Explanation:
Okay, so I need to find the value of $ a_1 $ in a strictly increasing sequence of positive integers. The sequence has some specific properties related to geometric and arithmetic subsequences. Let me try to parse the problem again.

The problem says that for every positive integer $ k $, the subsequence $ a_{2k-1} $, $ a_{2k} $, $ a_{2k+1} $ is geometric. Also, the subsequence $ a_{2k} $, $ a_{2k+1} $, $ a_{2k+2} $ is arithmetic. And we know that $ a_{13} = 2016 $. We need to find $ a_1 $.

First, let me recall what geometric and arithmetic sequences are. A geometric sequence has a common ratio between consecutive terms, so if three terms $ x $, $ y $, $ z $ are geometric, then $ y^2 = xz $. An arithmetic sequence has a common difference, so if three terms $ x $, $ y $, $ z $ are arithmetic, then $ 2y = x + z $.

Given that the sequence is strictly increasing, all terms are positive integers, and each term is larger than the previous one. So, for example, $ a_1 < a_2 < a_3 < \dots $.

Let me try to write down the conditions for the first few terms. Maybe I can find a pattern or recurrence relation.

Starting with $ k = 1 $:

For the geometric subsequence $ a_1 $, $ a_2 $, $ a_3 $:

So, $ a_2^2 = a_1 a_3 $.

For the arithmetic subsequence $ a_2 $, $ a_3 $, $ a_4 $:

So, $ 2a_3 = a_2 + a_4 $, which implies $ a_4 = 2a_3 - a_2 $.

Similarly, for $ k = 2 $:

Geometric subsequence $ a_3 $, $ a_4 $, $ a_5 $:

So, $ a_4^2 = a_3 a_5 $, which gives $ a_5 = \frac{a_4^2}{a_3} $.

Arithmetic subsequence $ a_4 $, $ a_5 $, $ a_6 $:

So, $ 2a_5 = a_4 + a_6 $, hence $ a_6 = 2a_5 - a_4 $.

Similarly, for $ k = 3 $:

Geometric subsequence $ a_5 $, $ a_6 $, $ a_7 $:

$ a_6^2 = a_5 a_7 $, so $ a_7 = \frac{a_6^2}{a_5} $.

Arithmetic subsequence $ a_6 $, $ a_7 $, $ a_8 $:

$ 2a_7 = a_6 + a_8 $, so $ a_8 = 2a_7 - a_6 $.

And so on. It seems like there's a pattern here where odd-indexed terms are part of geometric sequences with the next even term and the following odd term, and even-indexed terms are part of arithmetic sequences with the next odd term and the following even term.

But since the sequence is strictly increasing, all these relations must hold with the terms increasing. Also, since all terms are positive integers, the divisions in the geometric sequences must result in integers. So, for example, $ a_3 $ must divide $ a_4^2 $, and $ a_5 $ must divide $ a_6^2 $, etc.

This seems recursive. Maybe I can express terms in terms of previous terms and find a recurrence relation?

Alternatively, maybe there's a way to express the terms in terms of $ a_1 $ and some ratio or difference?

Let me try to write down the first few terms in terms of $ a_1 $ and $ a_2 $, since those are the starting points.

Let me denote $ a_1 = x $, $ a_2 = y $. Since the sequence is strictly increasing, $ x < y $, and both are positive integers.

From the first geometric sequence (k=1): $ a_1, a_2, a_3 $ is geometric. Therefore:

$ a_2^2 = a_1 a_3 \implies y^2 = x a_3 \implies a_3 = \frac{y^2}{x} $.

Since $ a_3 $ must be an integer, $ x $ divides $ y^2 $.

Also, since the sequence is strictly increasing, $ a_3 > a_2 $, so $ \frac{y^2}{x} > y \implies \frac{y}{x} > 1 \implies y > x $, which we already know.

Then, from the arithmetic sequence $ a_2, a_3, a_4 $:

$ 2a_3 = a_2 + a_4 \implies a_4 = 2a_3 - a_2 = 2 \cdot \frac{y^2}{x} - y $.

Since $ a_4 $ must be an integer, $ \frac{2y^2}{x} - y $ must be integer. Let me write that as:

$ a_4 = \frac{2y^2 - x y}{x} = \frac{y(2y - x)}{x} $.

Therefore, $ x $ divides $ y(2y - x) $.

But since $ x $ divides $ y^2 $ from the previous condition, maybe we can find some relations here.

Let me note that $ x $ divides $ y^2 $, so let me denote $ y = x \cdot m $, where $ m $ is a positive integer greater than 1 (since $ y > x $). Wait, but $ m $ might not necessarily be integer? Wait, no, since $ y $ is an integer and $ x $ divides $ y^2 $, but $ y $ could have factors in common with $ x $. However, if I set $ y = x \cdot m $, then $ x $ divides $ y^2 $, which is $ x^2 m^2 $, so that's true. But maybe there's a better way to parametrize.

Alternatively, since $ x $ divides $ y^2 $, let me denote $ y = x \cdot k $, where $ k $ is a rational number. But since $ y $ is an integer, $ k $ must be rational. However, maybe it's better to think in terms of divisors.

Alternatively, perhaps I can express $ a_3 $ as $ \frac{y^2}{x} $, and since $ a_3 $ must be an integer, $ x $ divides $ y^2 $. Let me suppose that $ x $ divides $ y^2 $, so let me write $ y = x \cdot t $, where $ t $ is a rational number. Wait, but since $ y $ is an integer, $ t $ must be rational. However, maybe I can think of $ t $ as a fraction in reduced form. Let me think.

Alternatively, maybe I can think of $ x $ and $ y $ as having some common factors. Let me suppose that $ \gcd(x, y) = d $, so $ x = d \cdot m $, $ y = d \cdot n $, where $ \gcd(m, n) = 1 $. Then, since $ x $ divides $ y^2 $, $ d m $ divides $ d^2 n^2 $, so $ m $ divides $ d n^2 $. Since $ \gcd(m, n) = 1 $, $ m $ divides $ d $. Therefore, $ d = m \cdot k $ for some integer $ k $. Therefore, $ x = d m = m^2 k $, $ y = d n = m k n $.

So, substituting back, $ x = m^2 k $, $ y = m k n $, with $ \gcd(m, n) = 1 $.

This might be helpful. Let me note that.

So, with this parametrization, $ a_3 = \frac{y^2}{x} = \frac{(m k n)^2}{m^2 k} = \frac{m^2 k^2 n^2}{m^2 k} = k n^2 $.

So, $ a_3 = k n^2 $.

Then, moving on to $ a_4 $:

$ a_4 = \frac{y(2y - x)}{x} = \frac{m k n (2 m k n - m^2 k)}{m^2 k} $.

Let me compute numerator:

$ m k n (2 m k n - m^2 k) = m k n \cdot m k (2 n - m) = m^2 k^2 n (2n - m) $.

Divide by $ m^2 k $:

$ \frac{m^2 k^2 n (2n - m)}{m^2 k} = k n (2n - m) $.

Therefore, $ a_4 = k n (2n - m) $.

Since $ a_4 $ must be an integer, which it is, given that $ k, n, m $ are integers.

Also, since the sequence is strictly increasing, $ a_4 > a_3 $. Let's check:

$ a_4 = k n (2n - m) $, $ a_3 = k n^2 $.

So, $ k n (2n - m) > k n^2 \implies 2n - m > n \implies n > m $.

Therefore, $ n > m $.

So, we have $ \gcd(m, n) = 1 $, $ n > m $, and $ a_1 = m^2 k $, $ a_2 = m k n $, $ a_3 = k n^2 $, $ a_4 = k n (2n - m) $.

Now, moving on to the next geometric sequence for $ k = 2 $: $ a_3, a_4, a_5 $ is geometric. Therefore:

$ a_4^2 = a_3 a_5 \implies a_5 = \frac{a_4^2}{a_3} $.

Let me compute that:

$ a_5 = \frac{[k n (2n - m)]^2}{k n^2} = \frac{k^2 n^2 (2n - m)^2}{k n^2} = k (2n - m)^2 $.

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

Then, the arithmetic sequence $ a_4, a_5, a_6 $:

$ 2a_5 = a_4 + a_6 \implies a_6 = 2a_5 - a_4 $.

Compute $ a_6 $:

$ 2a_5 - a_4 = 2k(2n - m)^2 - k n (2n - m) = k(2n - m)[2(2n - m) - n] = k(2n - m)(4n - 2m - n) = k(2n - m)(3n - 2m) $.

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

Now, since the sequence is strictly increasing, $ a_6 > a_5 $:

$ k(2n - m)(3n - 2m) > k(2n - m)^2 \implies (3n - 2m) > (2n - m) \implies 3n - 2m > 2n - m \implies n > m $, which we already have. So that's consistent.

Now, moving on to the next geometric sequence for $ k = 3 $: $ a_5, a_6, a_7 $ is geometric. Therefore:

$ a_6^2 = a_5 a_7 \implies a_7 = \frac{a_6^2}{a_5} $.

Let me compute that:

$ a_7 = \frac{[k(2n - m)(3n - 2m)]^2}{k(2n - m)^2} = \frac{k^2 (2n - m)^2 (3n - 2m)^2}{k (2n - m)^2} = k (3n - 2m)^2 $.

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

Then, the arithmetic sequence $ a_6, a_7, a_8 $:

$ 2a_7 = a_6 + a_8 \implies a_8 = 2a_7 - a_6 $.

Compute $ a_8 $:

$ 2k(3n - 2m)^2 - k(2n - m)(3n - 2m) = k(3n - 2m)[2(3n - 2m) - (2n - m)] $.

Let me compute the bracket:

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

Therefore, $ a_8 = k(3n - 2m)(4n - 3m) $.

Continuing this pattern, perhaps we can see a recursive structure here. Let me check the next term.

For $ k = 4 $, the geometric sequence $ a_7, a_8, a_9 $:

$ a_8^2 = a_7 a_9 \implies a_9 = \frac{a_8^2}{a_7} $.

Compute $ a_9 $:

$ \frac{[k(3n - 2m)(4n - 3m)]^2}{k(3n - 2m)^2} = \frac{k^2 (3n - 2m)^2 (4n - 3m)^2}{k (3n - 2m)^2} = k(4n - 3m)^2 $.

So, $ a_9 = k(4n - 3m)^2 $.

Then, the arithmetic sequence $ a_8, a_9, a_{10} $:

$ 2a_9 = a_8 + a_{10} \implies a_{10} = 2a_9 - a_8 $.

Compute $ a_{10} $:

$ 2k(4n - 3m)^2 - k(3n - 2m)(4n - 3m) = k(4n - 3m)[2(4n - 3m) - (3n - 2m)] $.

Compute the bracket:

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

Therefore, $ a_{10} = k(4n - 3m)(5n - 4m) $.

Hmm, I see a pattern here. Let me try to see if I can generalize this.

Looking at the terms:

- $ a_1 = m^2 k $
- $ a_2 = m k n $
- $ a_3 = k n^2 $
- $ a_4 = k n (2n - m) $
- $ a_5 = k (2n - m)^2 $
- $ a_6 = k(2n - m)(3n - 2m) $
- $ a_7 = k(3n - 2m)^2 $
- $ a_8 = k(3n - 2m)(4n - 3m) $
- $ a_9 = k(4n - 3m)^2 $
- $ a_{10} = k(4n - 3m)(5n - 4m) $
- ...

So, it seems that for odd indices $ 2k - 1 $, the term is $ k \times (\text{some linear expression in } n \text{ and } m)^2 $, and for even indices $ 2k $, it's $ k \times (\text{some linear expression}) \times (\text{next linear expression}) $.

Looking at the coefficients:

For $ a_1 $, which is index 1 (odd), it's $ m^2 k $.

For $ a_3 $, index 3, it's $ k n^2 $.

