n =5m +4 ⇒n=(5m +4)/6.

Solve 5m +4 ≡0 mod6 ⇒5m ≡-4 ≡2 mod6 ⇒Multiply both sides by 5:

m≡2*5=10≡4 mod6 ⇒ m=6t +4, t≥0.

Compute n=(5*(6t +4) +4)/6=(30t +20 +4)/6=(30t +24)/6=5t +4.

Therefore, n=5t +4, m=6t +4. Check m <n:

6t +4 <5t +4 ⇒ t <0. Not possible. Again, no solution. So Case 4 invalid.

Case 5: k=6, so 6n -5m=6, t=56.

Equation: 6n =5m +6 ⇒n=(5m +6)/6.

Solve 5m +6 ≡0 mod6 ⇒5m ≡-6≡0 mod6 ⇒5m≡0 mod6 ⇒m≡0 mod6 (since 5 and 6 are coprime). Therefore, m=6t, t≥1 (since m must be positive integer).

Compute n=(5*(6t) +6)/6=(30t +6)/6=5t +1.

Therefore, n=5t +1, m=6t. Check m <n?

6t <5t +1 ⇒ t <1. Since t≥1 (as m=6t must be positive integer, t≥1), then t=1 gives m=6*1=6, n=5*1 +1=6 ⇒ m=6, n=6 ⇒ m=n, which violates m <n. For t≥2: m=6t, n=5t +1. Check 6t <5t +1 ⇒ t <1. But t≥2. Therefore, no solution here. Therefore, Case 5 invalid.

Case 6: k=12, so 6n -5m=12, t=14.

Equation: 6n -5m=12 ⇒6n=5m +12 ⇒n=(5m +12)/6.

For n to be integer, 5m +12 must be divisible by 6 ⇒5m +12 ≡0 mod6 ⇒5m ≡-12 ≡-12 +12=0 mod6? Wait, -12 mod6 is 0 mod6? Let me compute:

Wait, 5m +12 ≡ 5m + 0 mod6 (since 12 is divisible by 6). Therefore, 5m ≡0 mod6 ⇒m≡0 mod6 (since 5 and 6 coprime). Therefore, m=6t, t≥1.

Compute n=(5*(6t) +12)/6=(30t +12)/6=5t +2.

So n=5t +2, m=6t. Check m <n ⇒6t <5t +2 ⇒t <2. Since t is a positive integer, t=1 or t=2.

t=1: m=6*1=6, n=5*1 +2=7. Here, m=6 < n=7. Good.

t=2: m=6*2=12, n=5*2 +2=12. So m=12, n=12. Not allowed since m <n. Therefore, only t=1 is valid here.

So in Case 6, we have a valid solution when t=1: m=6, n=7.

Therefore, the only possible case is Case 6 with t=1, m=6, n=7.

Therefore, let's check this case. So m=6, n=7, t=14? Wait, no. Wait in Case 6, t=14. Wait no. Wait, in the earlier notation, the value of t in the equation $ t (6n -5m)^2 = 2016 $. For Case 6, (6n -5m)^2 =144, which is 12², so t =2016 /144 =14. Yes.

So t=14, m=6, n=7. Therefore, we can compute a_1 = x = m² t = 6² *14 = 36*14. Let me compute that: 36*14 = (30*14) + (6*14) = 420 +84= 504.

But before accepting that, let me check whether all the terms are strictly increasing and satisfy all the given conditions. Because even though we have a solution for a_{13} = 2016, we need to make sure that all previous terms are integers and the sequence is strictly increasing.

So let me compute all terms up to, say, a_13 with m=6, n=7, t=14.

Let me recall:

Given m=6, n=7, t=14.

First, compute the terms:

a_1 = x = m² t = 6² *14 = 36*14 = 504.

a_2 = y = m t n = 6 *14 *7. Let's compute that: 6*14=84, 84*7=588. So a_2=588.

Check that a_1 < a_2: 504 < 588 ✔

a_3 = t n² =14*(7²)=14*49=686.

Check a_2 <a_3: 588 <686 ✔

a_4 = t n (2n -m) =14*7*(2*7 -6)=14*7*(14 -6)=14*7*8.

Compute that: 14*7=98, 98*8=784. So a_4=784. Check a_3 <a_4: 686 <784 ✔

a_5= t (2n -m)^2 =14*(14 -6)^2=14*(8)^2=14*64=896.

Check a_4 <a_5: 784 <896 ✔

a_6= t (3n -2m)(2n -m)=14*(3*7 -2*6)*(14 -6)=14*(21 -12)*8=14*9*8.

Compute that: 14*9=126, 126*8=1008. So a_6=1008. Check a_5 <a_6: 896 <1008 ✔

a_7= t (3n -2m)^2=14*(21 -12)^2=14*(9)^2=14*81=1134. Check a_6 <a_7: 1008 <1134 ✔

a_8= t (3n -2m)(4n -3m)=14*9*(4*7 -3*6)=14*9*(28 -18)=14*9*10.

Compute that: 14*9=126, 126*10=1260. Check a_7 <a_8: 1134 <1260 ✔

a_9= t (4n -3m)^2=14*(28 -18)^2=14*(10)^2=14*100=1400. Check a_8 <a_9: 1260 <1400 ✔

a_{10}=t (4n -3m)(5n -4m)=14*10*(5*7 -4*6)=14*10*(35 -24)=14*10*11.

Compute that: 14*10=140, 140*11=1540. Check a_9 <a_{10}: 1400 <1540 ✔

a_{11}= t (5n -4m)^2=14*(35 -24)^2=14*(11)^2=14*121=1694. Check a_{10} <a_{11}: 1540 <1694 ✔

a_{12}=t (5n -4m)(6n -5m)=14*11*(6*7 -5*6)=14*11*(42 -30)=14*11*12.

Compute that: 14*11=154, 154*12=1848. Check a_{11} <a_{12}: 1694 <1848 ✔

a_{13}= t (6n -5m)^2=14*(42 -30)^2=14*(12)^2=14*144=2016. Which matches the given value.

Therefore, all terms are strictly increasing, integers, and satisfy the given geometric and arithmetic conditions. Therefore, this is a valid solution.

Therefore, a_1 = 504. However, let me check if there are other possible solutions. In Case 6, only t=1 gives m=6, n=7. Other t's in Case 6 would lead to m=n or m>n. Hence, only one solution here.

But let me just confirm if there are other possible k values. Wait, earlier cases had no solutions, only Case 6. Therefore, the answer is a_1=504.

But let me check if there are other possibilities. For example, could there be other k's? But according to our earlier analysis, only Case 6 gives a valid solution. Therefore, the answer is 504.

Wait, but just to be thorough, let me check whether for Case 6, there could be other values of t. Wait, in Case 6, the equation was 6n -5m=12, and with m=6t, n=5t+2. But we found only t=1 gives m <n. For t=1, m=6, n=7. For t=2, m=12, n=12, which is invalid. Therefore, only one solution here.

Therefore, the answer is $ a_1 = 504 $.

But just to make sure, let me check with another approach or see if I can find another way. For example, maybe the sequence has a pattern where each term relates to the previous ones.

Alternatively, note that the sequence alternates between geometric and arithmetic sequences. Let me think recursively. For example, for each pair of terms, there's a relation.

But since we have already found a valid solution that fits all conditions, and the problem says "a strictly increasing sequence... suppose that a_{13}=2016", so there must be a unique solution, which is 504.

Therefore, the answer is \boxed{504}.

**Final Answer**
\boxed{504}

To solve the problem, we are given a strictly increasing sequence $ a_1, a_2, a_3, \dots $ of positive integers that satisfies the following conditions:

- For every $ k \in \mathbb{Z}^+ $, the subsequence $ a_{2k-1}, a_{2k}, a_{2k+1} $ is geometric.
- For every $ k \in \mathbb{Z}^+ $, the subsequence $ a_{2k}, a_{2k+1}, a_{2k+2} $ is arithmetic.

Given that $ a_{13} = 2016 $, the goal is to find $ a_1 $.

---

### Step 1: Analyze the Structure of the Sequence

Let’s define the first few terms using the given properties:

- $ a_1, a_2, a_3 $: geometric ⇒ $ a_2^2 = a_1 a_3 $
- $ a_2, a_3, a_4 $: arithmetic ⇒ $ 2a_3 = a_2 + a_4 \Rightarrow a_4 = 2a_3 - a_2 $
- $ a_3, a_4, a_5 $: geometric ⇒ $ a_4^2 = a_3 a_5 \Rightarrow a_5 = \frac{a_4^2}{a_3} $
- $ a_4, a_5, a_6 $: arithmetic ⇒ $ a_6 = 2a_5 - a_4 $

Through this process, we observe that the even and odd indexed terms follow predictable patterns.

Let’s assume:

- $ a_1 = m^2 t $
- $ a_2 = m t n $
- Where $ \gcd(m, n) = 1 $, and $ m < n $, ensuring the sequence is strictly increasing.

From the recurrence, it can be shown that:

- $ a_3 = t n^2 $
- $ a_5 = t(2n - m)^2 $
- $ a_7 = t(3n - 2m)^2 $
- $ a_9 = t(4n - 3m)^2 $
- ...
- $ a_{13} = t(6n - 5m)^2 $

---

### Step 2: Use Given Information

We are told that $ a_{13} = 2016 $. From above:

$$
a_{13} = t(6n - 5m)^2 = 2016
$$

Let us set $ (6n - 5m)^2 = d^2 $, then:

$$
t \cdot d^2 = 2016
$$

We factor $ 2016 $:

$$
2016 = 2^5 \cdot 3^2 \cdot 7
$$

The square divisors of 2016 are:

$$
1, 4, 9, 16, 36, 144
$$

Thus, the possible values of $ d $ (i.e., $ 6n - 5m $) are:

$$
d = 1, 2, 3, 4, 6, 12
$$

We now iterate through these to solve for $ m, n, t $ such that:

- $ m < n $
- $ \gcd(m, n) = 1 $
- $ t = \frac{2016}{d^2} $ is an integer

We test these values:

Let’s test $ d = 12 \Rightarrow (6n - 5m)^2 = 144 \Rightarrow 6n - 5m = 12 $. Then:

$$
t = \frac{2016}{144} = 14
$$

So we need to solve:

$$
6n - 5m = 12 \quad \text{(1)}
$$

Rewriting:

$$
6n = 5m + 12 \Rightarrow n = \frac{5m + 12}{6}
$$

Thus $ 5m + 12 $ must be divisible by 6. Try values for $ m $:

Try $ m = 6 \Rightarrow 5(6) + 12 = 42 \Rightarrow n = 42/6 = 7 $

Check:

- $ m = 6 $, $ n = 7 $
- $ \gcd(6, 7) = 1 $ ✔
- $ m < n $ ✔

Hence:

- $ t = 14 $
- $ a_1 = m^2 t = 6^2 \cdot 14 = 36 \cdot 14 = \boxed{504} $

---

### Step 3: Verify All Terms

To ensure this solution is valid, compute the first few terms:

- $ a_1 = 504 $
- $ a_2 = m t n = 6 \cdot 14 \cdot 7 = 588 $
- $ a_3 = t n^2 = 14 \cdot 49 = 686 $
- $ a_4 = t n(2n - m) = 14 \cdot 7 \cdot (14 - 6) = 14 \cdot 7 \cdot 8 = 784 $
- $ a_5 = t(2n - m)^2 = 14 \cdot 8^2 = 14 \cdot 64 = 896 $
- $ a_6 = t(2n - m)(3n - 2m) = 14 \cdot 8 \cdot 9 = 1008 $
- $ a_7 = t(3n - 2m)^2 = 14 \cdot 9^2 = 1134 $
- $ a_8 = t(3n - 2m)(4n - 3m) = 14 \cdot 9 \cdot 10 = 1260 $
- $ a_9 = t(4n - 3m)^2 = 14 \cdot 10^2 = 1400 $
- $ a_{10} = t(4n - 3m)(5n - 4m) = 14 \cdot 10 \cdot 11 = 1540 $
- $ a_{11} = t(5n - 4m)^2 = 14 \cdot 11^2 = 1694 $
- $ a_{12} = t(5n - 4m)(6n - 5m) = 14 \cdot 11 \cdot 12 = 1848 $
- $ a_{13} = t(6n - 5m)^2 = 14 \cdot 12^2 = 14 \cdot 144 = \boxed{2016} $

This confirms all conditions are satisfied and that the sequence is strictly increasing.

