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 to make sure I understand it correctly.

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 $. The goal is 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 $ a $, $ b $, $ c $ are geometric, then $ b^2 = a \cdot c $. An arithmetic sequence has a common difference, so if three terms $ a $, $ b $, $ c $ are arithmetic, then $ 2b = a + c $.

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 $.

Now, the problem mentions that for every positive integer $ k $, the subsequence $ a_{2k-1} $, $ a_{2k} $, $ a_{2k+1} $ is geometric. Let me note that for each $ k $, this gives us a geometric progression starting at position $ 2k - 1 $, then $ 2k $, then $ 2k + 1 $. Similarly, for each $ k $, the subsequence $ a_{2k} $, $ a_{2k+1} $, $ a_{2k+2} $ is arithmetic.

So, for example, when $ k = 1 $, the first geometric subsequence is $ a_1, a_2, a_3 $, and the first arithmetic subsequence is $ a_2, a_3, a_4 $. Similarly, for $ k = 2 $, the geometric subsequence is $ a_3, a_4, a_5 $, and the arithmetic subsequence is $ a_4, a_5, a_6 $, and so on.

Therefore, each odd-indexed term is part of a geometric progression with the next two terms, and each even-indexed term is part of an arithmetic progression with the next two terms. This seems like there's a recursive relationship here.

Let me try to write down equations based on these properties.

Starting with $ k = 1 $:

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

$ a_2^2 = a_1 \cdot a_3 $. (1)

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

$ 2a_3 = a_2 + a_4 $. (2)

Similarly, for $ k = 2 $:

Geometric sequence $ a_3, a_4, a_5 $:

$ a_4^2 = a_3 \cdot a_5 $. (3)

Arithmetic sequence $ a_4, a_5, a_6 $:

$ 2a_5 = a_4 + a_6 $. (4)

For $ k = 3 $:

Geometric sequence $ a_5, a_6, a_7 $:

$ a_6^2 = a_5 \cdot a_7 $. (5)

Arithmetic sequence $ a_6, a_7, a_8 $:

$ 2a_7 = a_6 + a_8 $. (6)

And so on. It seems like there's a pattern here. Each odd term is part of a geometric progression with the next two terms, and each even term is part of an arithmetic progression with the next two terms.

Since the sequence is strictly increasing, all the terms are positive integers, and each term is larger than the previous. Therefore, the common ratio in the geometric sequences must be greater than 1, and the common difference in the arithmetic sequences must be positive.

Let me try to find a recursive relationship between terms. Let me denote the terms as $ a_1, a_2, a_3, \dots $.

From equation (1): $ a_2^2 = a_1 a_3 $. Therefore, $ a_3 = \frac{a_2^2}{a_1} $. Since $ a_3 $ must be an integer, $ a_1 $ must divide $ a_2^2 $.

From equation (2): $ 2a_3 = a_2 + a_4 $. Therefore, $ a_4 = 2a_3 - a_2 $. Substituting $ a_3 $ from above, we get:

$ a_4 = 2 \cdot \frac{a_2^2}{a_1} - a_2 $.

Similarly, from equation (3): $ a_4^2 = a_3 a_5 $. Therefore, $ a_5 = \frac{a_4^2}{a_3} $.

From equation (4): $ 2a_5 = a_4 + a_6 $, so $ a_6 = 2a_5 - a_4 $.

Continuing this way, perhaps we can find a pattern or recurrence relation.

But since the problem gives us $ a_{13} = 2016 $, maybe we can find expressions for terms up to $ a_{13} $ in terms of $ a_1 $ and $ a_2 $, then set up an equation and solve for $ a_1 $.

But this might get complicated. Let me see if there's a pattern or if the sequence can be expressed in terms of a common ratio or something similar.

Alternatively, maybe there's a way to express each term in terms of the previous ones, leading to a recursive formula.

Alternatively, perhaps there's a relation between consecutive terms. Let me try to find relations between terms.

Let me consider the first few terms:

From equation (1): $ a_3 = \frac{a_2^2}{a_1} $.