For $ a_5 $, index 5, it's $ k (2n - m)^2 $.

For $ a_7 $, index 7, it's $ k (3n - 2m)^2 $.

For $ a_9 $, index 9, it's $ k (4n - 3m)^2 $.

So, the coefficients for the odd terms seem to follow a pattern where the coefficient is $ k \times ( (t)n - (t - 1)m )^2 $, where t is something. Let me check:

For $ a_1 $, t = 1? Wait, $ a_1 = m^2 k $. If I think of $ (1 \cdot n - 0 \cdot m)^2 = n^2 $, but $ a_1 $ is $ m^2 k $. Hmm, maybe not. Alternatively, maybe there's a different pattern.

Alternatively, looking at the coefficients for the odd terms:

- $ a_1 = m^2 k $
- $ a_3 = k n^2 $
- $ a_5 = k (2n - m)^2 $
- $ a_7 = k (3n - 2m)^2 $
- $ a_9 = k (4n - 3m)^2 $

So, if we think of the coefficients as starting with $ m $, then $ n $, then $ 2n - m $, then $ 3n - 2m $, then $ 4n - 3m $, etc. So, each time, the coefficient for the odd term is $ (t n - (t - 1)m) $, where t starts at 1? Wait:

For $ a_1 $, if t = 1, then $ 1 \cdot n - 0 \cdot m = n $, but $ a_1 $ is $ m^2 k $. Hmm, not quite. Wait, but maybe if we think of the coefficients for the odd terms as starting with $ m $, then $ n $, then $ 2n - m $, etc. So, perhaps there's a recursive relation here.

Alternatively, maybe we can think of the coefficients as following a linear recurrence. Let me denote $ c_t $ as the coefficient for the odd term at position $ 2t - 1 $. Then:

- $ c_1 = m $
- $ c_2 = n $
- $ c_3 = 2n - m $
- $ c_4 = 3n - 2m $
- $ c_5 = 4n - 3m $
- ...

So, each $ c_{t} = (t - 1)n - (t - 2)m $. Wait, for t=2: $ c_2 = 1 \cdot n - 0 \cdot m = n $. For t=3: $ 2n - m $. For t=4: 3n - 2m. Yes, that seems to hold. So, in general, $ c_t = (t - 1)n - (t - 2)m $. Therefore, for the odd term $ a_{2t - 1} $, it's $ k \cdot c_t^2 $.

Similarly, for the even terms:

- $ a_2 = m k n $
- $ a_4 = k n (2n - m) $
- $ a_6 = k(2n - m)(3n - 2m) $
- $ a_8 = k(3n - 2m)(4n - 3m) $
- $ a_{10} = k(4n - 3m)(5n - 4m) $
- ...

So, for even index $ 2t $, the term is $ k \times c_t \times c_{t + 1} $, where $ c_t $ is as above. For example, $ a_2 = k \times c_1 \times c_2 $, since $ c_1 = m $, $ c_2 = n $, so $ k \times m \times n $. Similarly, $ a_4 = k \times c_2 \times c_3 = k \times n \times (2n - m) $, etc.

Therefore, in general, for even index $ 2t $, $ a_{2t} = k \times c_t \times c_{t + 1} $, where $ c_t = (t - 1)n - (t - 2)m $.

Therefore, if I can find expressions for $ a_{13} $, which is an odd index, since 13 is odd. Let me check:

Since 13 is odd, it's of the form $ 2t - 1 $, so $ t = 7 $. Therefore, $ a_{13} = k \times c_7^2 $.

Let me compute $ c_7 $:

From the formula $ c_t = (t - 1)n - (t - 2)m $, so for t=7:

$ c_7 = 6n - 5m $.

Therefore, $ a_{13} = k \times (6n - 5m)^2 $.

But we are given that $ a_{13} = 2016 $. Therefore:

$ k \times (6n - 5m)^2 = 2016 $.

So, our goal is to find $ a_1 = m^2 k $, given that $ k \times (6n - 5m)^2 = 2016 $, with the constraints that $ \gcd(m, n) = 1 $, $ n > m $, and all terms are positive integers, and the sequence is strictly increasing.

So, we need to find integers $ m, n, k $ such that:

1. $ \gcd(m, n) = 1 $
2. $ n > m $
3. $ k \times (6n - 5m)^2 = 2016 $
4. All terms in the sequence are positive integers and strictly increasing.

Additionally, since $ a_1 = m^2 k $, and $ a_2 = m k n $, and the sequence is strictly increasing, we have $ a_1 < a_2 \implies m^2 k < m k n \implies m < n $, which is already given by condition 2.

Also, since $ a_3 = k n^2 $, and $ a_3 > a_2 \implies k n^2 > m k n \implies n > m $, again condition 2.

Similarly, other inequalities will be satisfied if $ n > m $, as we saw earlier.

So, our main equation is $ k \times (6n - 5m)^2 = 2016 $.

We need to find positive integers $ m, n, k $, with $ \gcd(m, n) = 1 $, $ n > m $, such that this equation holds.

Moreover, since $ 6n - 5m $ must be a positive integer (since $ a_{13} $ is positive, and $ k $ is positive), we have $ 6n - 5m > 0 \implies 6n > 5m \implies n > (5/6)m $.

But since $ n > m $, this is automatically satisfied because $ 5/6 m < m $, so $ n > m > 5/6 m $.

Therefore, the main equation is $ k \times (6n - 5m)^2 = 2016 $.

So, we need to find all possible factorizations of 2016 into $ k \times d^2 $, where $ d = 6n - 5m $, and then find corresponding $ m, n $ such that $ \gcd(m, n) = 1 $, $ n > m $, and $ d = 6n - 5m $.

First, let me factorize 2016.

2016 is a known number. Let me compute its prime factorization.

Divide by 2: 2016 / 2 = 1008

Again by 2: 1008 / 2 = 504

Again by 2: 504 / 2 = 252

Again by 2: 252 / 2 = 126

Again by 2: 126 / 2 = 63

Now 63 is 7 * 9 = 7 * 3^2.

So, total prime factors:

2^5 * 3^2 * 7^1.

Therefore, 2016 = 2^5 * 3^2 * 7.

So, the equation is $ k \times d^2 = 2^5 \times 3^2 \times 7 $, where $ d = 6n - 5m $.

Therefore, $ d^2 $ must divide 2016, so $ d^2 $ divides 2^5 * 3^2 * 7.

But since $ d^2 $ divides 2016, the exponents in the prime factorization of $ d^2 $ must be less than or equal to those in 2016.

Let me denote the prime factors of $ d $ as follows. Let me write $ d $ as $ 2^a \times 3^b \times 7^c $, where $ a, b, c $ are non-negative integers. Then $ d^2 = 2^{2a} \times 3^{2b} \times 7^{2c} $.

Since $ d^2 $ divides 2^5 * 3^2 * 7^1, we have:

For prime 2: 2a ≤ 5 ⇒ a ≤ 2 (since 2a ≤ 5 ⇒ a ≤ 2.5 ⇒ a ≤ 2)

For prime 3: 2b ≤ 2 ⇒ b ≤ 1

For prime 7: 2c ≤ 1 ⇒ c = 0 (since 2c ≤ 1 ⇒ c = 0)

Therefore, possible exponents:

a ∈ {0, 1, 2}

b ∈ {0, 1}

c = 0

Therefore, possible values of d:

Each d is of the form 2^a * 3^b, with a ≤ 2, b ≤ 1, and d > 0.

So, let me list all possible d:

For a=0:

- b=0: d=1
- b=1: d=3

For a=1:

- b=0: d=2
- b=1: d=2*3=6

For a=2:

- b=0: d=4
- b=1: d=4*3=12

So, possible d values: 1, 3, 2, 6, 4, 12.

Therefore, possible d: 1, 2, 3, 4, 6, 12.

Therefore, for each possible d, we can compute k = 2016 / d², and then check if there exist integers m, n with gcd(m, n) = 1, n > m, and d = 6n - 5m.

So, let me go through each possible d:

Case 1: d = 1

Then k = 2016 / 1² = 2016.

So, we have 6n - 5m = 1.

We need to solve for integers m, n with gcd(m, n) = 1, n > m.

Let me solve 6n - 5m = 1.

Rearranged: 6n = 5m + 1 ⇒ n = (5m + 1)/6.

Since n must be integer, 5m + 1 must be divisible by 6. So:

5m + 1 ≡ 0 mod 6 ⇒ 5m ≡ -1 mod 6 ⇒ 5m ≡ 5 mod 6 ⇒ m ≡ 1 mod 6 (since 5^{-1} mod 6 is 5, because 5*5=25≡1 mod6).

Therefore, m = 6t + 1 for some integer t ≥ 0.

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

Therefore, m = 6t + 1, n = 5t + 1.

Now, since n > m:

5t + 1 > 6t + 1 ⇒ 5t + 1 > 6t + 1 ⇒ -t > 0 ⇒ t < 0.

But t is a non-negative integer (since m must be positive, m = 6t + 1 ≥ 1 ⇒ t ≥ 0). Therefore, t < 0 is impossible. Therefore, no solutions in this case.

Therefore, d=1 is invalid.

Case 2: d = 2

Then k = 2016 / 2² = 2016 / 4 = 504.

Equation: 6n - 5m = 2.

Solve for integers m, n with gcd(m, n) = 1, n > m.

Rearranged: 6n = 5m + 2 ⇒ n = (5m + 2)/6.

So, 5m + 2 must be divisible by 6 ⇒ 5m ≡ -2 mod 6 ⇒ 5m ≡ 4 mod 6 ⇒ Multiply both sides by 5^{-1} mod 6, which is 5:

m ≡ 4*5 mod 6 ⇒ m ≡ 20 mod 6 ⇒ m ≡ 2 mod 6.

Therefore, m = 6t + 2 for some integer t ≥ 0.

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

Therefore, m = 6t + 2, n = 5t + 2.

Check n > m:

5t + 2 > 6t + 2 ⇒ -t > 0 ⇒ t < 0. Again, t ≥ 0, so no solution. Therefore, invalid.

Case 3: d = 3

k = 2016 / 3² = 2016 / 9 = 224.

Equation: 6n - 5m = 3.

Solve for integers m, n with gcd(m, n) = 1, n > m.

Rearranged: 6n = 5m + 3 ⇒ n = (5m + 3)/6.

So, 5m + 3 ≡ 0 mod 6 ⇒ 5m ≡ -3 mod 6 ⇒ 5m ≡ 3 mod 6 ⇒ Multiply by 5^{-1} = 5:

m ≡ 15 mod 6 ⇒ m ≡ 3 mod 6.

Therefore, m = 6t + 3 for t ≥ 0.

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

Check n > m:

5t + 3 > 6t + 3 ⇒ -t > 0 ⇒ t < 0. Again, no solution. Invalid.

Case 4: d = 4

k = 2016 / 4² = 2016 / 16 = 126.

Equation: 6n - 5m = 4.

Solve for integers m, n with gcd(m, n) = 1, n > m.

Rearranged: 6n = 5m + 4 ⇒ n = (5m + 4)/6.

So, 5m + 4 ≡ 0 mod 6 ⇒ 5m ≡ -4 mod 6 ⇒ 5m ≡ 2 mod 6 ⇒ Multiply by 5^{-1} = 5:

m ≡ 10 mod 6 ⇒ m ≡ 4 mod 6.

Therefore, m = 6t + 4 for t ≥ 0.

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

Check n > m:

5t + 4 > 6t + 4 ⇒ -t > 0 ⇒ t < 0. Again, no solution. Invalid.

Case 5: d = 6

k = 2016 / 6² = 2016 / 36 = 56.

Equation: 6n - 5m = 6.

Solve for integers m, n with gcd(m, n) = 1, n > m.