---

### Final Answer

$$
\boxed{504}
$$

[Assistant]
Explanation:
Okay, so I need to find the value of $ a_1 $ given that the sequence is strictly increasing, and for every positive integer $ k $, the subsequence $ a_{2k-1}, a_{2k}, a_{2k+1} $ is geometric, and the subsequence $ a_{2k}, a_{2k+1}, a_{2k+2} $ is arithmetic. Also, $ a_{13} = 2016 $.

First, let me make sure I understand the problem correctly. The sequence is strictly increasing, so each term is larger than the previous one. Then, for any $ k $, if I take the terms at positions $ 2k-1 $, $ 2k $, $ 2k+1 $, they form a geometric sequence. Similarly, the terms at positions $ 2k $, $ 2k+1 $, $ 2k+2 $ form an arithmetic sequence.

So, for example, when $ k = 1 $, the first three terms $ a_1, a_2, a_3 $ form a geometric sequence. Then, the terms $ a_2, a_3, a_4 $ form an arithmetic sequence. Similarly, for $ k = 2 $, the terms $ a_3, a_4, a_5 $ form a geometric sequence, and $ a_4, a_5, a_6 $ form an arithmetic sequence, and so on.

Since the sequence is strictly increasing, all the terms are positive integers and each term is larger than the previous.

Let me try to write down the relations for the first few terms and see if I can find a pattern or recurrence relation.

Starting with $ k = 1 $:

1. $ a_1, a_2, a_3 $ is geometric. So, $ a_2^2 = a_1 \cdot a_3 $.
2. $ a_2, a_3, a_4 $ is arithmetic. So, $ 2a_3 = a_2 + a_4 $. Therefore, $ a_4 = 2a_3 - a_2 $.

For $ k = 2 $:

1. $ a_3, a_4, a_5 $ is geometric. So, $ a_4^2 = a_3 \cdot a_5 $. Therefore, $ a_5 = \frac{a_4^2}{a_3} $.
2. $ a_4, a_5, a_6 $ is arithmetic. So, $ 2a_5 = a_4 + a_6 $. Therefore, $ a_6 = 2a_5 - a_4 $.

For $ k = 3 $:

1. $ a_5, a_6, a_7 $ is geometric. So, $ a_6^2 = a_5 \cdot a_7 $. Therefore, $ a_7 = \frac{a_6^2}{a_5} $.
2. $ a_6, a_7, a_8 $ is arithmetic. So, $ 2a_7 = a_6 + a_8 $. Therefore, $ a_8 = 2a_7 - a_6 $.

And so on. It seems like there's a pattern here. Each odd-indexed term (starting from $ a_1 $) is part of a geometric sequence with the next two terms, and then the following term is determined by the arithmetic sequence.

But since the sequence is strictly increasing, all these terms must satisfy $ a_1 < a_2 < a_3 < a_4 < \dots $. Also, since they are integers, the geometric sequences must have integer ratios, and the arithmetic sequences must have integer differences.

But let me think about how to express the terms in terms of previous terms.

Let me try to express $ a_3 $, $ a_4 $, $ a_5 $, etc., in terms of $ a_1 $ and $ a_2 $. Let me define variables for each term. Let me denote $ a_1 = x $, $ a_2 = y $. Then, since $ a_1, a_2, a_3 $ is geometric, $ a_3 = \frac{a_2^2}{a_1} = \frac{y^2}{x} $. But since $ a_3 $ must be an integer, $ x $ divides $ y^2 $.

Similarly, from the arithmetic sequence $ a_2, a_3, a_4 $, $ a_4 = 2a_3 - a_2 = 2 \cdot \frac{y^2}{x} - y $.

Then, moving to the next geometric sequence $ a_3, a_4, a_5 $. Since this is geometric, $ a_4^2 = a_3 \cdot a_5 $, so $ a_5 = \frac{a_4^2}{a_3} $. Let me compute that:

$ a_5 = \frac{(2 \cdot \frac{y^2}{x} - y)^2}{\frac{y^2}{x}} $

Let me simplify that:

First, compute numerator:

$ (2 \cdot \frac{y^2}{x} - y)^2 = \left( \frac{2y^2 - xy}{x} \right)^2 = \frac{(2y^2 - xy)^2}{x^2} $

Then divide by $ \frac{y^2}{x} $:

$ \frac{(2y^2 - xy)^2}{x^2} \div \frac{y^2}{x} = \frac{(2y^2 - xy)^2}{x^2} \cdot \frac{x}{y^2} = \frac{(2y^2 - xy)^2}{x y^2} $

Simplify numerator:

Factor out terms:

Let me factor $ 2y^2 - xy = y(2y - x) $

Therefore:

$ \frac{(y(2y - x))^2}{x y^2} = \frac{y^2 (2y - x)^2}{x y^2} = \frac{(2y - x)^2}{x} $

So $ a_5 = \frac{(2y - x)^2}{x} $

Since $ a_5 $ must be an integer, $ x $ divides $ (2y - x)^2 $.

Similarly, moving on, we can compute $ a_6 $ from the arithmetic sequence $ a_4, a_5, a_6 $:

$ a_6 = 2a_5 - a_4 = 2 \cdot \frac{(2y - x)^2}{x} - \left( 2 \cdot \frac{y^2}{x} - y \right) $

Let me compute this:

First term: $ 2 \cdot \frac{(2y - x)^2}{x} $

Second term: $ 2 \cdot \frac{y^2}{x} - y = \frac{2y^2 - xy}{x} $

Therefore:

$ a_6 = \frac{2(2y - x)^2}{x} - \frac{2y^2 - xy}{x} = \frac{2(2y - x)^2 - (2y^2 - xy)}{x} $

Let me expand $ 2(2y - x)^2 $:

First compute $ (2y - x)^2 = 4y^2 - 4xy + x^2 $

Multiply by 2: 8y² - 8xy + 2x²

Subtract $ (2y² - xy) $:

So total numerator:

$ 8y² - 8xy + 2x² - 2y² + xy = (8y² - 2y²) + (-8xy + xy) + 2x² = 6y² -7xy + 2x² $

Therefore, $ a_6 = \frac{6y² -7xy + 2x²}{x} = \frac{6y²}{x} -7y + 2x $

Hmm. So $ a_6 = \frac{6y^2 -7xy + 2x^2}{x} $. Let me check that again:

Wait, the numerator was 6y² -7xy +2x². So, dividing by x:

That's (6y²)/x - 7y + 2x. Correct.

Since $ a_6 $ must be an integer, so x divides 6y² -7xy +2x². Which is the same as x divides 6y² -7xy +2x².

But maybe we can factor this expression?

Let me try to factor 6y² -7xy +2x². Let me think of it as quadratic in y:

6y² -7x y +2x².

Let me try to factor:

Looking for two numbers a and b such that a*b = 6*2x² = 12x² and a + b = -7x. Wait, but maybe not. Alternatively, factor as (Ay + Bx)(Cy + Dx). Let me try:

Multiply out (Ay + Bx)(Cy + Dx) = AC y² + (AD + BC)xy + BD x². So we need AC = 6, BD = 2, and AD + BC = -7.

Possible A and C: factors of 6: 6 and 1, 3 and 2. Similarly, BD = 2: possible B and D: 2 and 1, 1 and 2, or negatives. Since coefficients are integers, and since the sequence is increasing, maybe positive coefficients?

Wait, but since the sequence is strictly increasing, and a1 < a2 < a3 < ..., so x < y < a3 < a4 < ... So x and y are positive integers, with y > x.

Let me try factors. Let me assume A and C positive, B and D positive.

Let me try AC = 6. Let's take A=3, C=2. Then BD = 2. Let me try B=2, D=1. Then AD + BC = 3*1 + 2*2 = 3 + 4 = 7. But we need AD + BC = -7. So if I take negative signs. Wait, but if we have BD=2, and if B and D are negative, but since y and x are positive, maybe the coefficients can be negative? But the overall expression is 6y² -7xy + 2x², which is positive? Not sure. Wait, if x and y are positive, then 6y² -7xy +2x² might be positive or negative? Well, since the sequence is increasing, so for example, a_6 must be larger than a_5, which must be larger than a_4, etc. So perhaps the expression 6y² -7xy +2x² is positive?

But regardless, let me check. Suppose that 6y² -7xy +2x² factors as (3y - 2x)(2y - x). Let me check:

(3y - 2x)(2y - x) = 6y² -3xy -4xy + 2x² = 6y² -7xy +2x². Yes! Perfect.

So 6y² -7xy +2x² = (3y -2x)(2y -x). Therefore, $ a_6 = \frac{(3y - 2x)(2y -x)}{x} $.

Therefore, $ a_6 = \frac{(3y -2x)(2y -x)}{x} $.

Since $ a_6 $ must be an integer, $ x $ divides $ (3y -2x)(2y -x) $.

But since earlier, we had that $ x $ divides $ y^2 $ from the geometric sequence $ a_1, a_2, a_3 $. Let me recall that $ a_3 = y^2 / x $, so x divides y². Let me denote that as x | y². So maybe we can write y = x * m, where m is a rational number? But since y and x are integers, m must be rational. However, since x divides y², then if we write m as a reduced fraction, say m = p/q where p and q are coprime integers, then x must be divisible by q².

Alternatively, maybe it's better to suppose that x divides y², so let me write y = kx, where k is a rational number. Wait, but since y is an integer, and x is an integer, then if x divides y², then y² = x * n for some integer n.

Alternatively, maybe we can let d = gcd(x, y), and write x = d * m, y = d * n, where m and n are coprime positive integers. Then since x divides y² = d² n², so x divides d² n². Since x = d m, then d m divides d² n², which implies m divides d n². Since m and n are coprime, m divides d. Let me denote d = m * t, so that x = d m = m^2 t, and y = d n = m t n. Therefore, since m and n are coprime, and t is a positive integer.

So, substituting back, we can write:

x = m² t

y = m t n

With m, n, t positive integers, gcd(m, n) = 1.

This substitution might help in simplifying expressions. Let me try this substitution.

So, let me define variables:

Let $ x = m^2 t $

$ y = m t n $

Where $ \gcd(m, n) = 1 $, and $ m, t, n $ are positive integers.

Now, since $ a_3 = y² / x = (m² t² n²) / (m² t) ) = t n² $. So $ a_3 = t n² $.

Similarly, $ a_4 = 2 a_3 - y = 2 t n² - m t n = t n (2n - m) $.

Then $ a_5 = (2y -x)^2 / x = (2 m t n - m² t)^2 / (m² t ) = [ m t (2n - m) ]² / (m² t ) = m² t² (2n - m)^2 / (m² t ) = t (2n - m)^2 $.