From equation (2): $ a_4 = 2a_3 - a_2 = 2 \cdot \frac{a_2^2}{a_1} - a_2 $.

From equation (3): $ a_5 = \frac{a_4^2}{a_3} $.

Let me compute $ a_5 $:

First, compute $ a_4 $:

$ a_4 = \frac{2a_2^2}{a_1} - a_2 = \frac{2a_2^2 - a_1 a_2}{a_1} = \frac{a_2(2a_2 - a_1)}{a_1} $.

Then $ a_5 = \frac{a_4^2}{a_3} = \frac{ \left( \frac{a_2(2a_2 - a_1)}{a_1} \right)^2 }{ \frac{a_2^2}{a_1} } = \frac{ a_2^2 (2a_2 - a_1)^2 }{ a_1^2 } \cdot \frac{a_1}{a_2^2} } = \frac{(2a_2 - a_1)^2}{a_1} $.

So $ a_5 = \frac{(2a_2 - a_1)^2}{a_1} $.

Similarly, from equation (4): $ a_6 = 2a_5 - a_4 $.

Let me compute $ a_6 $:

First, $ 2a_5 = 2 \cdot \frac{(2a_2 - a_1)^2}{a_1} $.

Then subtract $ a_4 $:

$ a_6 = \frac{2(2a_2 - a_1)^2}{a_1} - \frac{a_2(2a_2 - a_1)}{a_1} = \frac{2(2a_2 - a_1)^2 - a_2(2a_2 - a_1)}{a_1} $.

Factor out $ (2a_2 - a_1) $:

$ a_6 = \frac{(2a_2 - a_1)[2(2a_2 - a_1) - a_2]}{a_1} $.

Simplify the bracket:

$ 2(2a_2 - a_1) - a_2 = 4a_2 - 2a_1 - a_2 = 3a_2 - 2a_1 $.

Therefore, $ a_6 = \frac{(2a_2 - a_1)(3a_2 - 2a_1)}{a_1} $.

Now, moving on to equation (5): $ a_6^2 = a_5 a_7 $, so $ a_7 = \frac{a_6^2}{a_5} $.

Let me compute $ a_7 $:

First, compute $ a_6^2 $:

$ \left( \frac{(2a_2 - a_1)(3a_2 - 2a_1)}{a_1} \right)^2 $.

Then divide by $ a_5 = \frac{(2a_2 - a_1)^2}{a_1} $:

So $ a_7 = \frac{ \left( (2a_2 - a_1)^2 (3a_2 - 2a_1)^2 \right) / a_1^2 }{ (2a_2 - a_1)^2 / a_1 } } = \frac{(3a_2 - 2a_1)^2}{a_1} $.

Therefore, $ a_7 = \frac{(3a_2 - 2a_1)^2}{a_1} $.

Similarly, from equation (6): $ 2a_7 = a_6 + a_8 $, so $ a_8 = 2a_7 - a_6 $.

Let me compute $ a_8 $:

First, $ 2a_7 = 2 \cdot \frac{(3a_2 - 2a_1)^2}{a_1} $.

Subtract $ a_6 $:

$ a_8 = \frac{2(3a_2 - 2a_1)^2}{a_1} - \frac{(2a_2 - a_1)(3a_2 - 2a_1)}{a_1} $.

Factor out $ \frac{(3a_2 - 2a_1)}{a_1} $:

$ a_8 = \frac{(3a_2 - 2a_1)}{a_1} [ 2(3a_2 - 2a_1) - (2a_2 - a_1) ] $.

Compute the bracket:

$ 2(3a_2 - 2a_1) - (2a_2 - a_1) = 6a_2 - 4a_1 - 2a_2 + a_1 = 4a_2 - 3a_1 $.

Therefore, $ a_8 = \frac{(3a_2 - 2a_1)(4a_2 - 3a_1)}{a_1} $.

Hmm, interesting. So there seems to be a pattern here. Let me try to see if I can find a pattern in terms of coefficients.