Rearranged: 6n = 5m + 6 ⇒ n = (5m + 6)/6.

So, 5m + 6 ≡ 0 mod 6 ⇒ 5m ≡ -6 mod 6 ⇒ 5m ≡ 0 mod 6 ⇒ 5m ≡ 0 mod 6 ⇒ m ≡ 0 mod 6 (since 5 and 6 are coprime, so m ≡ 0 mod 6).

Therefore, m = 6t for t ≥ 0.

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

Check n > m:

5t + 1 > 6t ⇒ 1 > t ⇒ t < 1. Since t is a non-negative integer, t can be 0.

So, t = 0:

m = 0, but m must be positive integer. Wait, m = 6t, t ≥ 0. If t = 0, m = 0, which is invalid because m must be positive. Therefore, t ≥ 1?

Wait, but t must be such that m is positive. If t = 0, m = 0, which is invalid. So, t ≥ 1?

But for t = 1:

m = 6*1 = 6, n = 5*1 + 1 = 6. Then n = 6, m = 6. But n must be greater than m. However, n = 6 and m = 6, so n = m, which violates n > m. Therefore, invalid.

t = 0 gives m = 0 invalid, t ≥ 1 gives n = 5t + 1, m = 6t. For t ≥ 1, n = 5t + 1 < 6t? Let's check:

For t ≥ 1, 5t + 1 < 6t ⇒ 1 < t. So, for t ≥ 2, 5t + 1 < 6t ⇒ n < m. But we need n > m. Therefore, no solution here. Therefore, invalid.

Wait, but maybe I made a mistake here. Let me check again.

Equation: 6n - 5m = 6.

We have m = 6t, n = 5t + 1.

So, n - m = 5t + 1 - 6t = -t + 1. For n > m, we need -t + 1 > 0 ⇒ t < 1. Since t is integer ≥ 0, only t = 0. But t = 0 gives m = 0, which is invalid. Therefore, no solution. Hence, d=6 is invalid.

Case 6: d = 12

k = 2016 / 12² = 2016 / 144 = 14.

Equation: 6n - 5m = 12.

Solve for integers m, n with gcd(m, n) = 1, n > m.

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

So, 5m + 12 ≡ 0 mod 6 ⇒ 5m ≡ -12 mod 6 ⇒ 5m ≡ 0 mod 6 ⇒ 5m ≡ 0 mod 6 ⇒ m ≡ 0 mod 6 (since 5 and 6 are coprime).

Therefore, m = 6t for t ≥ 0.

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

Check n > m:

5t + 2 > 6t ⇒ 2 > t ⇒ t < 2. Since t is non-negative integer, t can be 0 or 1.

Check t = 0:

m = 0, invalid.

t = 1:

m = 6*1 = 6, n = 5*1 + 2 = 7. Check gcd(m, n) = gcd(6, 7) = 1. Good. Also, n = 7 > m = 6. So this is valid.

So, this is a valid solution: m = 6, n = 7, d = 12, k = 14.

Let me check if there are other t's. For t ≥ 2, n = 5t + 2, m = 6t. Then n - m = -t + 2. For t ≥ 2, n - m ≤ 0. So, only t=1 is valid.

Therefore, this case gives us a valid solution.

So, we have m = 6, n = 7, k = 14.

Therefore, let's compute $ a_1 = m^2 k = 6^2 * 14 = 36 * 14 $.

Compute 36 * 14: 36*10=360, 36*4=144, total 360+144=504. So, $ a_1 = 504 $.

But before finalizing, let me check if this is consistent with the rest of the sequence and that all terms are integers and strictly increasing.

Let me verify the terms step by step.

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

Compute:

a1 = m²k = 6² * 14 = 36 * 14 = 504.

a2 = m k n = 6 * 14 * 7 = 6 * 98 = 588.

Check a1 < a2: 504 < 588, yes.

a3 = k n² = 14 * 7² = 14 * 49 = 686.

Check a2 < a3: 588 < 686, yes.

a4 = k n (2n - m) = 14 * 7 * (14 - 6) = 14 * 7 * 8 = 14 * 56 = 784.

Check a3 < a4: 686 < 784, yes.

a5 = k (2n - m)^2 = 14 * (14 - 6)^2 = 14 * 8² = 14 * 64 = 896.

Check a4 < a5: 784 < 896, yes.

a6 = k (2n - m)(3n - 2m) = 14 * 8 * (21 - 12) = 14 * 8 * 9 = 14 * 72 = 1008.

Check a5 < a6: 896 < 1008, yes.

a7 = k (3n - 2m)^2 = 14 * (21 - 12)^2 = 14 * 9² = 14 * 81 = 1134.

Check a6 < a7: 1008 < 1134, yes.

a8 = k (3n - 2m)(4n - 3m) = 14 * 9 * (28 - 18) = 14 * 9 * 10 = 14 * 90 = 1260.

Check a7 < a8: 1134 < 1260, yes.

a9 = k (4n - 3m)^2 = 14 * (28 - 18)^2 = 14 * 10² = 14 * 100 = 1400.

Check a8 < a9: 1260 < 1400, yes.

a10 = k (4n - 3m)(5n - 4m) = 14 * 10 * (35 - 24) = 14 * 10 * 11 = 14 * 110 = 1540.

Check a9 < a10: 1400 < 1540, yes.

a11 = k (5n - 4m)^2 = 14 * (35 - 24)^2 = 14 * 11² = 14 * 121 = 1694.

Check a10 < a11: 1540 < 1694, yes.

a12 = k (5n - 4m)(6n - 5m) = 14 * 11 * (42 - 30) = 14 * 11 * 12 = 14 * 132 = 1848.

Check a11 < a12: 1694 < 1848, yes.

a13 = k (6n - 5m)^2 = 14 * (42 - 30)^2 = 14 * 12² = 14 * 144 = 2016. Which matches the given value. Perfect!

So, this works. Therefore, the value of $ a_1 $ is 504.

But let me check if there are other possible solutions. For example, in case d=12, we had only t=1. But maybe there are other d's?

Wait, we checked all possible d's (1,2,3,4,6,12) and only d=12 gave a valid solution. Therefore, this is the only solution.

But just to be thorough, let me check if there are other possible factorizations. Wait, earlier when I considered the possible d's, I considered only d's where d^2 divides 2016. However, is there a possibility that d^2 divides 2016 but d is not of the form 2^a 3^b? For example, could there be other prime factors?

But since 2016's prime factors are only 2, 3, 7, and in our equation, d^2 divides 2016, so d can only have primes 2, 3, 7. However, earlier, we saw that for prime 7, 2c ≤ 1 ⇒ c=0. Therefore, d cannot have 7 as a factor. Therefore, d must be of the form 2^a 3^b, with a ≤ 2, b ≤ 1. Therefore, our list of d's is complete. Therefore, only d=12 gives a valid solution.

Therefore, the answer is $ a_1 = 504 $.

But let me check once again if there are other possible values of d. For example, if d is negative? But since d = 6n - 5m, and n > m, and m, n positive integers, then 6n - 5m is positive? Let me check.

Since n > m, but 6n - 5m could be positive or negative? For example, if n = m + 1, then 6(m + 1) - 5m = 6m + 6 - 5m = m + 6, which is positive. So, since n > m, and m ≥ 1, then 6n - 5m ≥ 6(m + 1) - 5m = m + 6 ≥ 1 + 6 = 7. Wait, but in our case, d was 12. Wait, but in our previous cases, for example, when d=1, we had 6n - 5m = 1, which is possible if n and m are such that 6n - 5m = 1. However, in that case, we saw that there were no solutions. But in general, d must be positive. Therefore, our previous analysis holds.

Therefore, the only valid solution is when d=12, leading to a1=504.

Therefore, the answer is \boxed{504}.

**Final Answer**
\boxed{504}

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

- For every positive integer $ k $, the subsequence $ a_{2k-1}, a_{2k}, a_{2k+1} $ is **geometric**.
- For every positive integer $ k $, the subsequence $ a_{2k}, a_{2k+1}, a_{2k+2} $ is **arithmetic**.
- We are told that $ a_{13} = 2016 $, and we are to find $ a_1 $.

---

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

Let’s denote the first few terms in terms of $ a_1 = x $ and $ a_2 = y $, and use the given properties to derive a general pattern.

#### For $ k = 1 $:
- The subsequence $ a_1, a_2, a_3 $ is geometric:
  $$
  a_2^2 = a_1 a_3 \Rightarrow a_3 = \frac{y^2}{x}
  $$
- The subsequence $ a_2, a_3, a_4 $ is arithmetic:
  $$
  2a_3 = a_2 + a_4 \Rightarrow a_4 = 2a_3 - a_2 = \frac{2y^2 - xy}{x}
  $$

#### For $ k = 2 $:
- The subsequence $ a_3, a_4, a_5 $ is geometric:
  $$
  a_4^2 = a_3 a_5 \Rightarrow a_5 = \frac{a_4^2}{a_3}
  $$
- The subsequence $ a_4, a_5, a_6 $ is arithmetic:
  $$
  2a_5 = a_4 + a_6 \Rightarrow a_6 = 2a_5 - a_4
  $$

This pattern continues, and we can observe that the terms follow a recursive structure based on the previous terms.

---

### Step 2: General Pattern and Parametrization

By analyzing the recursive relations, we can derive a general form for the terms of the sequence. Let’s define:

- $ a_1 = m^2 k $
- $ a_2 = m k n $
- $ a_3 = k n^2 $
- $ a_4 = k n (2n - m) $
- $ a_5 = k (2n - m)^2 $
- $ a_6 = k (2n - m)(3n - 2m) $
- $ a_7 = k (3n - 2m)^2 $
- $ a_8 = k (3n - 2m)(4n - 3m) $
- $ a_9 = k (4n - 3m)^2 $
- $ a_{10} = k (4n - 3m)(5n - 4m) $
- $ a_{11} = k (5n - 4m)^2 $
- $ a_{12} = k (5n - 4m)(6n - 5m) $
- $ a_{13} = k (6n - 5m)^2 $

So, the general form for the odd-indexed terms is:
$$
a_{2t - 1} = k \cdot (t n - (t - 1)m)^2
$$

Given that $ a_{13} = 2016 $, and $ a_{13} = k (6n - 5m)^2 $, we have:
$$
k (6n - 5m)^2 = 2016
$$

---

### Step 3: Solve the Equation

We now solve:
$$
k (6n - 5m)^2 = 2016
$$

Let’s denote $ d = 6n - 5m $, so:
$$
k d^2 = 2016
$$

We need to find all integer factorizations of 2016 into $ k \cdot d^2 $, where $ d $ is a positive integer, and $ k $ is also a positive integer.

The prime factorization of 2016 is:
$$
2016 = 2^5 \cdot 3^2 \cdot 7
$$

We now find all possible values of $ d $ such that $ d^2 $ divides 2016. That is, $ d^2 \mid 2016 $, so $ d $ must be a divisor of $ \sqrt{2016} $, and $ d^2 $ must divide 2016.

Let’s list all such $ d $:

- $ d = 1 \Rightarrow d^2 = 1 \Rightarrow k = 2016 $
- $ d = 2 \Rightarrow d^2 = 4 \Rightarrow k = 504 $
- $ d = 3 \Rightarrow d^2 = 9 \Rightarrow k = 224 $
- $ d = 4 \Rightarrow d^2 = 16 \Rightarrow k = 126 $
- $ d = 6 \Rightarrow d^2 = 36 \Rightarrow k = 56 $
- $ d = 12 \Rightarrow d^2 = 144 \Rightarrow k = 14 $

Now, for each $ d $, we solve the equation:
$$
6n - 5m = d
$$
and check if there exist positive integers $ m, n $ such that $ \gcd(m, n) = 1 $ and $ n > m $.