So $ a_5 = t (2n - m)^2 $.

Then $ a_6 = (3y - 2x)(2y -x)/x $. Let me compute that:

First compute 3y - 2x = 3 m t n - 2 m² t = m t (3n - 2m)

Similarly, 2y - x = 2 m t n - m² t = m t (2n - m)

Therefore, (3y - 2x)(2y -x) = m t (3n - 2m) * m t (2n - m) = m² t² (3n - 2m)(2n - m)

Then divide by x = m² t:

So $ a_6 = [ m² t² (3n - 2m)(2n - m) ] / [ m² t ] = t (3n - 2m)(2n - m) $

Therefore, $ a_6 = t (3n - 2m)(2n - m) $

Similarly, let me check whether these expressions make sense. Since the sequence is strictly increasing, we can derive some inequalities.

First, since the sequence is strictly increasing:

$ a_1 < a_2 < a_3 < a_4 < a_5 < a_6 < ... $

So, $ x = m² t < y = m t n $. Therefore, m² t < m t n ⇒ m < n. Since m and n are positive integers, m < n.

Also, $ a_3 = t n² > y = m t n $ ⇒ t n² > m t n ⇒ n > m. Which is consistent with m < n.

Then $ a_4 = t n (2n - m) > a_3 = t n² $. Therefore:

$ t n (2n - m) > t n² ⇒ 2n - m > n ⇒ 2n - m > n ⇒ n > m $, which is again true.

Next, $ a_5 = t (2n - m)^2 $. Compare with $ a_4 = t n (2n - m) $. So:

$ t (2n - m)^2 > t n (2n - m) ⇒ (2n - m)^2 > n (2n - m) $, since 2n - m is positive? Let me check. Since the sequence is increasing, and a_4 is t n (2n - m), which must be greater than a_3 = t n². So 2n - m > n ⇒ n > m as before. Therefore 2n - m is positive. Thus, we can divide both sides by (2n - m):

$ (2n - m) > n ⇒ 2n - m > n ⇒ n > m $, which is true.

So that's okay.

Similarly, $ a_6 = t(3n - 2m)(2n - m) $. Compare with a_5 = t (2n - m)^2. So:

a_6 > a_5 ⇒ t(3n - 2m)(2n - m) > t (2n - m)^2 ⇒ (3n - 2m) > 2n - m ⇒ 3n - 2m > 2n - m ⇒ n > m. Which is true.

So all the inequalities so far hold as long as m < n.

So, with that substitution, we can perhaps find a pattern. Let me see if I can find expressions for subsequent terms.

But maybe instead of going further, I need to find a general formula for the terms. Let me see.

Alternatively, maybe there's a recursive relation here. Let me check. Let me note that each time, the odd terms are determined by the previous terms, and the even terms as well.

Alternatively, since the sequence alternates between geometric and arithmetic sequences, perhaps there's a recurrence relation.

But let me see. Let me try to compute $ a_7 $, which is part of the geometric sequence $ a_5, a_6, a_7 $. So:

$ a_6^2 = a_5 \cdot a_7 ⇒ a_7 = a_6² / a_5 $.

From above, $ a_5 = t (2n - m)^2 $, $ a_6 = t (3n - 2m)(2n - m) $.

Therefore, $ a_7 = [ t^2 (3n - 2m)^2 (2n - m)^2 ] / [ t (2n - m)^2 ] = t (3n - 2m)^2 $.

So $ a_7 = t (3n - 2m)^2 $.

Similarly, then $ a_8 $ is determined by the arithmetic sequence $ a_6, a_7, a_8 $. So:

$ 2a_7 = a_6 + a_8 ⇒ a_8 = 2a_7 - a_6 = 2 t (3n - 2m)^2 - t (3n - 2m)(2n - m) $.

Factor out t (3n - 2m):

$ a_8 = t (3n - 2m) [ 2(3n - 2m) - (2n - m) ] $

Compute inside the brackets:

$ 2(3n - 2m) - (2n - m) = 6n - 4m - 2n + m = 4n - 3m $

Thus, $ a_8 = t (3n - 2m)(4n - 3m) $

Similarly, moving on to the next geometric sequence $ a_7, a_8, a_9 $:

$ a_8^2 = a_7 \cdot a_9 ⇒ a_9 = a_8² / a_7 $

Let me compute that:

$ a_8 = t (3n - 2m)(4n - 3m) $

$ a_7 = t (3n - 2m)^2 $

Therefore:

$ a_9 = [ t^2 (3n - 2m)^2 (4n - 3m)^2 ] / [ t (3n - 2m)^2 ] = t (4n - 3m)^2 $

Similarly, moving on to the arithmetic sequence $ a_8, a_9, a_{10} $:

$ a_{10} = 2a_9 - a_8 = 2 t (4n - 3m)^2 - t (3n - 2m)(4n - 3m) $

Factor out t (4n - 3m):

$ a_{10} = t (4n - 3m) [ 2(4n - 3m) - (3n - 2m) ] $

Compute inside the brackets:

$ 2(4n - 3m) - (3n - 2m) = 8n - 6m - 3n + 2m = 5n - 4m $

Thus, $ a_{10} = t (4n - 3m)(5n - 4m) $

Hmm, I see a pattern here. It seems like for the odd terms, $ a_{2k+1} $, they are of the form $ t \times (\text{some linear expression in } n \text{ and } m)^2 $, and the even terms $ a_{2k} $ are of the form $ t \times (\text{some linear expression in } n \text{ and } m) \times (\text{another linear expression}) $.

Looking at the expressions for $ a_3, a_5, a_7, a_9 $:

- $ a_3 = t n^2 $
- $ a_5 = t (2n - m)^2 $
- $ a_7 = t (3n - 2m)^2 $
- $ a_9 = t (4n - 3m)^2 $

Similarly, even terms:

- $ a_4 = t n(2n - m) $
- $ a_6 = t(3n - 2m)(2n - m) $
- $ a_8 = t(3n - 2m)(4n - 3m) $
- $ a_{10} = t(4n - 3m)(5n - 4m) $

So, it seems like the odd-indexed terms follow a pattern where each term is $ t \times (kn - (k-1)m)^2 $, where k increases by 1 each time. Starting with k=1: $ a_3 = t(n)^2 $, then k=2: $ t(2n - m)^2 $, then k=3: $ t(3n - 2m)^2 $, etc.

Similarly, for even-indexed terms, they are products of consecutive expressions. For example, $ a_4 = t \times n \times (2n - m) $, $ a_6 = t \times (2n - m) \times (3n - 2m) $, $ a_8 = t \times (3n - 2m) \times (4n - 3m) $, $ a_{10} = t \times (4n - 3m) \times (5n - 4m) $, and so on.

This suggests that for the odd term $ a_{2k+1} $, it's $ t \times (kn - (k - 1)m)^2 $, and for the even term $ a_{2k} $, it's $ t \times (kn - (k - 1)m)( (k + 1)n - km ) $.

Let me check for $ k = 1 $:

- $ a_3 $, which is $ a_{2*1 + 1} $, so k = 1: $ t*(1n - 0m)^2 = t n² $. Correct.
- $ a_4 $, which is $ a_{2*1} $: t*(1n - 0m)*(2n - 1m) = t*n*(2n - m). Correct.

Similarly for $ k = 2 $:

- $ a_5 $: t*(2n - 1m)^2. Correct.
- $ a_6 $: t*(2n - m)*(3n - 2m). Correct.

So this seems like a general formula. Therefore, we can express $ a_{2k} $ as $ t \times (kn - (k - 1)m) \times ((k + 1)n - km) $, and $ a_{2k + 1} $ as $ t \times (kn - (k - 1)m)^2 $.

Therefore, if I can find expressions for $ a_{13} $, which is an odd-indexed term. Since 13 is odd, let me see. Let me figure out which term $ a_{13} $ corresponds to.

The odd indices are 1, 3, 5, 7, 9, 11, 13,... So $ a_{13} $ is the 7th odd term. Let me check:

For the formula $ a_{2k + 1} = t [k n - (k - 1)m]^2 $. Wait, when k=1, we get a_3. So if $ a_{2k + 1} $, then for k=1: a_3, k=2: a_5, k=3: a_7, k=4: a_9, k=5: a_{11}, k=6: a_{13}. Therefore, for $ a_{13} $, k = 6.

Therefore, $ a_{13} = t [6n - 5m]^2 $.

Given that $ a_{13} = 2016 $, so:

$ t (6n - 5m)^2 = 2016 $.

So, I need to find integers m, n, t with m < n, and $ \gcd(m, n) = 1 $, such that $ t (6n -5m)^2 = 2016 $. Also, since the sequence is strictly increasing, all the terms must be positive integers, and we have previous relations.

Additionally, since $ a_1 = x = m² t $, so once I find m, n, t, I can compute $ a_1 $.

Therefore, my task reduces to solving $ t (6n -5m)^2 = 2016 $, with the constraints that m, n, t are positive integers, m < n, $ \gcd(m, n) = 1 $, and also ensuring that all previous terms are integers and strictly increasing.

But first, let's factorize 2016 to see possible values for t and $ (6n -5m)^2 $.

2016 factors as:

Let me compute 2016:

Divide by 2: 2016 / 2 = 1008

/2 = 504

/2 = 252

/2 = 126

/2 = 63

So 2^5 * 63.

63 is 7*9 = 7*3².

So overall, prime factorization of 2016 is 2^5 * 3² * 7^1.

Therefore, 2016 = 2^5 * 3² * 7.

Now, since $ t \times (6n - 5m)^2 = 2016 $, so (6n -5m)^2 must be a divisor of 2016. Let me denote $ k = 6n -5m $, then $ k^2 $ divides 2016, so $ k^2 $ must be a square divisor of 2016.

Therefore, first, I need to find all square divisors of 2016.

The square divisors of 2016 are of the form (product of primes in exponents even numbers less than or equal to their exponents in 2016).

Given the prime factorization 2^5 * 3² * 7^1.

For square divisors, exponents for each prime must be even numbers:

- For prime 2: exponent can be 0, 2, 4 (since original exponent is 5, max even exponent less than or equal to 5 is 4)
- For prime 3: exponent can be 0, 2 (original exponent is 2)
- For prime 7: exponent must be 0 (since original exponent is 1, which is odd, so maximum even exponent is 0)

Therefore, square divisors of 2016 are:

- 2^0 * 3^0 * 7^0 = 1
- 2^2 * 3^0 * 7^0 = 4
- 2^4 * 3^0 * 7^0 = 16
- 2^0 * 3^2 * 7^0 = 9
- 2^2 * 3^2 * 7^0 = 4*9 = 36
- 2^4 * 3^2 * 7^0 = 16*9 = 144

Therefore, the square divisors of 2016 are 1, 4, 9, 16, 36, 144. Therefore, $ k^2 $ must be one of these, so k is 1, 2, 3, 4, 6, 12.

Therefore, possible values of k are 1,2,3,4,6,12. Therefore:

Case 1: $ k = 1 $, then $ (6n -5m)^2 = 1 $, so $ 6n -5m = 1 $, and t = 2016 / 1 = 2016.

Case 2: $ k = 2 $, then $ (6n -5m)^2 = 4 $, so $ 6n -5m = 2 $, t = 2016 / 4 = 504.

Case 3: $ k = 3 $, $ (6n -5m)^2 = 9 $, so $ 6n -5m = 3 $, t = 2016 / 9 = 224.

Case 4: $ k = 4 $, $ (6n -5m)^2 = 16 $, so $ 6n -5m = 4 $, t = 2016 / 16 = 126.