Looking at the expressions for $ a_3, a_5, a_7 $, they are of the form $ \frac{(m a_2 - n a_1)^2}{a_1} $, where m and n seem to follow some pattern.

Similarly, the even terms $ a_4, a_6, a_8 $ are products of terms like $ (m a_2 - n a_1) $ divided by $ a_1 $.

Let me try to see if there's a recursive pattern here. Let me note the coefficients:

For $ a_3 $: numerator is $ a_2^2 $, denominator $ a_1 $.

For $ a_5 $: numerator is $ (2a_2 - a_1)^2 $, denominator $ a_1 $.

For $ a_7 $: numerator is $ (3a_2 - 2a_1)^2 $, denominator $ a_1 $.

Similarly, for $ a_4 $: numerator is $ a_2(2a_2 - a_1) $, denominator $ a_1 $.

For $ a_6 $: numerator is $ (2a_2 - a_1)(3a_2 - 2a_1) $, denominator $ a_1 $.

For $ a_8 $: numerator is $ (3a_2 - 2a_1)(4a_2 - 3a_1) $, denominator $ a_1 $.

So, if I denote $ b_k = k a_2 - (k - 1)a_1 $, then:

- $ a_3 = \frac{(1 a_2 - 0 a_1)^2}{a_1} = \frac{a_2^2}{a_1} $
- $ a_5 = \frac{(2 a_2 - 1 a_1)^2}{a_1} $
- $ a_7 = \frac{(3 a_2 - 2 a_1)^2}{a_1} $
- Similarly, $ a_4 = \frac{a_2 (2a_2 - a_1)}{a_1} $
- $ a_6 = \frac{(2a_2 - a_1)(3a_2 - 2a_1)}{a_1} $
- $ a_8 = \frac{(3a_2 - 2a_1)(4a_2 - 3a_1)}{a_1} $

So, it seems like for odd indices $ a_{2k+1} $, the term is $ \frac{(k a_2 - (k - 1)a_1)^2}{a_1} $, and for even indices $ a_{2k} $, the term is $ \frac{( (k - 1)a_2 - (k - 2)a_1 ) (k a_2 - (k - 1)a_1 ) }{a_1} $.

Wait, let me check:

For $ a_4 $, which is $ k = 2 $, since $ a_{2k} $, so $ k = 2 $, then:

$ \frac{( (2 - 1)a_2 - (2 - 2)a_1 ) (2 a_2 - (2 - 1)a_1 ) }{a_1} = \frac{(a_2 - 0 a_1)(2a_2 - a_1)}{a_1} = \frac{a_2(2a_2 - a_1)}{a_1} $, which matches.

Similarly, for $ a_6 $, $ k = 3 $:

$ \frac{( (3 - 1)a_2 - (3 - 2)a_1 ) (3 a_2 - (3 - 1)a_1 ) }{a_1} = \frac{(2a_2 - a_1)(3a_2 - 2a_1)}{a_1} $, which matches.

Therefore, generalizing:

For each $ k \geq 1 $:

- $ a_{2k - 1} = \frac{( (k - 1)a_2 - (k - 2)a_1 )^2 }{a_1} $ ?

Wait, no. Wait for $ a_3 $, which is $ k = 2 $, since $ a_{2k - 1} = a_3 $ when $ k = 2 $. Let me check:

If $ a_{2k - 1} = \frac{( (k - 1)a_2 - (k - 2)a_1 )^2 }{a_1} $, then for $ k = 2 $, it would be $ \frac{(1 a_2 - 0 a_1)^2}{a_1} = \frac{a_2^2}{a_1} $, which is correct. For $ k = 3 $, $ a_5 = \frac{(2a_2 - 1a_1)^2}{a_1} $, which is correct. Similarly, for $ k = 4 $, $ a_7 = \frac{(3a_2 - 2a_1)^2}{a_1} $, which is correct. So yes, the general formula for $ a_{2k - 1} $ is $ \frac{( (k - 1)a_2 - (k - 2)a_1 )^2 }{a_1} $.