After testing all possible values, the only valid solution is:

- $ d = 12 $
- $ k = 14 $
- Solving $ 6n - 5m = 12 $, we find $ m = 6 $, $ n = 7 $, and $ \gcd(6, 7) = 1 $

---

### Step 4: Compute $ a_1 $

Recall:
$$
a_1 = m^2 k = 6^2 \cdot 14 = 36 \cdot 14 = 504
$$

---

### Final Answer

$$
\boxed{504}
$$

[Assistant]
Explanation:
Okay, so I need to find the value of $ a_1 $ in a strictly increasing sequence of positive integers. The sequence has some specific properties related to geometric and arithmetic subsequences. Let me try to parse the problem again.

The problem says that for every positive integer $ k $, the subsequence $ a_{2k-1} $, $ a_{2k} $, $ a_{2k+1} $ is geometric. Also, the subsequence $ a_{2k} $, $ a_{2k+1} $, $ a_{2k+2} $ is arithmetic. And we know that $ a_{13} = 2016 $. We need to find $ a_1 $.

First, let me recall what geometric and arithmetic sequences are. A geometric sequence has a common ratio between consecutive terms, so if three terms $ x $, $ y $, $ z $ are geometric, then $ y^2 = xz $. An arithmetic sequence has a common difference, so if three terms $ x $, $ y $, $ z $ are arithmetic, then $ 2y = x + z $.

Given that the sequence is strictly increasing, all terms are positive integers, and each term is larger than the previous one. So, for example, $ a_1 < a_2 < a_3 < \dots $.

Let me try to write down the conditions for the first few terms. Maybe I can find a pattern or recurrence relation.

Starting with $ k = 1 $:

For the geometric subsequence $ a_1 $, $ a_2 $, $ a_3 $:

So, $ a_2^2 = a_1 a_3 $.

For the arithmetic subsequence $ a_2 $, $ a_3 $, $ a_4 $:

So, $ 2a_3 = a_2 + a_4 $, which implies $ a_4 = 2a_3 - a_2 $.

Similarly, for $ k = 2 $:

Geometric subsequence $ a_3 $, $ a_4 $, $ a_5 $:

So, $ a_4^2 = a_3 a_5 $, which gives $ a_5 = \frac{a_4^2}{a_3} $.

Arithmetic subsequence $ a_4 $, $ a_5 $, $ a_6 $:

So, $ 2a_5 = a_4 + a_6 $, hence $ a_6 = 2a_5 - a_4 $.

Similarly, for $ k = 3 $:

Geometric subsequence $ a_5 $, $ a_6 $, $ a_7 $:

$ a_6^2 = a_5 a_7 $, so $ a_7 = \frac{a_6^2}{a_5} $.

Arithmetic subsequence $ a_6 $, $ a_7 $, $ a_8 $:

$ 2a_7 = a_6 + a_8 $, so $ a_8 = 2a_7 - a_6 $.

And so on. It seems like there's a pattern here where odd-indexed terms are part of geometric sequences with the next even term and the following odd term, and even-indexed terms are part of arithmetic sequences with the next odd term and the following even term.

But since the sequence is strictly increasing, all these relations must hold with the terms increasing. Also, since all terms are positive integers, the divisions in the geometric sequences must result in integers. So, for example, $ a_3 $ must divide $ a_4^2 $, and $ a_5 $ must divide $ a_6^2 $, etc.

This seems recursive. Maybe I can express terms in terms of previous terms and find a recurrence relation?

Alternatively, maybe there's a way to express the terms in terms of $ a_1 $ and some ratio or difference?

Let me try to write down the first few terms in terms of $ a_1 $ and $ a_2 $, since those are the starting points.

Let me denote $ a_1 = x $, $ a_2 = y $. Since the sequence is strictly increasing, $ x < y $, and both are positive integers.

From the first geometric sequence (k=1): $ a_1, a_2, a_3 $ is geometric. Therefore:

$ a_2^2 = a_1 a_3 \implies y^2 = x a_3 \implies a_3 = \frac{y^2}{x} $.

Since $ a_3 $ must be an integer, $ x $ divides $ y^2 $.

Also, since the sequence is strictly increasing, $ a_3 > a_2 $, so $ \frac{y^2}{x} > y \implies \frac{y}{x} > 1 \implies y > x $, which we already know.

Then, from the arithmetic sequence $ a_2, a_3, a_4 $:

$ 2a_3 = a_2 + a_4 \implies a_4 = 2a_3 - a_2 = 2 \cdot \frac{y^2}{x} - y $.

Since $ a_4 $ must be an integer, $ \frac{2y^2}{x} - y $ must be integer. Let me write that as:

$ a_4 = \frac{2y^2 - x y}{x} = \frac{y(2y - x)}{x} $.

Therefore, $ x $ divides $ y(2y - x) $.

But since $ x $ divides $ y^2 $ from the previous condition, maybe we can find some relations here.

Let me note that $ x $ divides $ y^2 $, so let me denote $ y = x \cdot m $, where $ m $ is a positive integer greater than 1 (since $ y > x $). Wait, but $ m $ might not necessarily be integer? Wait, no, since $ y $ is an integer and $ x $ divides $ y^2 $, but $ y $ could have factors in common with $ x $. However, if I set $ y = x \cdot m $, then $ x $ divides $ y^2 $, which is $ x^2 m^2 $, so that's true. But maybe there's a better way to parametrize.

Alternatively, since $ x $ divides $ y^2 $, let me denote $ y = x \cdot k $, where $ k $ is a rational number. But since $ y $ is an integer, $ k $ must be rational. However, maybe it's better to think in terms of divisors.

Alternatively, perhaps I can express $ a_3 $ as $ \frac{y^2}{x} $, and since $ a_3 $ must be an integer, $ x $ divides $ y^2 $. Let me suppose that $ x $ divides $ y^2 $, so let me write $ y = x \cdot t $, where $ t $ is a rational number. Wait, but since $ y $ is an integer, $ t $ must be rational. However, maybe I can think of $ t $ as a fraction in reduced form. Let me think.

Alternatively, maybe I can think of $ x $ and $ y $ as having some common factors. Let me suppose that $ \gcd(x, y) = d $, so $ x = d \cdot m $, $ y = d \cdot n $, where $ \gcd(m, n) = 1 $. Then, since $ x $ divides $ y^2 $, $ d m $ divides $ d^2 n^2 $, so $ m $ divides $ d n^2 $. Since $ \gcd(m, n) = 1 $, $ m $ divides $ d $. Therefore, $ d = m \cdot k $ for some integer $ k $. Therefore, $ x = d m = m^2 k $, $ y = d n = m k n $.

So, substituting back, $ x = m^2 k $, $ y = m k n $, with $ \gcd(m, n) = 1 $.

This might be helpful. Let me note that.

So, with this parametrization, $ a_3 = \frac{y^2}{x} = \frac{(m k n)^2}{m^2 k} = \frac{m^2 k^2 n^2}{m^2 k} = k n^2 $.

So, $ a_3 = k n^2 $.

Then, moving on to $ a_4 $:

$ a_4 = \frac{y(2y - x)}{x} = \frac{m k n (2 m k n - m^2 k)}{m^2 k} $.

Let me compute numerator:

$ m k n (2 m k n - m^2 k) = m k n \cdot m k (2 n - m) = m^2 k^2 n (2n - m) $.

Divide by $ m^2 k $:

$ \frac{m^2 k^2 n (2n - m)}{m^2 k} = k n (2n - m) $.

Therefore, $ a_4 = k n (2n - m) $.

Since $ a_4 $ must be an integer, which it is, given that $ k, n, m $ are integers.

Also, since the sequence is strictly increasing, $ a_4 > a_3 $. Let's check:

$ a_4 = k n (2n - m) $, $ a_3 = k n^2 $.

So, $ k n (2n - m) > k n^2 \implies 2n - m > n \implies n > m $.

Therefore, $ n > m $.

So, we have $ \gcd(m, n) = 1 $, $ n > m $, and $ a_1 = m^2 k $, $ a_2 = m k n $, $ a_3 = k n^2 $, $ a_4 = k n (2n - m) $.

Now, moving on to the next geometric sequence for $ k = 2 $: $ a_3, a_4, a_5 $ is geometric. Therefore:

$ a_4^2 = a_3 a_5 \implies a_5 = \frac{a_4^2}{a_3} $.

Let me compute that:

$ a_5 = \frac{[k n (2n - m)]^2}{k n^2} = \frac{k^2 n^2 (2n - m)^2}{k n^2} = k (2n - m)^2 $.

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

Then, the arithmetic sequence $ a_4, a_5, a_6 $:

$ 2a_5 = a_4 + a_6 \implies a_6 = 2a_5 - a_4 $.

Compute $ a_6 $:

$ 2a_5 - a_4 = 2k(2n - m)^2 - k n (2n - m) = k(2n - m)[2(2n - m) - n] = k(2n - m)(4n - 2m - n) = k(2n - m)(3n - 2m) $.

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

Now, since the sequence is strictly increasing, $ a_6 > a_5 $:

$ k(2n - m)(3n - 2m) > k(2n - m)^2 \implies (3n - 2m) > (2n - m) \implies 3n - 2m > 2n - m \implies n > m $, which we already have. So that's consistent.

Now, moving on to the next geometric sequence for $ k = 3 $: $ a_5, a_6, a_7 $ is geometric. Therefore:

$ a_6^2 = a_5 a_7 \implies a_7 = \frac{a_6^2}{a_5} $.

Let me compute that:

$ a_7 = \frac{[k(2n - m)(3n - 2m)]^2}{k(2n - m)^2} = \frac{k^2 (2n - m)^2 (3n - 2m)^2}{k (2n - m)^2} = k (3n - 2m)^2 $.

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

Then, the arithmetic sequence $ a_6, a_7, a_8 $:

$ 2a_7 = a_6 + a_8 \implies a_8 = 2a_7 - a_6 $.

Compute $ a_8 $:

$ 2k(3n - 2m)^2 - k(2n - m)(3n - 2m) = k(3n - 2m)[2(3n - 2m) - (2n - m)] $.

Let me compute the bracket:

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

Therefore, $ a_8 = k(3n - 2m)(4n - 3m) $.

Continuing this pattern, perhaps we can see a recursive structure here. Let me check the next term.

For $ k = 4 $, the geometric sequence $ a_7, a_8, a_9 $:

$ a_8^2 = a_7 a_9 \implies a_9 = \frac{a_8^2}{a_7} $.

Compute $ a_9 $:

$ \frac{[k(3n - 2m)(4n - 3m)]^2}{k(3n - 2m)^2} = \frac{k^2 (3n - 2m)^2 (4n - 3m)^2}{k (3n - 2m)^2} = k(4n - 3m)^2 $.

So, $ a_9 = k(4n - 3m)^2 $.

Then, the arithmetic sequence $ a_8, a_9, a_{10} $:

$ 2a_9 = a_8 + a_{10} \implies a_{10} = 2a_9 - a_8 $.

Compute $ a_{10} $:

$ 2k(4n - 3m)^2 - k(3n - 2m)(4n - 3m) = k(4n - 3m)[2(4n - 3m) - (3n - 2m)] $.

Compute the bracket:

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

Therefore, $ a_{10} = k(4n - 3m)(5n - 4m) $.

Hmm, I see a pattern here. Let me try to see if I can generalize this.

Looking at the terms:

- $ a_1 = m^2 k $
- $ a_2 = m k n $
- $ a_3 = k n^2 $
- $ a_4 = k n (2n - m) $
- $ a_5 = k (2n - m)^2 $
- $ a_6 = k(2n - m)(3n - 2m) $
- $ a_7 = k(3n - 2m)^2 $
- $ a_8 = k(3n - 2m)(4n - 3m) $
- $ a_9 = k(4n - 3m)^2 $
- $ a_{10} = k(4n - 3m)(5n - 4m) $
- ...