Case 5: $ k = 6 $, $ (6n -5m)^2 = 36 $, so $ 6n -5m = 6 $, t = 2016 / 36 = 56.

Case 6: $ k = 12 $, $ (6n -5m)^2 = 144 $, so $ 6n -5m = 12 $, t = 2016 / 144 = 14.

Therefore, these are the possible cases. Now, for each case, I need to check if there exist positive integers m, n with m < n, gcd(m,n) = 1, and also ensuring that the previous terms are strictly increasing and satisfy the required properties.

Let me analyze each case one by one.

Case 1: k = 1, so 6n -5m = 1, t = 2016

So, equation: 6n -5m = 1 ⇒ 6n = 5m +1 ⇒ n = (5m +1)/6. Since n must be an integer, 5m +1 must be divisible by 6. Let me solve for m:

5m +1 ≡ 0 mod 6 ⇒ 5m ≡ -1 mod 6 ⇒ 5m ≡ 5 mod 6 ⇒ m ≡ 1 mod 6 (since 5 is invertible modulo 6, inverse of 5 mod 6 is 5, because 5*5=25≡1 mod6). Therefore, m = 6t +1 for some integer t ≥ 0. Since m must be a positive integer, t ≥ 0. Let me check for t=0: m=1, n=(5*1 +1)/6 = 6/6 =1. But m=1, n=1. However, m < n is required. But here m = n = 1, which violates m < n. Therefore, t=0 is invalid. Next, t=1: m=7, n=(5*7 +1)/6 = (35 +1)/6=36/6=6. So m=7, n=6. But m=7, n=6 ⇒ m > n, which is invalid. Wait, that can't be. Wait, m = 6t +1, for t=1: m=6*1 +1=7. Then n=(5*7 +1)/6=36/6=6. So m=7, n=6. But m > n, which violates m < n. Therefore, invalid. Next t=2: m=6*2 +1=13, then n=(5*13 +1)/6=(65 +1)/6=66/6=11. So m=13, n=11 ⇒ m > n. Same problem. Similarly, t=3: m=19, n=(5*19 +1)/6=(95 +1)/6=96/6=16, m=19 > n=16. Not good. Wait, so this seems like for all t ≥0, m = 6t +1 and n=(5m +1)/6. Let me check if n can be greater than m? Let me write the equation 6n -5m =1. So, rearranged as 6n =5m +1 ⇒ n = (5m +1)/6. For n > m, we have:

(5m +1)/6 > m ⇒ 5m +1 >6m ⇒ 1 > m ⇒ m < 1. But m is a positive integer, so m ≥1. Therefore, m <1 is impossible. Therefore, n cannot be greater than m. Therefore, in this case, there are no solutions where m < n. Therefore, Case 1 is impossible.

Case 2: k=2, so 6n -5m =2, t=504.

Equation: 6n -5m =2 ⇒ 6n =5m +2 ⇒ n=(5m +2)/6. For n to be integer, 5m +2 must be divisible by 6. Let me solve:

5m +2 ≡0 mod6 ⇒5m ≡-2 ≡4 mod6 ⇒Multiply both sides by inverse of 5 mod6, which is 5 again:

m ≡4*5 ≡20 ≡2 mod6 ⇒ m ≡2 mod6. So m=6t +2, t≥0.

Let me compute n for m=6t +2:

n=(5*(6t +2) +2)/6=(30t +10 +2)/6=(30t +12)/6=5t +2.

Therefore, n=5t +2, m=6t +2. Now, m <n ?

Check m < n ⇒ 6t +2 <5t +2 ⇒ t <0. But t ≥0. Therefore, no solution. Therefore, in this case, m ≥n for all t ≥0. Therefore, no solution here. Case 2 is invalid.

Case 3: k=3, so 6n -5m =3, t=224.

Equation: 6n -5m =3 ⇒ 6n =5m +3 ⇒n=(5m +3)/6. So 5m +3 must be divisible by 6.

Let me solve 5m +3 ≡0 mod6 ⇒5m ≡ -3 ≡3 mod6 ⇒Multiply both sides by inverse of 5 mod6 (which is 5):

m ≡3*5=15≡3 mod6 ⇒ m≡3 mod6. Therefore, m=6t +3 for t ≥0.

Compute n=(5*(6t +3)+3)/6=(30t +15 +3)/6=(30t +18)/6=5t +3.

So n=5t +3, m=6t +3. Then check m <n:

6t +3 <5t +3 ⇒ t <0. Not possible. Therefore, again, no solution here. Case 3 invalid.

Case 4: k=4, 6n -5m=4, t=126.

Equation: 6n =5m +4 ⇒n=(5m +4)/6.

Solve 5m +4 ≡0 mod6 ⇒5m ≡-4 ≡2 mod6 ⇒Multiply both sides by 5:

m≡2*5=10≡4 mod6 ⇒ m=6t +4, t≥0.

Compute n=(5*(6t +4) +4)/6=(30t +20 +4)/6=(30t +24)/6=5t +4.

Therefore, n=5t +4, m=6t +4. Check m <n:

6t +4 <5t +4 ⇒ t <0. Not possible. Again, no solution. So Case 4 invalid.

Case 5: k=6, so 6n -5m=6, t=56.

Equation: 6n =5m +6 ⇒n=(5m +6)/6.

Solve 5m +6 ≡0 mod6 ⇒5m ≡-6≡0 mod6 ⇒5m≡0 mod6 ⇒m≡0 mod6 (since 5 and 6 are coprime). Therefore, m=6t, t≥1 (since m must be positive integer).

Compute n=(5*(6t) +6)/6=(30t +6)/6=5t +1.

Therefore, n=5t +1, m=6t. Check m <n?

6t <5t +1 ⇒ t <1. Since t≥1 (as m=6t must be positive integer, t≥1), then t=1 gives m=6*1=6, n=5*1 +1=6 ⇒ m=6, n=6 ⇒ m=n, which violates m <n. For t≥2: m=6t, n=5t +1. Check 6t <5t +1 ⇒ t <1. But t≥2. Therefore, no solution here. Therefore, Case 5 invalid.

Case 6: k=12, so 6n -5m=12, t=14.

Equation: 6n -5m=12 ⇒6n=5m +12 ⇒n=(5m +12)/6.

For n to be integer, 5m +12 must be divisible by 6 ⇒5m +12 ≡0 mod6 ⇒5m ≡-12 ≡-12 +12=0 mod6? Wait, -12 mod6 is 0 mod6? Let me compute:

Wait, 5m +12 ≡ 5m + 0 mod6 (since 12 is divisible by 6). Therefore, 5m ≡0 mod6 ⇒m≡0 mod6 (since 5 and 6 coprime). Therefore, m=6t, t≥1.

Compute n=(5*(6t) +12)/6=(30t +12)/6=5t +2.

So n=5t +2, m=6t. Check m <n ⇒6t <5t +2 ⇒t <2. Since t is a positive integer, t=1 or t=2.

t=1: m=6*1=6, n=5*1 +2=7. Here, m=6 < n=7. Good.

t=2: m=6*2=12, n=5*2 +2=12. So m=12, n=12. Not allowed since m <n. Therefore, only t=1 is valid here.

So in Case 6, we have a valid solution when t=1: m=6, n=7.

Therefore, the only possible case is Case 6 with t=1, m=6, n=7.

Therefore, let's check this case. So m=6, n=7, t=14? Wait, no. Wait in Case 6, t=14. Wait no. Wait, in the earlier notation, the value of t in the equation $ t (6n -5m)^2 = 2016 $. For Case 6, (6n -5m)^2 =144, which is 12², so t =2016 /144 =14. Yes.

So t=14, m=6, n=7. Therefore, we can compute a_1 = x = m² t = 6² *14 = 36*14. Let me compute that: 36*14 = (30*14) + (6*14) = 420 +84= 504.

But before accepting that, let me check whether all the terms are strictly increasing and satisfy all the given conditions. Because even though we have a solution for a_{13} = 2016, we need to make sure that all previous terms are integers and the sequence is strictly increasing.

So let me compute all terms up to, say, a_13 with m=6, n=7, t=14.

Let me recall:

Given m=6, n=7, t=14.

First, compute the terms:

a_1 = x = m² t = 6² *14 = 36*14 = 504.

a_2 = y = m t n = 6 *14 *7. Let's compute that: 6*14=84, 84*7=588. So a_2=588.

Check that a_1 < a_2: 504 < 588 ✔

a_3 = t n² =14*(7²)=14*49=686.

Check a_2 <a_3: 588 <686 ✔

a_4 = t n (2n -m) =14*7*(2*7 -6)=14*7*(14 -6)=14*7*8.

Compute that: 14*7=98, 98*8=784. So a_4=784. Check a_3 <a_4: 686 <784 ✔

a_5= t (2n -m)^2 =14*(14 -6)^2=14*(8)^2=14*64=896.

Check a_4 <a_5: 784 <896 ✔

a_6= t (3n -2m)(2n -m)=14*(3*7 -2*6)*(14 -6)=14*(21 -12)*8=14*9*8.

Compute that: 14*9=126, 126*8=1008. So a_6=1008. Check a_5 <a_6: 896 <1008 ✔

a_7= t (3n -2m)^2=14*(21 -12)^2=14*(9)^2=14*81=1134. Check a_6 <a_7: 1008 <1134 ✔

a_8= t (3n -2m)(4n -3m)=14*9*(4*7 -3*6)=14*9*(28 -18)=14*9*10.

Compute that: 14*9=126, 126*10=1260. Check a_7 <a_8: 1134 <1260 ✔

a_9= t (4n -3m)^2=14*(28 -18)^2=14*(10)^2=14*100=1400. Check a_8 <a_9: 1260 <1400 ✔

a_{10}=t (4n -3m)(5n -4m)=14*10*(5*7 -4*6)=14*10*(35 -24)=14*10*11.

Compute that: 14*10=140, 140*11=1540. Check a_9 <a_{10}: 1400 <1540 ✔

a_{11}= t (5n -4m)^2=14*(35 -24)^2=14*(11)^2=14*121=1694. Check a_{10} <a_{11}: 1540 <1694 ✔

a_{12}=t (5n -4m)(6n -5m)=14*11*(6*7 -5*6)=14*11*(42 -30)=14*11*12.

Compute that: 14*11=154, 154*12=1848. Check a_{11} <a_{12}: 1694 <1848 ✔

a_{13}= t (6n -5m)^2=14*(42 -30)^2=14*(12)^2=14*144=2016. Which matches the given value.

Therefore, all terms are strictly increasing, integers, and satisfy the given geometric and arithmetic conditions. Therefore, this is a valid solution.

Therefore, a_1 = 504. However, let me check if there are other possible solutions. In Case 6, only t=1 gives m=6, n=7. Other t's in Case 6 would lead to m=n or m>n. Hence, only one solution here.

But let me just confirm if there are other possible k values. Wait, earlier cases had no solutions, only Case 6. Therefore, the answer is a_1=504.

But let me check if there are other possibilities. For example, could there be other k's? But according to our earlier analysis, only Case 6 gives a valid solution. Therefore, the answer is 504.

Wait, but just to be thorough, let me check whether for Case 6, there could be other values of t. Wait, in Case 6, the equation was 6n -5m=12, and with m=6t, n=5t+2. But we found only t=1 gives m <n. For t=1, m=6, n=7. For t=2, m=12, n=12, which is invalid. Therefore, only one solution here.

Therefore, the answer is $ a_1 = 504 $.

But just to make sure, let me check with another approach or see if I can find another way. For example, maybe the sequence has a pattern where each term relates to the previous ones.

Alternatively, note that the sequence alternates between geometric and arithmetic sequences. Let me think recursively. For example, for each pair of terms, there's a relation.

But since we have already found a valid solution that fits all conditions, and the problem says "a strictly increasing sequence... suppose that a_{13}=2016", so there must be a unique solution, which is 504.

Therefore, the answer is \boxed{504}.

**Final Answer**
\boxed{504}

To solve the problem, we are given a strictly increasing sequence $ a_1, a_2, a_3, \dots $ of positive integers that satisfies the following conditions:

- For every $ k \in \mathbb{Z}^+ $, the subsequence $ a_{2k-1}, a_{2k}, a_{2k+1} $ is geometric.
- For every $ k \in \mathbb{Z}^+ $, the subsequence $ a_{2k}, a_{2k+1}, a_{2k+2} $ is arithmetic.

Given that $ a_{13} = 2016 $, the goal is to find $ a_1 $.

---

### Step 1: Analyze the Structure of the Sequence

Let’s define the first few terms using the given properties:

- $ a_1, a_2, a_3 $: geometric ⇒ $ a_2^2 = a_1 a_3 $
- $ a_2, a_3, a_4 $: arithmetic ⇒ $ 2a_3 = a_2 + a_4 \Rightarrow a_4 = 2a_3 - a_2 $
- $ a_3, a_4, a_5 $: geometric ⇒ $ a_4^2 = a_3 a_5 \Rightarrow a_5 = \frac{a_4^2}{a_3} $
- $ a_4, a_5, a_6 $: arithmetic ⇒ $ a_6 = 2a_5 - a_4 $

Through this process, we observe that the even and odd indexed terms follow predictable patterns.

Let’s assume:

- $ a_1 = m^2 t $
- $ a_2 = m t n $
- Where $ \gcd(m, n) = 1 $, and $ m < n $, ensuring the sequence is strictly increasing.

From the recurrence, it can be shown that:

- $ a_3 = t n^2 $
- $ a_5 = t(2n - m)^2 $
- $ a_7 = t(3n - 2m)^2 $
- $ a_9 = t(4n - 3m)^2 $
- ...
- $ a_{13} = t(6n - 5m)^2 $

---

### Step 2: Use Given Information

We are told that $ a_{13} = 2016 $. From above:

$$
a_{13} = t(6n - 5m)^2 = 2016
$$

Let us set $ (6n - 5m)^2 = d^2 $, then:

$$
t \cdot d^2 = 2016
$$

We factor $ 2016 $:

$$
2016 = 2^5 \cdot 3^2 \cdot 7
$$

The square divisors of 2016 are:

$$
1, 4, 9, 16, 36, 144
$$

Thus, the possible values of $ d $ (i.e., $ 6n - 5m $) are:

$$
d = 1, 2, 3, 4, 6, 12
$$

We now iterate through these to solve for $ m, n, t $ such that:

- $ m < n $
- $ \gcd(m, n) = 1 $
- $ t = \frac{2016}{d^2} $ is an integer

We test these values:

Let’s test $ d = 12 \Rightarrow (6n - 5m)^2 = 144 \Rightarrow 6n - 5m = 12 $. Then:

$$
t = \frac{2016}{144} = 14
$$

So we need to solve:

$$
6n - 5m = 12 \quad \text{(1)}
$$

Rewriting:

$$
6n = 5m + 12 \Rightarrow n = \frac{5m + 12}{6}
$$

Thus $ 5m + 12 $ must be divisible by 6. Try values for $ m $:

Try $ m = 6 \Rightarrow 5(6) + 12 = 42 \Rightarrow n = 42/6 = 7 $

Check:

- $ m = 6 $, $ n = 7 $
- $ \gcd(6, 7) = 1 $ ✔
- $ m < n $ ✔

Hence:

- $ t = 14 $
- $ a_1 = m^2 t = 6^2 \cdot 14 = 36 \cdot 14 = \boxed{504} $

---

### Step 3: Verify All Terms

To ensure this solution is valid, compute the first few terms:

- $ a_1 = 504 $
- $ a_2 = m t n = 6 \cdot 14 \cdot 7 = 588 $
- $ a_3 = t n^2 = 14 \cdot 49 = 686 $
- $ a_4 = t n(2n - m) = 14 \cdot 7 \cdot (14 - 6) = 14 \cdot 7 \cdot 8 = 784 $
- $ a_5 = t(2n - m)^2 = 14 \cdot 8^2 = 14 \cdot 64 = 896 $
- $ a_6 = t(2n - m)(3n - 2m) = 14 \cdot 8 \cdot 9 = 1008 $
- $ a_7 = t(3n - 2m)^2 = 14 \cdot 9^2 = 1134 $
- $ a_8 = t(3n - 2m)(4n - 3m) = 14 \cdot 9 \cdot 10 = 1260 $
- $ a_9 = t(4n - 3m)^2 = 14 \cdot 10^2 = 1400 $
- $ a_{10} = t(4n - 3m)(5n - 4m) = 14 \cdot 10 \cdot 11 = 1540 $
- $ a_{11} = t(5n - 4m)^2 = 14 \cdot 11^2 = 1694 $
- $ a_{12} = t(5n - 4m)(6n - 5m) = 14 \cdot 11 \cdot 12 = 1848 $
- $ a_{13} = t(6n - 5m)^2 = 14 \cdot 12^2 = 14 \cdot 144 = \boxed{2016} $

This confirms all conditions are satisfied and that the sequence is strictly increasing.

---

### Final Answer

$$
\boxed{504}
$$

Solution:
Let $\frac{a_2}{a_1} = \frac{b}{a},$ where $a$ and $b$ are relatively prime positive integers, and $a < b.$  Then $a_2 = \frac{b}{a} \cdot a_1,$ and
\[a_3 = \frac{a_2^2}{a_1} = \frac{(b/a \cdot a_1)^2}{a_1} = \frac{b^2}{a^2} \cdot a_1.\]This implies $a_1$ is divisible by $a^2.$  Let $a_1 = ca^2$; then $a_2 = cab,$ $a_3 = cb^2,$
\begin{align*}
a_4 &= 2a_3 - a_2 = 2cb^2 - cab = cb(2b - a), \\
a_5 &= \frac{a_4^2}{a_3} = \frac{[cb(2b - a)]^2}{(cb^2)} = c(2b - 2a)^2, \\
a_6 &= 2a_5 - a_4 = 2c(2b - a)^2 - cb(2b - a) = c(2b - a)(3b - 2a), \\
a_7 &= \frac{a_6^2}{a_5} = \frac{[c(2b - a)(3b - 2a)]^2}{c(2b - a)^2} = c(3b - 2a)^2, \\
a_8 &= 2a_7 - a_6 = 2c(3b - 2a)^2 - c(2b - a)(3b - 2a) = c(3b - 2a)(4b - 3a), \\
a_9 &= \frac{a_8^2}{a_7} = \frac{[c(3b - 2a)(4b - 3a)]^2}{[c(3b - 2a)^2} = c(4b - 3a)^2,
\end{align*}and so on.

More generally, we can prove by induction that
\begin{align*}
a_{2k} &= c[(k - 1)b - (k - 2)a][kb - (k - 1)a], \\
a_{2k + 1} &= c[kb - (k - 1)a]^2,
\end{align*}for all positive integers $k.$

Hence, from $a_{13} = 2016,$
\[c(6b - 5a)^2 = 2016 = 2^5 \cdot 3^2 \cdot 7 = 14 \cdot 12^2.\]Thus, $6b - 5a$ must be a factor of 12.

Let $n = 6b - 5a.$  Then $a < a + 6(b - a) = n,$ and
\[n - a = 6b - 6a = 6(b - a),\]so $n - a$ is a multiple of 6.  Hence,
\[6 < a + 6 \le n \le 12,\]and the only solution is $(a,b,n) = (6,7,12).$  Then $c = 14,$ and $a_1 = 14 \cdot 6^2 = \boxed{504}.$


Explanation:
Okay, so I need to simplify the expression (1 + √2)/(2 + √3). The problem says that the simplified form can be written as A(1 + √B) - (√C + √D), where A, B, C, D are positive integers. Then I need to find A + B + C + D. Alright, let me think about how to approach this.

First, I remember that when you have a fraction with radicals in the denominator, you can rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator. The conjugate of a binomial with radicals is the same binomial but with the sign between the terms flipped. So, since the denominator is 2 + √3, its conjugate would be 2 - √3. Let me try that.

So, multiply numerator and denominator by (2 - √3):

[(1 + √2)(2 - √3)] / [(2 + √3)(2 - √3)]

Let me compute the denominator first because that might be simpler. Using the difference of squares formula, (a + b)(a - b) = a² - b². So here, a is 2 and b is √3. Therefore:

Denominator = 2² - (√3)² = 4 - 3 = 1. Wait, that's nice! The denominator simplifies to 1? That seems too easy. Let me check again.

Yes, 2 squared is 4, and (√3) squared is 3. So 4 - 3 is 1. So the denominator is 1. That means the entire expression simplifies to just the numerator after multiplying by the conjugate. So that means the original fraction simplifies to (1 + √2)(2 - √3). Let me compute that.

Expanding the numerator:

First, multiply 1 by (2 - √3): that's 2 - √3.

Then multiply √2 by (2 - √3): that's 2√2 - √2 * √3.

So combining those two parts:

2 - √3 + 2√2 - √6.

Wait, so the entire expression simplifies to 2 - √3 + 2√2 - √6. Let me write that out:

2 + 2√2 - √3 - √6.

Hmm, okay. Now the problem says that this can be written in the form A(1 + √B) - (√C + √D). Let me see how to rearrange the terms I have here.

Looking at my simplified expression: 2 + 2√2 - √3 - √6.

I need to express this as something like A(1 + √B) minus (√C + √D). Let me think.

First, maybe I can group terms. Let me see:

The terms with √2 and the constants. Let's look at the constants and the radicals:

The constant term is 2. Then there is 2√2. Then we have -√3 and -√6.

Wait, perhaps I can factor something out. Let me try to group the constants and the radicals. Let me think: If I can write 2 + 2√2 as A(1 + √B), then maybe?

Let me suppose that 2 + 2√2 is equal to A(1 + √B). Let me solve for A and B. Let me suppose that A(1 + √B) = 2 + 2√2. Let me assume A is an integer. Let me see, if A is 2, then 2(1 + √B) = 2 + 2√B. Comparing to 2 + 2√2, that would mean √B is √2, so B is 2. So then, that works. Therefore, 2 + 2√2 is 2(1 + √2). So that's a good start.

So, then our expression is 2(1 + √2) - √3 - √6. Now, the problem wants the form A(1 + √B) - (√C + √D). So that seems to fit if we take A = 2, B = 2, and then the rest is - (√3 + √6). Therefore, C is 3 and D is 6? Wait, but let me check if that's acceptable. Because the problem says that the form is A(1 + √B) - (√C + √D). So yes, if we have 2(1 + √2) - (√3 + √6), then A = 2, B = 2, C = 3, D = 6. Therefore, A + B + C + D would be 2 + 2 + 3 + 6 = 13? Wait, is that correct?

But let me check again if that's the correct way to represent the expression. Let me verify that my original expression equals 2(1 + √2) - √3 - √6. Let me compute 2(1 + √2) - √3 - √6:

First, 2*1 is 2, 2*√2 is 2√2. Then minus √3 minus √6. So that's indeed 2 + 2√2 - √3 - √6. Which matches exactly the expression I had earlier. Therefore, this seems to fit. So then A is 2, B is 2, C is 3, D is 6. Therefore, adding them: 2 + 2 + 3 + 6 = 13. Is 13 the answer?