So, it seems that for odd indices $ 2k - 1 $, the term is $ k \times (\text{some linear expression in } n \text{ and } m)^2 $, and for even indices $ 2k $, it's $ k \times (\text{some linear expression}) \times (\text{next linear expression}) $.

Looking at the coefficients:

For $ a_1 $, which is index 1 (odd), it's $ m^2 k $.

For $ a_3 $, index 3, it's $ k n^2 $.

For $ a_5 $, index 5, it's $ k (2n - m)^2 $.

For $ a_7 $, index 7, it's $ k (3n - 2m)^2 $.

For $ a_9 $, index 9, it's $ k (4n - 3m)^2 $.

So, the coefficients for the odd terms seem to follow a pattern where the coefficient is $ k \times ( (t)n - (t - 1)m )^2 $, where t is something. Let me check:

For $ a_1 $, t = 1? Wait, $ a_1 = m^2 k $. If I think of $ (1 \cdot n - 0 \cdot m)^2 = n^2 $, but $ a_1 $ is $ m^2 k $. Hmm, maybe not. Alternatively, maybe there's a different pattern.

Alternatively, looking at the coefficients for the odd terms:

- $ a_1 = m^2 k $
- $ a_3 = k n^2 $
- $ a_5 = k (2n - m)^2 $
- $ a_7 = k (3n - 2m)^2 $
- $ a_9 = k (4n - 3m)^2 $

So, if we think of the coefficients as starting with $ m $, then $ n $, then $ 2n - m $, then $ 3n - 2m $, then $ 4n - 3m $, etc. So, each time, the coefficient for the odd term is $ (t n - (t - 1)m) $, where t starts at 1? Wait:

For $ a_1 $, if t = 1, then $ 1 \cdot n - 0 \cdot m = n $, but $ a_1 $ is $ m^2 k $. Hmm, not quite. Wait, but maybe if we think of the coefficients for the odd terms as starting with $ m $, then $ n $, then $ 2n - m $, etc. So, perhaps there's a recursive relation here.

Alternatively, maybe we can think of the coefficients as following a linear recurrence. Let me denote $ c_t $ as the coefficient for the odd term at position $ 2t - 1 $. Then:

- $ c_1 = m $
- $ c_2 = n $
- $ c_3 = 2n - m $
- $ c_4 = 3n - 2m $
- $ c_5 = 4n - 3m $
- ...

So, each $ c_{t} = (t - 1)n - (t - 2)m $. Wait, for t=2: $ c_2 = 1 \cdot n - 0 \cdot m = n $. For t=3: $ 2n - m $. For t=4: 3n - 2m. Yes, that seems to hold. So, in general, $ c_t = (t - 1)n - (t - 2)m $. Therefore, for the odd term $ a_{2t - 1} $, it's $ k \cdot c_t^2 $.

Similarly, for the even terms:

- $ a_2 = m k n $
- $ a_4 = k n (2n - m) $
- $ a_6 = k(2n - m)(3n - 2m) $
- $ a_8 = k(3n - 2m)(4n - 3m) $
- $ a_{10} = k(4n - 3m)(5n - 4m) $
- ...

So, for even index $ 2t $, the term is $ k \times c_t \times c_{t + 1} $, where $ c_t $ is as above. For example, $ a_2 = k \times c_1 \times c_2 $, since $ c_1 = m $, $ c_2 = n $, so $ k \times m \times n $. Similarly, $ a_4 = k \times c_2 \times c_3 = k \times n \times (2n - m) $, etc.

Therefore, in general, for even index $ 2t $, $ a_{2t} = k \times c_t \times c_{t + 1} $, where $ c_t = (t - 1)n - (t - 2)m $.

Therefore, if I can find expressions for $ a_{13} $, which is an odd index, since 13 is odd. Let me check:

Since 13 is odd, it's of the form $ 2t - 1 $, so $ t = 7 $. Therefore, $ a_{13} = k \times c_7^2 $.

Let me compute $ c_7 $:

From the formula $ c_t = (t - 1)n - (t - 2)m $, so for t=7:

$ c_7 = 6n - 5m $.

Therefore, $ a_{13} = k \times (6n - 5m)^2 $.

But we are given that $ a_{13} = 2016 $. Therefore:

$ k \times (6n - 5m)^2 = 2016 $.

So, our goal is to find $ a_1 = m^2 k $, given that $ k \times (6n - 5m)^2 = 2016 $, with the constraints that $ \gcd(m, n) = 1 $, $ n > m $, and all terms are positive integers, and the sequence is strictly increasing.

So, we need to find integers $ m, n, k $ such that:

1. $ \gcd(m, n) = 1 $
2. $ n > m $
3. $ k \times (6n - 5m)^2 = 2016 $
4. All terms in the sequence are positive integers and strictly increasing.

Additionally, since $ a_1 = m^2 k $, and $ a_2 = m k n $, and the sequence is strictly increasing, we have $ a_1 < a_2 \implies m^2 k < m k n \implies m < n $, which is already given by condition 2.

Also, since $ a_3 = k n^2 $, and $ a_3 > a_2 \implies k n^2 > m k n \implies n > m $, again condition 2.

Similarly, other inequalities will be satisfied if $ n > m $, as we saw earlier.

So, our main equation is $ k \times (6n - 5m)^2 = 2016 $.

We need to find positive integers $ m, n, k $, with $ \gcd(m, n) = 1 $, $ n > m $, such that this equation holds.

Moreover, since $ 6n - 5m $ must be a positive integer (since $ a_{13} $ is positive, and $ k $ is positive), we have $ 6n - 5m > 0 \implies 6n > 5m \implies n > (5/6)m $.

But since $ n > m $, this is automatically satisfied because $ 5/6 m < m $, so $ n > m > 5/6 m $.

Therefore, the main equation is $ k \times (6n - 5m)^2 = 2016 $.

So, we need to find all possible factorizations of 2016 into $ k \times d^2 $, where $ d = 6n - 5m $, and then find corresponding $ m, n $ such that $ \gcd(m, n) = 1 $, $ n > m $, and $ d = 6n - 5m $.

First, let me factorize 2016.

2016 is a known number. Let me compute its prime factorization.

Divide by 2: 2016 / 2 = 1008

Again by 2: 1008 / 2 = 504

Again by 2: 504 / 2 = 252

Again by 2: 252 / 2 = 126

Again by 2: 126 / 2 = 63

Now 63 is 7 * 9 = 7 * 3^2.

So, total prime factors:

2^5 * 3^2 * 7^1.

Therefore, 2016 = 2^5 * 3^2 * 7.

So, the equation is $ k \times d^2 = 2^5 \times 3^2 \times 7 $, where $ d = 6n - 5m $.

Therefore, $ d^2 $ must divide 2016, so $ d^2 $ divides 2^5 * 3^2 * 7.

But since $ d^2 $ divides 2016, the exponents in the prime factorization of $ d^2 $ must be less than or equal to those in 2016.

Let me denote the prime factors of $ d $ as follows. Let me write $ d $ as $ 2^a \times 3^b \times 7^c $, where $ a, b, c $ are non-negative integers. Then $ d^2 = 2^{2a} \times 3^{2b} \times 7^{2c} $.

Since $ d^2 $ divides 2^5 * 3^2 * 7^1, we have:

For prime 2: 2a ≤ 5 ⇒ a ≤ 2 (since 2a ≤ 5 ⇒ a ≤ 2.5 ⇒ a ≤ 2)

For prime 3: 2b ≤ 2 ⇒ b ≤ 1

For prime 7: 2c ≤ 1 ⇒ c = 0 (since 2c ≤ 1 ⇒ c = 0)

Therefore, possible exponents:

a ∈ {0, 1, 2}

b ∈ {0, 1}

c = 0

Therefore, possible values of d:

Each d is of the form 2^a * 3^b, with a ≤ 2, b ≤ 1, and d > 0.

So, let me list all possible d:

For a=0:

- b=0: d=1
- b=1: d=3

For a=1:

- b=0: d=2
- b=1: d=2*3=6

For a=2:

- b=0: d=4
- b=1: d=4*3=12

So, possible d values: 1, 3, 2, 6, 4, 12.

Therefore, possible d: 1, 2, 3, 4, 6, 12.

Therefore, for each possible d, we can compute k = 2016 / d², and then check if there exist integers m, n with gcd(m, n) = 1, n > m, and d = 6n - 5m.

So, let me go through each possible d:

Case 1: d = 1

Then k = 2016 / 1² = 2016.

So, we have 6n - 5m = 1.

We need to solve for integers m, n with gcd(m, n) = 1, n > m.

Let me solve 6n - 5m = 1.

Rearranged: 6n = 5m + 1 ⇒ n = (5m + 1)/6.

Since n must be integer, 5m + 1 must be divisible by 6. So:

5m + 1 ≡ 0 mod 6 ⇒ 5m ≡ -1 mod 6 ⇒ 5m ≡ 5 mod 6 ⇒ m ≡ 1 mod 6 (since 5^{-1} mod 6 is 5, because 5*5=25≡1 mod6).

Therefore, m = 6t + 1 for some integer t ≥ 0.

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

Therefore, m = 6t + 1, n = 5t + 1.

Now, since n > m:

5t + 1 > 6t + 1 ⇒ 5t + 1 > 6t + 1 ⇒ -t > 0 ⇒ t < 0.

But t is a non-negative integer (since m must be positive, m = 6t + 1 ≥ 1 ⇒ t ≥ 0). Therefore, t < 0 is impossible. Therefore, no solutions in this case.

Therefore, d=1 is invalid.

Case 2: d = 2

Then k = 2016 / 2² = 2016 / 4 = 504.

Equation: 6n - 5m = 2.

Solve for integers m, n with gcd(m, n) = 1, n > m.

Rearranged: 6n = 5m + 2 ⇒ n = (5m + 2)/6.

So, 5m + 2 must be divisible by 6 ⇒ 5m ≡ -2 mod 6 ⇒ 5m ≡ 4 mod 6 ⇒ Multiply both sides by 5^{-1} mod 6, which is 5:

m ≡ 4*5 mod 6 ⇒ m ≡ 20 mod 6 ⇒ m ≡ 2 mod 6.

Therefore, m = 6t + 2 for some integer t ≥ 0.

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

Therefore, m = 6t + 2, n = 5t + 2.

Check n > m:

5t + 2 > 6t + 2 ⇒ -t > 0 ⇒ t < 0. Again, t ≥ 0, so no solution. Therefore, invalid.

Case 3: d = 3

k = 2016 / 3² = 2016 / 9 = 224.

Equation: 6n - 5m = 3.

Solve for integers m, n with gcd(m, n) = 1, n > m.

Rearranged: 6n = 5m + 3 ⇒ n = (5m + 3)/6.

So, 5m + 3 ≡ 0 mod 6 ⇒ 5m ≡ -3 mod 6 ⇒ 5m ≡ 3 mod 6 ⇒ Multiply by 5^{-1} = 5:

m ≡ 15 mod 6 ⇒ m ≡ 3 mod 6.

Therefore, m = 6t + 3 for t ≥ 0.

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

Check n > m:

5t + 3 > 6t + 3 ⇒ -t > 0 ⇒ t < 0. Again, no solution. Invalid.

Case 4: d = 4

k = 2016 / 4² = 2016 / 16 = 126.

Equation: 6n - 5m = 4.

Solve for integers m, n with gcd(m, n) = 1, n > m.

Rearranged: 6n = 5m + 4 ⇒ n = (5m + 4)/6.

So, 5m + 4 ≡ 0 mod 6 ⇒ 5m ≡ -4 mod 6 ⇒ 5m ≡ 2 mod 6 ⇒ Multiply by 5^{-1} = 5:

m ≡ 10 mod 6 ⇒ m ≡ 4 mod 6.

Therefore, m = 6t + 4 for t ≥ 0.

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

Check n > m:

5t + 4 > 6t + 4 ⇒ -t > 0 ⇒ t < 0. Again, no solution. Invalid.

Case 5: d = 6

k = 2016 / 6² = 2016 / 36 = 56.

Equation: 6n - 5m = 6.

Solve for integers m, n with gcd(m, n) = 1, n > m.

Rearranged: 6n = 5m + 6 ⇒ n = (5m + 6)/6.

So, 5m + 6 ≡ 0 mod 6 ⇒ 5m ≡ -6 mod 6 ⇒ 5m ≡ 0 mod 6 ⇒ 5m ≡ 0 mod 6 ⇒ m ≡ 0 mod 6 (since 5 and 6 are coprime, so m ≡ 0 mod 6).

Therefore, m = 6t for t ≥ 0.

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

Check n > m:

5t + 1 > 6t ⇒ 1 > t ⇒ t < 1. Since t is a non-negative integer, t can be 0.

So, t = 0:

m = 0, but m must be positive integer. Wait, m = 6t, t ≥ 0. If t = 0, m = 0, which is invalid because m must be positive. Therefore, t ≥ 1?

Wait, but t must be such that m is positive. If t = 0, m = 0, which is invalid. So, t ≥ 1?

But for t = 1:

m = 6*1 = 6, n = 5*1 + 1 = 6. Then n = 6, m = 6. But n must be greater than m. However, n = 6 and m = 6, so n = m, which violates n > m. Therefore, invalid.

t = 0 gives m = 0 invalid, t ≥ 1 gives n = 5t + 1, m = 6t. For t ≥ 1, n = 5t + 1 < 6t? Let's check:

For t ≥ 1, 5t + 1 < 6t ⇒ 1 < t. So, for t ≥ 2, 5t + 1 < 6t ⇒ n < m. But we need n > m. Therefore, no solution here. Therefore, invalid.

Wait, but maybe I made a mistake here. Let me check again.

Equation: 6n - 5m = 6.

We have m = 6t, n = 5t + 1.

So, n - m = 5t + 1 - 6t = -t + 1. For n > m, we need -t + 1 > 0 ⇒ t < 1. Since t is integer ≥ 0, only t = 0. But t = 0 gives m = 0, which is invalid. Therefore, no solution. Hence, d=6 is invalid.

Case 6: d = 12

k = 2016 / 12² = 2016 / 144 = 14.

Equation: 6n - 5m = 12.

Solve for integers m, n with gcd(m, n) = 1, n > m.

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

So, 5m + 12 ≡ 0 mod 6 ⇒ 5m ≡ -12 mod 6 ⇒ 5m ≡ 0 mod 6 ⇒ 5m ≡ 0 mod 6 ⇒ m ≡ 0 mod 6 (since 5 and 6 are coprime).

Therefore, m = 6t for t ≥ 0.

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

Check n > m:

5t + 2 > 6t ⇒ 2 > t ⇒ t < 2. Since t is non-negative integer, t can be 0 or 1.

Check t = 0:

m = 0, invalid.

t = 1:

m = 6*1 = 6, n = 5*1 + 2 = 7. Check gcd(m, n) = gcd(6, 7) = 1. Good. Also, n = 7 > m = 6. So this is valid.

So, this is a valid solution: m = 6, n = 7, d = 12, k = 14.

Let me check if there are other t's. For t ≥ 2, n = 5t + 2, m = 6t. Then n - m = -t + 2. For t ≥ 2, n - m ≤ 0. So, only t=1 is valid.

Therefore, this case gives us a valid solution.

So, we have m = 6, n = 7, k = 14.

Therefore, let's compute $ a_1 = m^2 k = 6^2 * 14 = 36 * 14 $.

Compute 36 * 14: 36*10=360, 36*4=144, total 360+144=504. So, $ a_1 = 504 $.

But before finalizing, let me check if this is consistent with the rest of the sequence and that all terms are integers and strictly increasing.

Let me verify the terms step by step.

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

Compute:

a1 = m²k = 6² * 14 = 36 * 14 = 504.

a2 = m k n = 6 * 14 * 7 = 6 * 98 = 588.

Check a1 < a2: 504 < 588, yes.

a3 = k n² = 14 * 7² = 14 * 49 = 686.

Check a2 < a3: 588 < 686, yes.

a4 = k n (2n - m) = 14 * 7 * (14 - 6) = 14 * 7 * 8 = 14 * 56 = 784.

Check a3 < a4: 686 < 784, yes.

a5 = k (2n - m)^2 = 14 * (14 - 6)^2 = 14 * 8² = 14 * 64 = 896.

Check a4 < a5: 784 < 896, yes.

a6 = k (2n - m)(3n - 2m) = 14 * 8 * (21 - 12) = 14 * 8 * 9 = 14 * 72 = 1008.

Check a5 < a6: 896 < 1008, yes.

a7 = k (3n - 2m)^2 = 14 * (21 - 12)^2 = 14 * 9² = 14 * 81 = 1134.

Check a6 < a7: 1008 < 1134, yes.

a8 = k (3n - 2m)(4n - 3m) = 14 * 9 * (28 - 18) = 14 * 9 * 10 = 14 * 90 = 1260.

Check a7 < a8: 1134 < 1260, yes.

a9 = k (4n - 3m)^2 = 14 * (28 - 18)^2 = 14 * 10² = 14 * 100 = 1400.

Check a8 < a9: 1260 < 1400, yes.

a10 = k (4n - 3m)(5n - 4m) = 14 * 10 * (35 - 24) = 14 * 10 * 11 = 14 * 110 = 1540.

Check a9 < a10: 1400 < 1540, yes.

a11 = k (5n - 4m)^2 = 14 * (35 - 24)^2 = 14 * 11² = 14 * 121 = 1694.

Check a10 < a11: 1540 < 1694, yes.

a12 = k (5n - 4m)(6n - 5m) = 14 * 11 * (42 - 30) = 14 * 11 * 12 = 14 * 132 = 1848.

Check a11 < a12: 1694 < 1848, yes.

a13 = k (6n - 5m)^2 = 14 * (42 - 30)^2 = 14 * 12² = 14 * 144 = 2016. Which matches the given value. Perfect!

So, this works. Therefore, the value of $ a_1 $ is 504.

But let me check if there are other possible solutions. For example, in case d=12, we had only t=1. But maybe there are other d's?

Wait, we checked all possible d's (1,2,3,4,6,12) and only d=12 gave a valid solution. Therefore, this is the only solution.

But just to be thorough, let me check if there are other possible factorizations. Wait, earlier when I considered the possible d's, I considered only d's where d^2 divides 2016. However, is there a possibility that d^2 divides 2016 but d is not of the form 2^a 3^b? For example, could there be other prime factors?

But since 2016's prime factors are only 2, 3, 7, and in our equation, d^2 divides 2016, so d can only have primes 2, 3, 7. However, earlier, we saw that for prime 7, 2c ≤ 1 ⇒ c=0. Therefore, d cannot have 7 as a factor. Therefore, d must be of the form 2^a 3^b, with a ≤ 2, b ≤ 1. Therefore, our list of d's is complete. Therefore, only d=12 gives a valid solution.

Therefore, the answer is $ a_1 = 504 $.

But let me check once again if there are other possible values of d. For example, if d is negative? But since d = 6n - 5m, and n > m, and m, n positive integers, then 6n - 5m is positive? Let me check.

Since n > m, but 6n - 5m could be positive or negative? For example, if n = m + 1, then 6(m + 1) - 5m = 6m + 6 - 5m = m + 6, which is positive. So, since n > m, and m ≥ 1, then 6n - 5m ≥ 6(m + 1) - 5m = m + 6 ≥ 1 + 6 = 7. Wait, but in our case, d was 12. Wait, but in our previous cases, for example, when d=1, we had 6n - 5m = 1, which is possible if n and m are such that 6n - 5m = 1. However, in that case, we saw that there were no solutions. But in general, d must be positive. Therefore, our previous analysis holds.

Therefore, the only valid solution is when d=12, leading to a1=504.

Therefore, the answer is \boxed{504}.

**Final Answer**
\boxed{504}

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

- For every positive integer $ k $, the subsequence $ a_{2k-1}, a_{2k}, a_{2k+1} $ is **geometric**.
- For every positive integer $ k $, the subsequence $ a_{2k}, a_{2k+1}, a_{2k+2} $ is **arithmetic**.
- We are told that $ a_{13} = 2016 $, and we are to find $ a_1 $.

---

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

Let’s denote the first few terms in terms of $ a_1 = x $ and $ a_2 = y $, and use the given properties to derive a general pattern.

#### For $ k = 1 $:
- The subsequence $ a_1, a_2, a_3 $ is geometric:
  $$
  a_2^2 = a_1 a_3 \Rightarrow a_3 = \frac{y^2}{x}
  $$
- The subsequence $ a_2, a_3, a_4 $ is arithmetic:
  $$
  2a_3 = a_2 + a_4 \Rightarrow a_4 = 2a_3 - a_2 = \frac{2y^2 - xy}{x}
  $$

#### For $ k = 2 $:
- The subsequence $ a_3, a_4, a_5 $ is geometric:
  $$
  a_4^2 = a_3 a_5 \Rightarrow a_5 = \frac{a_4^2}{a_3}
  $$
- The subsequence $ a_4, a_5, a_6 $ is arithmetic:
  $$
  2a_5 = a_4 + a_6 \Rightarrow a_6 = 2a_5 - a_4
  $$

This pattern continues, and we can observe that the terms follow a recursive structure based on the previous terms.

---

### Step 2: General Pattern and Parametrization

By analyzing the recursive relations, we can derive a general form for the terms of the sequence. Let’s define:

- $ a_1 = m^2 k $
- $ a_2 = m k n $
- $ a_3 = k n^2 $
- $ a_4 = k n (2n - m) $
- $ a_5 = k (2n - m)^2 $
- $ a_6 = k (2n - m)(3n - 2m) $
- $ a_7 = k (3n - 2m)^2 $
- $ a_8 = k (3n - 2m)(4n - 3m) $
- $ a_9 = k (4n - 3m)^2 $
- $ a_{10} = k (4n - 3m)(5n - 4m) $
- $ a_{11} = k (5n - 4m)^2 $
- $ a_{12} = k (5n - 4m)(6n - 5m) $
- $ a_{13} = k (6n - 5m)^2 $

So, the general form for the odd-indexed terms is:
$$
a_{2t - 1} = k \cdot (t n - (t - 1)m)^2
$$

Given that $ a_{13} = 2016 $, and $ a_{13} = k (6n - 5m)^2 $, we have:
$$
k (6n - 5m)^2 = 2016
$$

---

### Step 3: Solve the Equation

We now solve:
$$
k (6n - 5m)^2 = 2016
$$

Let’s denote $ d = 6n - 5m $, so:
$$
k d^2 = 2016
$$

We need to find all integer factorizations of 2016 into $ k \cdot d^2 $, where $ d $ is a positive integer, and $ k $ is also a positive integer.

The prime factorization of 2016 is:
$$
2016 = 2^5 \cdot 3^2 \cdot 7
$$

We now find all possible values of $ d $ such that $ d^2 $ divides 2016. That is, $ d^2 \mid 2016 $, so $ d $ must be a divisor of $ \sqrt{2016} $, and $ d^2 $ must divide 2016.

Let’s list all such $ d $:

- $ d = 1 \Rightarrow d^2 = 1 \Rightarrow k = 2016 $
- $ d = 2 \Rightarrow d^2 = 4 \Rightarrow k = 504 $
- $ d = 3 \Rightarrow d^2 = 9 \Rightarrow k = 224 $
- $ d = 4 \Rightarrow d^2 = 16 \Rightarrow k = 126 $
- $ d = 6 \Rightarrow d^2 = 36 \Rightarrow k = 56 $
- $ d = 12 \Rightarrow d^2 = 144 \Rightarrow k = 14 $

Now, for each $ d $, we solve the equation:
$$
6n - 5m = d
$$
and check if there exist positive integers $ m, n $ such that $ \gcd(m, n) = 1 $ and $ n > m $.