But wait, let me check again if there might be another way to represent the expression that would result in different numbers for A, B, C, D. Because maybe there's another way to factor or combine terms. Let me think.

Alternatively, perhaps there's a different grouping. Let me check if there's a different expression. Let me consider if I can write the expression in terms of different radicals. For example, maybe √6 can be written as √2 * √3. But I don't know if that helps. Alternatively, maybe grouping terms differently. Let me see.

Original simplified expression: 2 + 2√2 - √3 - √6. Let me think if there's a different way to group terms. For example, can I combine the -√3 - √6 into something else? Let me see:

Is there a way to factor out something from -√3 - √6? For example, -√3 - √6 = -√3(1 + √2). Wait, let's check:

-√3(1 + √2) = -√3 - √3*√2 = -√3 - √6. Yes! That's exactly the same. So that means the expression can be written as 2 + 2√2 - √3(1 + √2). Then, 2 + 2√2 is 2(1 + √2), so we can write:

2(1 + √2) - √3(1 + √2) = [2 - √3](1 + √2). Wait, but that would be factoring out (1 + √2), but that's different from the required form. The problem wants it in the form A(1 + √B) - (√C + √D). So if we write it as [2 - √3](1 + √2), that is a product, not the required form. So perhaps that's not helpful.

But maybe if I consider that [2 - √3](1 + √2) is equivalent to my previous expression, but the problem wants it as A(1 + √B) - (√C + √D). So maybe I need to stick with the earlier approach where I have 2(1 + √2) - √3 - √6. But then, if I write that as 2(1 + √2) - (√3 + √6), then that's exactly the form required. So then A is 2, B is 2, C is 3, D is 6, so their sum is 2 + 2 + 3 + 6 = 13. But let me check again if that's correct. Maybe there's a mistake here.

Wait, but the problem says "Your solution can be converted to the form A(1 + √B) - (√C + √D)". So maybe there's another way to express this? Let me check if there's another possible combination. Let me think. For instance, maybe trying to combine √6 with something else. For example, is there a way to have different values for A, B, C, D? Let me think.

Alternatively, suppose I didn't rationalize the denominator in the beginning. Let me see. But that seems more complicated. Let me try. Wait, I think my first approach is correct. Let me check my calculation again step by step.

Original expression: (1 + √2)/(2 + √3). Multiply numerator and denominator by (2 - √3):

Numerator: (1 + √2)(2 - √3). Let me compute this again:

Multiply 1 by 2: 2.

Multiply 1 by (-√3): -√3.

Multiply √2 by 2: 2√2.

Multiply √2 by (-√3): -√2*√3 = -√6.

So adding them up: 2 - √3 + 2√2 - √6. Yes, that's correct. So numerator is 2 + 2√2 - √3 - √6, denominator is 1. So that's the simplified expression. So then, as I said before, 2 + 2√2 - √3 - √6 can be written as 2(1 + √2) - (√3 + √6). Therefore, A = 2, B = 2, C = 3, D = 6. Therefore, A + B + C + D = 2 + 2 + 3 + 6 = 13.

But wait, let me check if there's another way. For example, maybe if we take A(1 + √B) - (√C + √D) and try to match coefficients?

Suppose I think of the expression as follows: Let me assume that the expression is equal to A(1 + √B) - √C - √D. Let me write that:

A + A√B - √C - √D.

Comparing this to the expression we have: 2 + 2√2 - √3 - √6. Therefore, we can set up equations:

For the constant term: A = 2.

For the √B term: A√B = 2√2. Since A is 2, then √B = √2 => B = 2.

Then for the other radicals: -√C - √D = -√3 - √6. Which implies that √C = √3 and √D = √6, so C = 3, D = 6. Alternatively, maybe √C and √D can be arranged differently? For example, if we have -√C - √D, then maybe C and D can be 6 and 3? But since addition is commutative, it doesn't matter. The problem says "positive integers", so C and D are positive integers regardless of order. Therefore, C=3 and D=6, or vice versa. But since they are in the form (√C + √D), order doesn't matter. So regardless, the sum A + B + C + D is 2 + 2 + 3 + 6 = 13.

Is there any other possible way? Let me think. Suppose someone tried to write the expression as something else. For example, can I combine the terms differently? For instance, if I think that √6 is √2*√3, but not sure if that helps. Let me see:

Suppose I want to write the expression as A(1 + √B) - (√C + √D). If I tried to set up different equations. Let me suppose that instead of grouping 2 + 2√2, maybe there's another way. But unless there is a different way to factor, I don't see another way. For example, could I write 2 + 2√2 - √3 - √6 as, say, (something else)?

Alternatively, let me think if there is a mistake in the problem statement. For example, if the answer is different? Let me check my calculation again. Let me recompute the numerator:

(1 + √2)(2 - √3). Let me use distributive property again:

First, 1*(2 - √3) = 2 - √3.

Then √2*(2 - √3) = 2√2 - √2*√3. √2*√3 is √6. So that term is 2√2 - √6.

So adding both parts: 2 - √3 + 2√2 - √6. Yes, that's correct. So 2 + 2√2 - √3 - √6. Correct.

Alternatively, maybe there's another way to rationalize or manipulate the original expression? Let me think. Wait, maybe I made a mistake in assuming that multiplying by the conjugate is the only way? Let me check. Suppose instead of multiplying by 2 - √3, I did something else? For example, perhaps multiply numerator and denominator by something else? But the standard method is to multiply by the conjugate, and since denominator is 2 + √3, multiplying by 2 - √3 is the right way. So that should be correct.

Alternatively, maybe the problem expects a different form? Let me check. The problem says "Your solution can be converted to the form A(1 + √B) - (√C + √D)". So maybe there's a different way? Let me try to think.

Let me consider that 2 + 2√2 - √3 - √6. Maybe I can write this as 2(1 + √2) - √3 - √6. But perhaps there's another way to group terms. For example, is there a way to write -√3 - √6 as -√3(1 + √2)? Wait, yes, as I thought earlier. Let me check:

-√3(1 + √2) = -√3 - √3*√2 = -√3 - √6. Correct. So then, the entire expression becomes 2(1 + √2) - √3(1 + √2) = (2 - √3)(1 + √2). But that would be the product form, not the required form. But the required form is A(1 + √B) - (√C + √D). So unless I can write (2 - √3)(1 + √2) in that form, but that seems like it would require expanding again.

Alternatively, maybe if I take A = (2 - √3) and then multiply by (1 + √2). But that's not helpful. So perhaps the initial approach is the correct one.

Alternatively, could I have different values for A, B, C, D? Let me think. Suppose I take A = 1, then 1*(1 + √B) - (√C + √D). Let me see if that's possible. Then the expression would be 1 + √B - √C - √D. Comparing to our expression 2 + 2√2 - √3 - √6. Then, equating:

1 + √B - √C - √D = 2 + 2√2 - √3 - √6. Therefore, 1 + √B - √C - √D = 2 + 2√2 - √3 - √6. Therefore, √B - √C - √D = 1 + 2√2 - √3 - √6. But this seems complicated. It would require matching terms. For example, √B would have to account for the 2√2 term. But √B is a single radical, so this seems impossible. Therefore, A can't be 1. Similarly, trying A = 3? Probably not. Therefore, likely A = 2 is the correct value.

Alternatively, if I take A = something else. Let me think. For example, if A = 1 and B = something else. But I think it's not possible. So I think A = 2, B = 2, C = 3, D = 6 is the correct answer. Then A + B + C + D = 2 + 2 + 3 + 6 = 13.

But let me check with another approach. Suppose I didn't rationalize the denominator first. Let me see if I can do something else.

Original problem: (1 + √2)/(2 + √3). Maybe I can split the fraction into two terms? Let me try:

(1 + √2)/(2 + √3) = 1/(2 + √3) + √2/(2 + √3). Then I can rationalize each term separately.

First term: 1/(2 + √3). Multiply numerator and denominator by 2 - √3:

[1*(2 - √3)] / [(2 + √3)(2 - √3)] = (2 - √3)/1 = 2 - √3.

Second term: √2/(2 + √3). Multiply numerator and denominator by 2 - √3:

[√2*(2 - √3)] / 1 = 2√2 - √6, same as before.

Therefore, adding both terms: (2 - √3) + (2√2 - √6) = 2 + 2√2 - √3 - √6. Same as before. Therefore, same expression. So again, leading to same conclusion.

Therefore, seems like 13 is the answer. But just to be thorough, let me check if the problem expects different values for A, B, C, D. Maybe I need to check if the form A(1 + √B) - (√C + √D) has constraints on the values. For example, maybe B, C, D are square-free or something? Wait, the problem says they are positive integers, no other constraints. So √C and √D can be any radicals, even if they have squares. For example, √6 is okay, even though 6 isn't square-free. So in that case, 3 and 6 are acceptable as C and D. So then, I think the answer is 13.

But just to check, maybe there's another way? For example, suppose we have different grouping. For example, suppose I thought of 2 + 2√2 - √3 - √6 as something like 2(1 + √2) - √3 - √6, which is the same as before. Alternatively, is there a way to write that as, say, something with √6? Let me think. For example, if I had written 2(1 + √2) - √3 - √6, maybe I can write √3 + √6 as √3 + √(2*3) = √3 + √2√3 = √3(1 + √2). Which is exactly what I did earlier. So then, that would make the entire expression as 2(1 + √2) - √3(1 + √2) = (2 - √3)(1 + √2). But again, that is a product, not the required form. Therefore, if I wanted to write it as A(1 + √B) - (√C + √D), then I can't combine those terms into a single factor. Therefore, the initial approach is the way to go.

Alternatively, suppose someone thought that maybe B is different? Let me think. For example, is there a way to have B = something else? Suppose I tried to write 2 + 2√2 as something else. For example, if A is 1, then 1*(something) would need to have 2 + 2√2. But 1*(anything) would have to be 2 + 2√2. So if B is 8, then 1 + √8 = 1 + 2√2, so multiplying by 2 would give 2 + 4√2. Which isn't matching. Alternatively, if B is 2, then A(1 + √2) = 2 + 2√2 implies A = 2, which is the same as before. So no other B seems possible. For example, if B was 8, then A*(1 + √8) = A*(1 + 2√2). To get 2 + 2√2, A would need to be 2/ (1 + 2√2), which is not an integer. So that's not acceptable. Therefore, B must be 2, and A must be 2. Therefore, I think my answer is correct.

Therefore, the answer is 13.

But let me check once again with another approach. Let me assume that the expression is in the form A(1 + √B) - (√C + √D), and let me try to find A, B, C, D by equating terms.

Let me write:

A(1 + √B) - √C - √D = A + A√B - √C - √D.

Compare to our expression: 2 + 2√2 - √3 - √6. Therefore, the constants must match: A = 2. Then, the coefficients of the radicals: A√B must equal 2√2. Since A = 2, then √B = √2, so B = 2. Then, the remaining terms: -√C - √D must equal -√3 - √6. Therefore, √C and √D must be √3 and √6, so C = 3 and D = 6. Therefore, A + B + C + D = 2 + 2 + 3 + 6 = 13. So that confirms it again.

Alternatively, maybe the problem expects the radicals to be in a certain order or something else? For example, maybe C and D need to be square-free? But even then, √6 is not square-free, but the problem didn't specify that. The problem just says positive integers. Therefore, even if they aren't square-free, it's acceptable. So, I think the answer is 13.