After testing all possible values, the only valid solution is:

- $ d = 12 $
- $ k = 14 $
- Solving $ 6n - 5m = 12 $, we find $ m = 6 $, $ n = 7 $, and $ \gcd(6, 7) = 1 $

---

### Step 4: Compute $ a_1 $

Recall:
$$
a_1 = m^2 k = 6^2 \cdot 14 = 36 \cdot 14 = 504
$$

---

### 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, and then 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 like (a + b) is (a - b), right? So in this case, the denominator is 2 + √3, so its conjugate would be 2 - √3. Let me try that.

So, multiplying 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. The denominator is a difference of squares: (2)^2 - (√3)^2 = 4 - 3 = 1. Wait, that's nice! The denominator simplifies to 1? That would make the whole expression just equal to the numerator. Wait, let me check that again.

Yes, (2 + √3)(2 - √3) = 2^2 - (√3)^2 = 4 - 3 = 1. So the denominator is 1, so the entire expression simplifies to just the numerator: (1 + √2)(2 - √3). Hmm, so maybe I need to expand that numerator?

Let me do that. Let's multiply (1 + √2)(2 - √3). Using the distributive property (FOIL):

First: 1 * 2 = 2

Outer: 1 * (-√3) = -√3

Inner: √2 * 2 = 2√2

Last: √2 * (-√3) = -√6

So combining all those terms: 2 - √3 + 2√2 - √6. So the entire expression simplifies to 2 - √3 + 2√2 - √6. Wait, but the problem says the simplified form can be written as A(1 + √B) - (√C + √D). Let me see if I can rearrange the terms I have here to match that form.

So, the expression is 2 - √3 + 2√2 - √6. Let me group the terms. Let me see:

First, maybe group the constants and the radicals. But there's a 2, then -√3, then +2√2, then -√6. Let me think. The target form is A(1 + √B) - (√C + √D). Let me think about how to express this.

Alternatively, maybe I made a mistake in the simplification? Let me check again. Wait, the original expression was (1 + √2)/(2 + √3). After rationalizing, we multiplied numerator and denominator by 2 - √3, denominator becomes 1, numerator is (1 + √2)(2 - √3). Let me check that multiplication again.

Yes, expanding (1 + √2)(2 - √3):

1*2 = 2

1*(-√3) = -√3

√2*2 = 2√2

√2*(-√3) = -√6

So total is 2 - √3 + 2√2 - √6. That seems correct.

So, the simplified expression is 2 - √3 + 2√2 - √6. Now, how can I write this in the form A(1 + √B) - (√C + √D)?

Let me think. Let me try to group terms. Let me see:

Looking at the expression: 2 - √3 + 2√2 - √6. Let me see if I can factor something out. For example, maybe group the 2 and 2√2 together? Let me try:

2 + 2√2 - √3 - √6. Then, factor 2 from the first two terms: 2(1 + √2) - (√3 + √6). Wait, that seems promising. Let me check:

Yes, 2(1 + √2) - (√3 + √6). But the problem says the form is A(1 + √B) - (√C + √D). So here, A would be 2, B would be 2, and then √C + √D would be √3 + √6. But then C and D would be 3 and 6. However, the problem says that A, B, C, D are positive integers. So that would make A=2, B=2, C=3, D=6. Then A+B+C+D would be 2+2+3+6=13. But wait, let me check if that's correct.

Wait, but let me check if the original expression is equal to 2(1 + √2) - (√3 + √6). Let me compute that:

2(1 + √2) = 2 + 2√2

Then subtract (√3 + √6): 2 + 2√2 - √3 - √6. Which is exactly the same as the simplified expression. So that seems correct. Therefore, A=2, B=2, C=3, D=6. Then A+B+C+D=2+2+3+6=13. Is that the answer? Wait, but let me check again if there's another way to write it, maybe with different A, B, C, D? Because sometimes there might be multiple forms, but the problem says "the form A(1 + √B) - (√C + √D)", so maybe this is the only way? Let me check.

Alternatively, maybe there's a different grouping? Let me think. Suppose I tried to write the expression as something else. For example, maybe combining the -√3 and -√6? But that would be -√3 - √6, which is the same as -(√3 + √6). So that's the same as before. So the expression is 2 + 2√2 - √3 - √6, which is 2(1 + √2) - (√3 + √6). So that seems to fit the required form. Therefore, A=2, B=2, C=3, D=6. Then A+B+C+D=2+2+3+6=13. But let me check again if that's correct.

Wait, but let me check if the original problem had a different answer. Wait, maybe I made a mistake in rationalizing? Let me check again. Original expression: (1 + √2)/(2 + √3). Multiply numerator and denominator by 2 - √3:

Numerator: (1 + √2)(2 - √3) = 2 - √3 + 2√2 - √6. Denominator: 1. So yes, that's correct. So the simplified expression is 2 - √3 + 2√2 - √6. Which is equal to 2(1 + √2) - (√3 + √6). Therefore, A=2, B=2, C=3, D=6. So sum is 13. But let me check if there's another way to write this expression. For example, maybe combining √6 with something else?

Wait, √6 can be written as √2*√3. But I don't know if that helps. Let me think. Suppose I want to write the expression as A(1 + √B) - (√C + √D). Maybe there's another way? Let me think. For example, if I can write 2 - √3 + 2√2 - √6 as something else. Let me see:

Alternatively, maybe factor something else. Let me think. Suppose I take 2 - √3 and 2√2 - √6. Let me see:

2 - √3 is as is. Then 2√2 - √6. Let me factor √2 from that: √2*(2 - √3). So then the entire expression would be (2 - √3) + √2*(2 - √3) = (2 - √3)(1 + √2). Wait, that's interesting. So that would be (2 - √3)(1 + √2). But that's the original numerator before simplifying. Wait, but that's equal to the original expression, which is (1 + √2)/(2 + √3) multiplied by (2 - √3)/(2 - √3) which is 1. So that's the same as before. So that gives us (2 - √3)(1 + √2). But how does that help?

Wait, but if I write it as (2 - √3)(1 + √2), then maybe that's another way to write it. But the problem wants it in the form A(1 + √B) - (√C + √D). So if I have (2 - √3)(1 + √2), that's not directly in that form. However, if I expand it, I get back to 2(1 + √2) - √3(1 + √2) = 2(1 + √2) - √3 - √3√2 = 2(1 + √2) - √3 - √6. Which is the same as before. So that's the same expression. Therefore, the previous conclusion seems correct.

Alternatively, maybe there's a different way to rationalize? Wait, but the denominator was 2 + √3, and multiplying by its conjugate gives 1. So that's the simplest way. So I think my answer is correct. Therefore, A=2, B=2, C=3, D=6. Therefore, A+B+C+D=2+2+3+6=13. Wait, but let me check again if the problem says "converted to the form A(1 + √B) - (√C + √D)". So in my case, it's 2(1 + √2) - (√3 + √6). So that's exactly the form. Therefore, A=2, B=2, C=3, D=6. So the sum is 13. But let me check if there's another possible way. For example, maybe if I can write √6 as √something else? But √6 is already simplified. Similarly, √3 is simplified. So I think that's the only way.

Wait, but let me check if the problem requires that C and D are square-free or something? The problem says they are positive integers, so they can be any positive integers. So even if they are not square-free, like 6, that's okay. So 6 is allowed. Therefore, I think my answer is correct. Therefore, the answer is 13. Wait, but let me check again with actual numbers. Let me compute the original expression and my simplified expression numerically to check if they are equal.

Original expression: (1 + √2)/(2 + √3). Let me compute approximate decimal values.

First, √2 ≈ 1.4142, √3 ≈ 1.732.

So numerator: 1 + 1.4142 ≈ 2.4142.

Denominator: 2 + 1.732 ≈ 3.732.

So original expression ≈ 2.4142 / 3.732 ≈ Let me compute that. Let me divide 2.4142 by 3.732. Let me approximate:

Well, 3.732 * 0.6 = 2.2392. 2.4142 - 2.2392 = 0.175. So 0.6 + 0.175 / 3.732 ≈ 0.6 + 0.0469 ≈ 0.6469. So approximately 0.6469.

Now, my simplified expression: 2(1 + √2) - (√3 + √6). Let me compute that:

First, 1 + √2 ≈ 1 + 1.4142 ≈ 2.4142. Multiply by 2: ≈ 4.8284.

Then, √3 + √6 ≈ 1.732 + 2.449 ≈ 4.181.

Subtracting: 4.8284 - 4.181 ≈ 0.6474. Which is very close to the original value of approximately 0.6469. The slight difference is due to rounding errors in my calculations. Therefore, this confirms that my simplified expression is correct.

Therefore, the answer is A + B + C + D = 2 + 2 + 3 + 6 = 13. Wait, but let me check again if there's another possible way. For example, maybe if I factor differently?

Suppose I tried to write the expression as something else. Let me think. For example, suppose I have 2 - √3 + 2√2 - √6. Maybe I can write this as (2 + 2√2) - (√3 + √6). Then factor 2 from the first group: 2(1 + √2) - (√3 + √6). Which is the same as before. So that's the same as before. Therefore, same result.

Alternatively, if I tried to write it as A(1 + √B) - √C - √D. So that's the same as A(1 + √B) - (√C + √D). So that's the same form. Therefore, I think that's the only way. Therefore, the answer is 13.

But just to be thorough, let me check if there's another possible combination. For example, maybe if I can write √3 + √6 as something else? For example, √3 + √6 = √3 + √(2*3) = √3 + √2√3 = √3(1 + √2). Therefore, then the expression would be 2(1 + √2) - √3(1 + √2) = (2 - √3)(1 + √2). Which is the original numerator. But that's not helpful for the required form. Because the required form is A(1 + √B) - (√C + √D). So if I write it as (2 - √3)(1 + √2), that's not in the required form. However, if I expand it, I get back to the same expression. Therefore, I think my original answer is correct.

Alternatively, maybe there's a different A, B, C, D? Let me think. Suppose someone tried to write the expression as something else. For example, maybe A is 1? Let me see. If A is 1, then 1*(1 + √B) - (√C + √D) = 1 + √B - √C - √D. Comparing with our expression 2 - √3 + 2√2 - √6. So we need 1 + √B - √C - √D = 2 - √3 + 2√2 - √6. Therefore, equating terms:

1 + √B - √C - √D = 2 - √3 + 2√2 - √6.

So, rearranged:

(1 - 2) + √B + (-√C - √D) = -√3 + 2√2 - √6.

Which is:

-1 + √B - √C - √D = -√3 + 2√2 - √6.

Therefore, √B - √C - √D = -√3 + 2√2 - √6 + 1. Which seems complicated. It's not clear how to assign B, C, D here. So maybe A=1 is not possible. Similarly, trying A=3? Probably not. So A=2 seems the only possible. Therefore, I think my original answer is correct.

Therefore, the answer is 13. Wait, but let me check again if the problem says "converted to the form A(1 + √B) - (√C + √D)". So in my case, it's 2(1 + √2) - (√3 + √6). So A=2, B=2, C=3, D=6. Therefore, A+B+C+D=2+2+3+6=13. So I think that's correct.

But just to be 100% sure, let me check if there's another way. For example, maybe if I rationalize differently? Wait, but the denominator was 2 + √3, and multiplying by its conjugate is the standard way. There's no other way to rationalize it. So I think that's the only way. Therefore, I think my answer is correct.