But let me check with the original problem once again. The problem says "Simplify (1 + √2)/(2 + √3). Your solution can be converted to the form A(1 + √B) - (√C + √D)... What is A + B + C + D?" So my answer is 13. But to make sure, maybe I can check with approximate decimal values.

Compute original expression: (1 + √2)/(2 + √3). Let me compute that numerically:

First, √2 ≈ 1.4142, so numerator ≈ 1 + 1.4142 ≈ 2.4142.

Denominator: √3 ≈ 1.732, so denominator ≈ 2 + 1.732 ≈ 3.732.

So the value is approximately 2.4142 / 3.732 ≈ 0.647.

Now, let me compute 2(1 + √2) - √3 - √6:

Compute 2(1 + √2) ≈ 2*(1 + 1.4142) ≈ 2*2.4142 ≈ 4.8284.

Then subtract √3 ≈ 1.732 and √6 ≈ 2.449. So total is 4.8284 - 1.732 - 2.449 ≈ 4.8284 - 4.181 ≈ 0.6474. Which matches the original value. So that's correct.

Therefore, the answer must be 13. Therefore, I think that is correct.

**Final Answer**
\boxed{13}

To simplify the expression
$$
\frac{1 + \sqrt{2}}{2 + \sqrt{3}},
$$
we begin by rationalizing the denominator. The conjugate of $2 + \sqrt{3}$ is $2 - \sqrt{3}$, so we multiply both the numerator and the denominator by this conjugate:

$$
\frac{1 + \sqrt{2}}{2 + \sqrt{3}} \cdot \frac{2 - \sqrt{3}}{2 - \sqrt{3}}.
$$

---

### Step 1: Simplify the Denominator

Using the identity $(a + b)(a - b) = a^2 - b^2$, we compute the denominator:

$$
(2 + \sqrt{3})(2 - \sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1.
$$

So the denominator becomes 1, and the expression simplifies to just the numerator:

$$
(1 + \sqrt{2})(2 - \sqrt{3}).
$$

---

### Step 2: Expand the Numerator

Now expand the numerator using the distributive property:

$$
(1 + \sqrt{2})(2 - \sqrt{3}) = 1 \cdot (2 - \sqrt{3}) + \sqrt{2} \cdot (2 - \sqrt{3}),
$$

$$
= (2 - \sqrt{3}) + (2\sqrt{2} - \sqrt{6}) = 2 + 2\sqrt{2} - \sqrt{3} - \sqrt{6}.
$$

So, the simplified form of the expression is:

$$
2 + 2\sqrt{2} - \sqrt{3} - \sqrt{6}.
$$

---

### Step 3: Match the Given Form

We aim to express this in the form:

$$
A(1 + \sqrt{B}) - (\sqrt{C} + \sqrt{D}).
$$

Let’s look at the constant and radical terms:

- The term $2 + 2\sqrt{2}$ can be written as:
  $$
  2(1 + \sqrt{2}),
  $$
  so $A = 2$, $B = 2$.

- The remaining terms are:
  $$
  - \sqrt{3} - \sqrt{6} = - (\sqrt{3} + \sqrt{6}),
  $$
  so $C = 3$, $D = 6$.

Thus, the simplified expression can indeed be written as:

$$
2(1 + \sqrt{2}) - (\sqrt{3} + \sqrt{6}).
$$

---

### Final Step: Compute the Sum

We are given that the expression is of the form
$$
A(1 + \sqrt{B}) - (\sqrt{C} + \sqrt{D}),
$$
and we are to find $A + B + C + D$.

From above:

- $A = 2$
- $B = 2$
- $C = 3$
- $D = 6$

So,

$$
A + B + C + D = 2 + 2 + 3 + 6 = \boxed{13}.
$$

[Assistant]
Explanation:
Okay, so I need to simplify the expression (1 + √2)/(2 + √3). The problem says that the simplified form can be written as A(1 + √B) - (√C + √D), where A, B, C, D are positive integers. Then I need to find A + B + C + D. Alright, let me think about how to approach this.

First, I remember that when you have a fraction with radicals in the denominator, you can rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator. The conjugate of a binomial with radicals is the same binomial but with the sign between the terms flipped. So, since the denominator is 2 + √3, its conjugate would be 2 - √3. Let me try that.

So, multiply numerator and denominator by (2 - √3):

[(1 + √2)(2 - √3)] / [(2 + √3)(2 - √3)]

Let me compute the denominator first because that might be simpler. Using the difference of squares formula, (a + b)(a - b) = a² - b². So here, a is 2 and b is √3. Therefore:

Denominator = 2² - (√3)² = 4 - 3 = 1. Wait, that's nice! The denominator simplifies to 1? That seems too easy. Let me check again.

Yes, 2 squared is 4, and (√3) squared is 3. So 4 - 3 is 1. So the denominator is 1. That means the entire expression simplifies to just the numerator after multiplying by the conjugate. So that means the original fraction simplifies to (1 + √2)(2 - √3). Let me compute that.

Expanding the numerator:

First, multiply 1 by (2 - √3): that's 2 - √3.

Then multiply √2 by (2 - √3): that's 2√2 - √2 * √3.

So combining those two parts:

2 - √3 + 2√2 - √6.

Wait, so the entire expression simplifies to 2 - √3 + 2√2 - √6. Let me write that out:

2 + 2√2 - √3 - √6.

Hmm, okay. Now the problem says that this can be written in the form A(1 + √B) - (√C + √D). Let me see how to rearrange the terms I have here.

Looking at my simplified expression: 2 + 2√2 - √3 - √6.

I need to express this as something like A(1 + √B) minus (√C + √D). Let me think.

First, maybe I can group terms. Let me see:

The terms with √2 and the constants. Let's look at the constants and the radicals:

The constant term is 2. Then there is 2√2. Then we have -√3 and -√6.

Wait, perhaps I can factor something out. Let me try to group the constants and the radicals. Let me think: If I can write 2 + 2√2 as A(1 + √B), then maybe?

Let me suppose that 2 + 2√2 is equal to A(1 + √B). Let me solve for A and B. Let me suppose that A(1 + √B) = 2 + 2√2. Let me assume A is an integer. Let me see, if A is 2, then 2(1 + √B) = 2 + 2√B. Comparing to 2 + 2√2, that would mean √B is √2, so B is 2. So then, that works. Therefore, 2 + 2√2 is 2(1 + √2). So that's a good start.

So, then our expression is 2(1 + √2) - √3 - √6. Now, the problem wants the form A(1 + √B) - (√C + √D). So that seems to fit if we take A = 2, B = 2, and then the rest is - (√3 + √6). Therefore, C is 3 and D is 6? Wait, but let me check if that's acceptable. Because the problem says that the form is A(1 + √B) - (√C + √D). So yes, if we have 2(1 + √2) - (√3 + √6), then A = 2, B = 2, C = 3, D = 6. Therefore, A + B + C + D would be 2 + 2 + 3 + 6 = 13? Wait, is that correct?

But let me check again if that's the correct way to represent the expression. Let me verify that my original expression equals 2(1 + √2) - √3 - √6. Let me compute 2(1 + √2) - √3 - √6:

First, 2*1 is 2, 2*√2 is 2√2. Then minus √3 minus √6. So that's indeed 2 + 2√2 - √3 - √6. Which matches exactly the expression I had earlier. Therefore, this seems to fit. So then A is 2, B is 2, C is 3, D is 6. Therefore, adding them: 2 + 2 + 3 + 6 = 13. Is 13 the answer?

But wait, let me check again if there might be another way to represent the expression that would result in different numbers for A, B, C, D. Because maybe there's another way to factor or combine terms. Let me think.

Alternatively, perhaps there's a different grouping. Let me check if there's a different expression. Let me consider if I can write the expression in terms of different radicals. For example, maybe √6 can be written as √2 * √3. But I don't know if that helps. Alternatively, maybe grouping terms differently. Let me see.

Original simplified expression: 2 + 2√2 - √3 - √6. Let me think if there's a different way to group terms. For example, can I combine the -√3 - √6 into something else? Let me see:

Is there a way to factor out something from -√3 - √6? For example, -√3 - √6 = -√3(1 + √2). Wait, let's check:

-√3(1 + √2) = -√3 - √3*√2 = -√3 - √6. Yes! That's exactly the same. So that means the expression can be written as 2 + 2√2 - √3(1 + √2). Then, 2 + 2√2 is 2(1 + √2), so we can write:

2(1 + √2) - √3(1 + √2) = [2 - √3](1 + √2). Wait, but that would be factoring out (1 + √2), but that's different from the required form. The problem wants it in the form A(1 + √B) - (√C + √D). So if we write it as [2 - √3](1 + √2), that is a product, not the required form. So perhaps that's not helpful.

But maybe if I consider that [2 - √3](1 + √2) is equivalent to my previous expression, but the problem wants it as A(1 + √B) - (√C + √D). So maybe I need to stick with the earlier approach where I have 2(1 + √2) - √3 - √6. But then, if I write that as 2(1 + √2) - (√3 + √6), then that's exactly the form required. So then A is 2, B is 2, C is 3, D is 6, so their sum is 2 + 2 + 3 + 6 = 13. But let me check again if that's correct. Maybe there's a mistake here.

Wait, but the problem says "Your solution can be converted to the form A(1 + √B) - (√C + √D)". So maybe there's another way to express this? Let me check if there's another possible combination. Let me think. For instance, maybe trying to combine √6 with something else. For example, is there a way to have different values for A, B, C, D? Let me think.

Alternatively, suppose I didn't rationalize the denominator in the beginning. Let me see. But that seems more complicated. Let me try. Wait, I think my first approach is correct. Let me check my calculation again step by step.

Original expression: (1 + √2)/(2 + √3). Multiply numerator and denominator by (2 - √3):

Numerator: (1 + √2)(2 - √3). Let me compute this again:

Multiply 1 by 2: 2.

Multiply 1 by (-√3): -√3.

Multiply √2 by 2: 2√2.

Multiply √2 by (-√3): -√2*√3 = -√6.

So adding them up: 2 - √3 + 2√2 - √6. Yes, that's correct. So numerator is 2 + 2√2 - √3 - √6, denominator is 1. So that's the simplified expression. So then, as I said before, 2 + 2√2 - √3 - √6 can be written as 2(1 + √2) - (√3 + √6). Therefore, A = 2, B = 2, C = 3, D = 6. Therefore, A + B + C + D = 2 + 2 + 3 + 6 = 13.

But wait, let me check if there's another way. For example, maybe if we take A(1 + √B) - (√C + √D) and try to match coefficients?

Suppose I think of the expression as follows: Let me assume that the expression is equal to A(1 + √B) - √C - √D. Let me write that:

A + A√B - √C - √D.

Comparing this to the expression we have: 2 + 2√2 - √3 - √6. Therefore, we can set up equations:

For the constant term: A = 2.

For the √B term: A√B = 2√2. Since A is 2, then √B = √2 => B = 2.

Then for the other radicals: -√C - √D = -√3 - √6. Which implies that √C = √3 and √D = √6, so C = 3, D = 6. Alternatively, maybe √C and √D can be arranged differently? For example, if we have -√C - √D, then maybe C and D can be 6 and 3? But since addition is commutative, it doesn't matter. The problem says "positive integers", so C and D are positive integers regardless of order. Therefore, C=3 and D=6, or vice versa. But since they are in the form (√C + √D), order doesn't matter. So regardless, the sum A + B + C + D is 2 + 2 + 3 + 6 = 13.