**Final Answer**
\boxed{13}

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

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

---

### Step 1: Simplify the Denominator

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

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

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

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

---

### Step 2: Expand the Numerator

Use the distributive property (FOIL) to expand:

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

So the simplified expression is:

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

---

### Step 3: Match the Target Form

We are told the expression can be written in the form:

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

Let’s group and factor the expression:

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

This matches the required form exactly, with:

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

---

### Step 4: Compute the Final Answer

Now, sum the values:

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

[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, and then 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 like (a + b) is (a - b), right? So in this case, the denominator is 2 + √3, so its conjugate would be 2 - √3. Let me try that.

So, multiplying 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. The denominator is a difference of squares: (2)^2 - (√3)^2 = 4 - 3 = 1. Wait, that's nice! The denominator simplifies to 1? That would make the whole expression just equal to the numerator. Wait, let me check that again.

Yes, (2 + √3)(2 - √3) = 2^2 - (√3)^2 = 4 - 3 = 1. So the denominator is 1, so the entire expression simplifies to just the numerator: (1 + √2)(2 - √3). Hmm, so maybe I need to expand that numerator?

Let me do that. Let's multiply (1 + √2)(2 - √3). Using the distributive property (FOIL):

First: 1 * 2 = 2

Outer: 1 * (-√3) = -√3

Inner: √2 * 2 = 2√2

Last: √2 * (-√3) = -√6

So combining all those terms: 2 - √3 + 2√2 - √6. So the entire expression simplifies to 2 - √3 + 2√2 - √6. Wait, but the problem says the simplified form can be written as A(1 + √B) - (√C + √D). Let me see if I can rearrange the terms I have here to match that form.

So, the expression is 2 - √3 + 2√2 - √6. Let me group the terms. Let me see:

First, maybe group the constants and the radicals. But there's a 2, then -√3, then +2√2, then -√6. Let me think. The target form is A(1 + √B) - (√C + √D). Let me think about how to express this.

Alternatively, maybe I made a mistake in the simplification? Let me check again. Wait, the original expression was (1 + √2)/(2 + √3). After rationalizing, we multiplied numerator and denominator by 2 - √3, denominator becomes 1, numerator is (1 + √2)(2 - √3). Let me check that multiplication again.

Yes, expanding (1 + √2)(2 - √3):

1*2 = 2

1*(-√3) = -√3

√2*2 = 2√2

√2*(-√3) = -√6

So total is 2 - √3 + 2√2 - √6. That seems correct.

So, the simplified expression is 2 - √3 + 2√2 - √6. Now, how can I write this in the form A(1 + √B) - (√C + √D)?

Let me think. Let me try to group terms. Let me see:

Looking at the expression: 2 - √3 + 2√2 - √6. Let me see if I can factor something out. For example, maybe group the 2 and 2√2 together? Let me try:

2 + 2√2 - √3 - √6. Then, factor 2 from the first two terms: 2(1 + √2) - (√3 + √6). Wait, that seems promising. Let me check:

Yes, 2(1 + √2) - (√3 + √6). But the problem says the form is A(1 + √B) - (√C + √D). So here, A would be 2, B would be 2, and then √C + √D would be √3 + √6. But then C and D would be 3 and 6. However, the problem says that A, B, C, D are positive integers. So that would make A=2, B=2, C=3, D=6. Then A+B+C+D would be 2+2+3+6=13. But wait, let me check if that's correct.

Wait, but let me check if the original expression is equal to 2(1 + √2) - (√3 + √6). Let me compute that:

2(1 + √2) = 2 + 2√2

Then subtract (√3 + √6): 2 + 2√2 - √3 - √6. Which is exactly the same as the simplified expression. So that seems correct. Therefore, A=2, B=2, C=3, D=6. Then A+B+C+D=2+2+3+6=13. Is that the answer? Wait, but let me check again if there's another way to write it, maybe with different A, B, C, D? Because sometimes there might be multiple forms, but the problem says "the form A(1 + √B) - (√C + √D)", so maybe this is the only way? Let me check.

Alternatively, maybe there's a different grouping? Let me think. Suppose I tried to write the expression as something else. For example, maybe combining the -√3 and -√6? But that would be -√3 - √6, which is the same as -(√3 + √6). So that's the same as before. So the expression is 2 + 2√2 - √3 - √6, which is 2(1 + √2) - (√3 + √6). So that seems to fit the required form. Therefore, A=2, B=2, C=3, D=6. Then A+B+C+D=2+2+3+6=13. But let me check again if that's correct.

Wait, but let me check if the original problem had a different answer. Wait, maybe I made a mistake in rationalizing? Let me check again. Original expression: (1 + √2)/(2 + √3). Multiply numerator and denominator by 2 - √3:

Numerator: (1 + √2)(2 - √3) = 2 - √3 + 2√2 - √6. Denominator: 1. So yes, that's correct. So the simplified expression is 2 - √3 + 2√2 - √6. Which is equal to 2(1 + √2) - (√3 + √6). Therefore, A=2, B=2, C=3, D=6. So sum is 13. But let me check if there's another way to write this expression. For example, maybe combining √6 with something else?

Wait, √6 can be written as √2*√3. But I don't know if that helps. Let me think. Suppose I want to write the expression as A(1 + √B) - (√C + √D). Maybe there's another way? Let me think. For example, if I can write 2 - √3 + 2√2 - √6 as something else. Let me see:

Alternatively, maybe factor something else. Let me think. Suppose I take 2 - √3 and 2√2 - √6. Let me see:

2 - √3 is as is. Then 2√2 - √6. Let me factor √2 from that: √2*(2 - √3). So then the entire expression would be (2 - √3) + √2*(2 - √3) = (2 - √3)(1 + √2). Wait, that's interesting. So that would be (2 - √3)(1 + √2). But that's the original numerator before simplifying. Wait, but that's equal to the original expression, which is (1 + √2)/(2 + √3) multiplied by (2 - √3)/(2 - √3) which is 1. So that's the same as before. So that gives us (2 - √3)(1 + √2). But how does that help?

Wait, but if I write it as (2 - √3)(1 + √2), then maybe that's another way to write it. But the problem wants it in the form A(1 + √B) - (√C + √D). So if I have (2 - √3)(1 + √2), that's not directly in that form. However, if I expand it, I get back to 2(1 + √2) - √3(1 + √2) = 2(1 + √2) - √3 - √3√2 = 2(1 + √2) - √3 - √6. Which is the same as before. So that's the same expression. Therefore, the previous conclusion seems correct.

Alternatively, maybe there's a different way to rationalize? Wait, but the denominator was 2 + √3, and multiplying by its conjugate gives 1. So that's the simplest way. So I think my answer is correct. Therefore, A=2, B=2, C=3, D=6. Therefore, A+B+C+D=2+2+3+6=13. Wait, but let me check again if the problem says "converted to the form A(1 + √B) - (√C + √D)". So in my case, it's 2(1 + √2) - (√3 + √6). So that's exactly the form. Therefore, A=2, B=2, C=3, D=6. So the sum is 13. But let me check if there's another possible way. For example, maybe if I can write √6 as √something else? But √6 is already simplified. Similarly, √3 is simplified. So I think that's the only way.

Wait, but let me check if the problem requires that C and D are square-free or something? The problem says they are positive integers, so they can be any positive integers. So even if they are not square-free, like 6, that's okay. So 6 is allowed. Therefore, I think my answer is correct. Therefore, the answer is 13. Wait, but let me check again with actual numbers. Let me compute the original expression and my simplified expression numerically to check if they are equal.

Original expression: (1 + √2)/(2 + √3). Let me compute approximate decimal values.

First, √2 ≈ 1.4142, √3 ≈ 1.732.

So numerator: 1 + 1.4142 ≈ 2.4142.

Denominator: 2 + 1.732 ≈ 3.732.

So original expression ≈ 2.4142 / 3.732 ≈ Let me compute that. Let me divide 2.4142 by 3.732. Let me approximate:

Well, 3.732 * 0.6 = 2.2392. 2.4142 - 2.2392 = 0.175. So 0.6 + 0.175 / 3.732 ≈ 0.6 + 0.0469 ≈ 0.6469. So approximately 0.6469.

Now, my simplified expression: 2(1 + √2) - (√3 + √6). Let me compute that:

First, 1 + √2 ≈ 1 + 1.4142 ≈ 2.4142. Multiply by 2: ≈ 4.8284.

Then, √3 + √6 ≈ 1.732 + 2.449 ≈ 4.181.

Subtracting: 4.8284 - 4.181 ≈ 0.6474. Which is very close to the original value of approximately 0.6469. The slight difference is due to rounding errors in my calculations. Therefore, this confirms that my simplified expression is correct.

Therefore, the answer is A + B + C + D = 2 + 2 + 3 + 6 = 13. Wait, but let me check again if there's another possible way. For example, maybe if I factor differently?

Suppose I tried to write the expression as something else. Let me think. For example, suppose I have 2 - √3 + 2√2 - √6. Maybe I can write this as (2 + 2√2) - (√3 + √6). Then factor 2 from the first group: 2(1 + √2) - (√3 + √6). Which is the same as before. So that's the same as before. Therefore, same result.

Alternatively, if I tried to write it as A(1 + √B) - √C - √D. So that's the same as A(1 + √B) - (√C + √D). So that's the same form. Therefore, I think that's the only way. Therefore, the answer is 13.

But just to be thorough, let me check if there's another possible combination. For example, maybe if I can write √3 + √6 as something else? For example, √3 + √6 = √3 + √(2*3) = √3 + √2√3 = √3(1 + √2). Therefore, then the expression would be 2(1 + √2) - √3(1 + √2) = (2 - √3)(1 + √2). Which is the original numerator. But that's not helpful for the required form. Because the required form is A(1 + √B) - (√C + √D). So if I write it as (2 - √3)(1 + √2), that's not in the required form. However, if I expand it, I get back to the same expression. Therefore, I think my original answer is correct.

Alternatively, maybe there's a different A, B, C, D? Let me think. Suppose someone tried to write the expression as something else. For example, maybe A is 1? Let me see. If A is 1, then 1*(1 + √B) - (√C + √D) = 1 + √B - √C - √D. Comparing with our expression 2 - √3 + 2√2 - √6. So we need 1 + √B - √C - √D = 2 - √3 + 2√2 - √6. Therefore, equating terms:

1 + √B - √C - √D = 2 - √3 + 2√2 - √6.

So, rearranged:

(1 - 2) + √B + (-√C - √D) = -√3 + 2√2 - √6.

Which is:

-1 + √B - √C - √D = -√3 + 2√2 - √6.

Therefore, √B - √C - √D = -√3 + 2√2 - √6 + 1. Which seems complicated. It's not clear how to assign B, C, D here. So maybe A=1 is not possible. Similarly, trying A=3? Probably not. So A=2 seems the only possible. Therefore, I think my original answer is correct.

Therefore, the answer is 13. Wait, but let me check again if the problem says "converted to the form A(1 + √B) - (√C + √D)". So in my case, it's 2(1 + √2) - (√3 + √6). So A=2, B=2, C=3, D=6. Therefore, A+B+C+D=2+2+3+6=13. So I think that's correct.

But just to be 100% sure, let me check if there's another way. For example, maybe if I rationalize differently? Wait, but the denominator was 2 + √3, and multiplying by its conjugate is the standard way. There's no other way to rationalize it. So I think that's the only way. Therefore, I think my answer is correct.

**Final Answer**
\boxed{13}

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

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

---

### Step 1: Simplify the Denominator

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

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

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

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

---

### Step 2: Expand the Numerator

Use the distributive property (FOIL) to expand:

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

So the simplified expression is:

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

---

### Step 3: Match the Target Form

We are told the expression can be written in the form:

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

Let’s group and factor the expression:

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

This matches the required form exactly, with:

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

---

### Step 4: Compute the Final Answer

Now, sum the values:

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

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