Is there any other possible way? Let me think. Suppose someone tried to write the expression as something else. For example, can I combine the terms differently? For instance, if I think that √6 is √2*√3, but not sure if that helps. Let me see:

Suppose I want to write the expression as A(1 + √B) - (√C + √D). If I tried to set up different equations. Let me suppose that instead of grouping 2 + 2√2, maybe there's another way. But unless there is a different way to factor, I don't see another way. For example, could I write 2 + 2√2 - √3 - √6 as, say, (something else)?

Alternatively, let me think if there is a mistake in the problem statement. For example, if the answer is different? Let me check my calculation again. Let me recompute the numerator:

(1 + √2)(2 - √3). Let me use distributive property again:

First, 1*(2 - √3) = 2 - √3.

Then √2*(2 - √3) = 2√2 - √2*√3. √2*√3 is √6. So that term is 2√2 - √6.

So adding both parts: 2 - √3 + 2√2 - √6. Yes, that's correct. So 2 + 2√2 - √3 - √6. Correct.

Alternatively, maybe there's another way to rationalize or manipulate the original expression? Let me think. Wait, maybe I made a mistake in assuming that multiplying by the conjugate is the only way? Let me check. Suppose instead of multiplying by 2 - √3, I did something else? For example, perhaps multiply numerator and denominator by something else? But the standard method is to multiply by the conjugate, and since denominator is 2 + √3, multiplying by 2 - √3 is the right way. So that should be correct.

Alternatively, maybe the problem expects a different form? Let me check. The problem says "Your solution can be converted to the form A(1 + √B) - (√C + √D)". So maybe there's a different way? Let me try to think.

Let me consider that 2 + 2√2 - √3 - √6. Maybe I can write this as 2(1 + √2) - √3 - √6. But perhaps there's another way to group terms. For example, is there a way to write -√3 - √6 as -√3(1 + √2)? Wait, yes, as I thought earlier. Let me check:

-√3(1 + √2) = -√3 - √3*√2 = -√3 - √6. Correct. So then, the entire expression becomes 2(1 + √2) - √3(1 + √2) = (2 - √3)(1 + √2). But that would be the product form, not the required form. But the required form is A(1 + √B) - (√C + √D). So unless I can write (2 - √3)(1 + √2) in that form, but that seems like it would require expanding again.

Alternatively, maybe if I take A = (2 - √3) and then multiply by (1 + √2). But that's not helpful. So perhaps the initial approach is the correct one.

Alternatively, could I have different values for A, B, C, D? Let me think. Suppose I take A = 1, then 1*(1 + √B) - (√C + √D). Let me see if that's possible. Then the expression would be 1 + √B - √C - √D. Comparing to our expression 2 + 2√2 - √3 - √6. Then, equating:

1 + √B - √C - √D = 2 + 2√2 - √3 - √6. Therefore, 1 + √B - √C - √D = 2 + 2√2 - √3 - √6. Therefore, √B - √C - √D = 1 + 2√2 - √3 - √6. But this seems complicated. It would require matching terms. For example, √B would have to account for the 2√2 term. But √B is a single radical, so this seems impossible. Therefore, A can't be 1. Similarly, trying A = 3? Probably not. Therefore, likely A = 2 is the correct value.

Alternatively, if I take A = something else. Let me think. For example, if A = 1 and B = something else. But I think it's not possible. So I think A = 2, B = 2, C = 3, D = 6 is the correct answer. Then A + B + C + D = 2 + 2 + 3 + 6 = 13.

But let me check with another approach. Suppose I didn't rationalize the denominator first. Let me see if I can do something else.

Original problem: (1 + √2)/(2 + √3). Maybe I can split the fraction into two terms? Let me try:

(1 + √2)/(2 + √3) = 1/(2 + √3) + √2/(2 + √3). Then I can rationalize each term separately.

First term: 1/(2 + √3). Multiply numerator and denominator by 2 - √3:

[1*(2 - √3)] / [(2 + √3)(2 - √3)] = (2 - √3)/1 = 2 - √3.

Second term: √2/(2 + √3). Multiply numerator and denominator by 2 - √3:

[√2*(2 - √3)] / 1 = 2√2 - √6, same as before.

Therefore, adding both terms: (2 - √3) + (2√2 - √6) = 2 + 2√2 - √3 - √6. Same as before. Therefore, same expression. So again, leading to same conclusion.

Therefore, seems like 13 is the answer. But just to be thorough, let me check if the problem expects different values for A, B, C, D. Maybe I need to check if the form A(1 + √B) - (√C + √D) has constraints on the values. For example, maybe B, C, D are square-free or something? Wait, the problem says they are positive integers, no other constraints. So √C and √D can be any radicals, even if they have squares. For example, √6 is okay, even though 6 isn't square-free. So in that case, 3 and 6 are acceptable as C and D. So then, I think the answer is 13.

But just to check, maybe there's another way? For example, suppose we have different grouping. For example, suppose I thought of 2 + 2√2 - √3 - √6 as something like 2(1 + √2) - √3 - √6, which is the same as before. Alternatively, is there a way to write that as, say, something with √6? Let me think. For example, if I had written 2(1 + √2) - √3 - √6, maybe I can write √3 + √6 as √3 + √(2*3) = √3 + √2√3 = √3(1 + √2). Which is exactly what I did earlier. So then, that would make the entire expression as 2(1 + √2) - √3(1 + √2) = (2 - √3)(1 + √2). But again, that is a product, not the required form. Therefore, if I wanted to write it as A(1 + √B) - (√C + √D), then I can't combine those terms into a single factor. Therefore, the initial approach is the way to go.

Alternatively, suppose someone thought that maybe B is different? Let me think. For example, is there a way to have B = something else? Suppose I tried to write 2 + 2√2 as something else. For example, if A is 1, then 1*(something) would need to have 2 + 2√2. But 1*(anything) would have to be 2 + 2√2. So if B is 8, then 1 + √8 = 1 + 2√2, so multiplying by 2 would give 2 + 4√2. Which isn't matching. Alternatively, if B is 2, then A(1 + √2) = 2 + 2√2 implies A = 2, which is the same as before. So no other B seems possible. For example, if B was 8, then A*(1 + √8) = A*(1 + 2√2). To get 2 + 2√2, A would need to be 2/ (1 + 2√2), which is not an integer. So that's not acceptable. Therefore, B must be 2, and A must be 2. Therefore, I think my answer is correct.

Therefore, the answer is 13.

But let me check once again with another approach. Let me assume that the expression is in the form A(1 + √B) - (√C + √D), and let me try to find A, B, C, D by equating terms.

Let me write:

A(1 + √B) - √C - √D = A + A√B - √C - √D.

Compare to our expression: 2 + 2√2 - √3 - √6. Therefore, the constants must match: A = 2. Then, the coefficients of the radicals: A√B must equal 2√2. Since A = 2, then √B = √2, so B = 2. Then, the remaining terms: -√C - √D must equal -√3 - √6. Therefore, √C and √D must be √3 and √6, so C = 3 and D = 6. Therefore, A + B + C + D = 2 + 2 + 3 + 6 = 13. So that confirms it again.

Alternatively, maybe the problem expects the radicals to be in a certain order or something else? For example, maybe C and D need to be square-free? But even then, √6 is not square-free, but the problem didn't specify that. The problem just says positive integers. Therefore, even if they aren't square-free, it's acceptable. So, I think the answer is 13.

But let me check with the original problem once again. The problem says "Simplify (1 + √2)/(2 + √3). Your solution can be converted to the form A(1 + √B) - (√C + √D)... What is A + B + C + D?" So my answer is 13. But to make sure, maybe I can check with approximate decimal values.

Compute original expression: (1 + √2)/(2 + √3). Let me compute that numerically:

First, √2 ≈ 1.4142, so numerator ≈ 1 + 1.4142 ≈ 2.4142.

Denominator: √3 ≈ 1.732, so denominator ≈ 2 + 1.732 ≈ 3.732.

So the value is approximately 2.4142 / 3.732 ≈ 0.647.

Now, let me compute 2(1 + √2) - √3 - √6:

Compute 2(1 + √2) ≈ 2*(1 + 1.4142) ≈ 2*2.4142 ≈ 4.8284.

Then subtract √3 ≈ 1.732 and √6 ≈ 2.449. So total is 4.8284 - 1.732 - 2.449 ≈ 4.8284 - 4.181 ≈ 0.6474. Which matches the original value. So that's correct.

Therefore, the answer must be 13. Therefore, I think that is correct.

**Final Answer**
\boxed{13}

To simplify the expression
$$
\frac{1 + \sqrt{2}}{2 + \sqrt{3}},
$$
we begin by rationalizing the denominator. The conjugate of $2 + \sqrt{3}$ is $2 - \sqrt{3}$, so we multiply both the numerator and the denominator by this conjugate:

$$
\frac{1 + \sqrt{2}}{2 + \sqrt{3}} \cdot \frac{2 - \sqrt{3}}{2 - \sqrt{3}}.
$$

---

### Step 1: Simplify the Denominator

Using the identity $(a + b)(a - b) = a^2 - b^2$, we compute the denominator:

$$
(2 + \sqrt{3})(2 - \sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1.
$$

So the denominator becomes 1, and the expression simplifies to just the numerator:

$$
(1 + \sqrt{2})(2 - \sqrt{3}).
$$

---

### Step 2: Expand the Numerator

Now expand the numerator using the distributive property:

$$
(1 + \sqrt{2})(2 - \sqrt{3}) = 1 \cdot (2 - \sqrt{3}) + \sqrt{2} \cdot (2 - \sqrt{3}),
$$

$$
= (2 - \sqrt{3}) + (2\sqrt{2} - \sqrt{6}) = 2 + 2\sqrt{2} - \sqrt{3} - \sqrt{6}.
$$

So, the simplified form of the expression is:

$$
2 + 2\sqrt{2} - \sqrt{3} - \sqrt{6}.
$$

---

### Step 3: Match the Given Form

We aim to express this in the form:

$$
A(1 + \sqrt{B}) - (\sqrt{C} + \sqrt{D}).
$$

Let’s look at the constant and radical terms:

- The term $2 + 2\sqrt{2}$ can be written as:
  $$
  2(1 + \sqrt{2}),
  $$
  so $A = 2$, $B = 2$.

- The remaining terms are:
  $$
  - \sqrt{3} - \sqrt{6} = - (\sqrt{3} + \sqrt{6}),
  $$
  so $C = 3$, $D = 6$.

Thus, the simplified expression can indeed be written as:

$$
2(1 + \sqrt{2}) - (\sqrt{3} + \sqrt{6}).
$$

---

### Final Step: Compute the Sum

We are given that the expression is of the form
$$
A(1 + \sqrt{B}) - (\sqrt{C} + \sqrt{D}),
$$
and we are to find $A + B + C + D$.

From above:

- $A = 2$
- $B = 2$
- $C = 3$
- $D = 6$

So,

$$
A + B + C + D = 2 + 2 + 3 + 6 = \boxed{13}.
$$

Solution: Multiplying top and bottom by the conjugate, we have $\frac{1+\sqrt{2}}{2+\sqrt{3}} = \frac{(1+\sqrt{2})(2-\sqrt{3})}{(2+\sqrt{3})((2-\sqrt{3}))} = \frac{2-\sqrt{3}+2\sqrt{2}-\sqrt{6}}{4-3} = 2-\sqrt{3}+2\sqrt{2}-\sqrt{6}$. So, we obtain $A=2, B=2, C=3$ and $D=6$ ($C$ and $D$ are interchangeable). So $A+B+C+D = 2+2+3+6 = \boxed{13}$.