Similarly, for even terms $ a_{2k} $, the formula is $ \frac{ ( (k - 1)a_2 - (k - 2)a_1 ) (k a_2 - (k - 1)a_1 ) }{a_1} $.

Therefore, if I can find a general formula for $ a_n $, maybe I can express $ a_{13} $ in terms of $ a_1 $ and $ a_2 $, then solve for $ a_1 $.

But first, let me check if this pattern continues. Let me compute $ a_9 $, for example.

From equation (7): For $ k = 4 $, the geometric sequence $ a_7, a_8, a_9 $:

So $ a_8^2 = a_7 a_9 $, so $ a_9 = \frac{a_8^2}{a_7} $.

From previous expressions:

$ a_7 = \frac{(3a_2 - 2a_1)^2}{a_1} $

$ a_8 = \frac{(3a_2 - 2a_1)(4a_2 - 3a_1)}{a_1} $

Therefore, $ a_9 = \frac{ [ (3a_2 - 2a_1)(4a_2 - 3a_1) / a_1 ]^2 }{ (3a_2 - 2a_1)^2 / a_1 } } = \frac{ (4a_2 - 3a_1)^2 }{a_1 } $.

Which fits the pattern: for $ a_{2k - 1} $, when $ k = 5 $, $ a_9 = \frac{(4a_2 - 3a_1)^2}{a_1} $, which is consistent with the formula $ \frac{( (k - 1)a_2 - (k - 2)a_1 )^2 }{a_1} $, since $ k = 5 $, $ (5 - 1)a_2 - (5 - 2)a_1 = 4a_2 - 3a_1 $.

Similarly, $ a_{10} $ would be $ 2a_9 - a_8 $, which would be:

$ 2 \cdot \frac{(4a_2 - 3a_1)^2}{a_1} - \frac{(3a_2 - 2a_1)(4a_2 - 3a_1)}{a_1} = \frac{2(4a_2 - 3a_1)^2 - (3a_2 - 2a_1)(4a_2 - 3a_1)}{a_1} $.

Factor out $ (4a_2 - 3a_1) $:

$ \frac{(4a_2 - 3a_1)[2(4a_2 - 3a_1) - (3a_2 - 2a_1)]}{a_1} $.

Compute the bracket:

$ 2(4a_2 - 3a_1) - (3a_2 - 2a_1) = 8a_2 - 6a_1 - 3a_2 + 2a_1 = 5a_2 - 4a_1 $.

Therefore, $ a_{10} = \frac{(4a_2 - 3a_1)(5a_2 - 4a_1)}{a_1} $, which fits the pattern for even terms.

So, the pattern seems consistent. Therefore, we can generalize:

For each $ n $, if $ n $ is odd, say $ n = 2k - 1 $, then:

$ a_n = \frac{( (k - 1)a_2 - (k - 2)a_1 )^2 }{a_1} $.

If $ n $ is even, say $ n = 2k $, then:

$ a_n = \frac{ ( (k - 1)a_2 - (k - 2)a_1 ) (k a_2 - (k - 1)a_1 ) }{a_1} $.

Alternatively, if we let $ m = k - 1 $, then for odd indices $ n = 2m + 1 $:

$ a_{2m + 1} = \frac{(m a_2 - (m - 1)a_1)^2}{a_1} $.

Similarly, for even indices $ n = 2m $:

$ a_{2m} = \frac{( (m - 1)a_2 - (m - 2)a_1 ) (m a_2 - (m - 1)a_1 ) }{a_1} $.

Therefore, if we can express $ a_{13} $ in terms of $ a_1 $ and $ a_2 $, then we can set up an equation.

Since 13 is odd, $ n = 13 = 2m + 1 $, so $ m = 6 $. Therefore:

$ a_{13} = \frac{(6 a_2 - 5 a_1)^2}{a_1} $.

Given that $ a_{13} = 2016 $, we have:

$ \frac{(6 a_2 - 5 a_1)^2}{a_1} = 2016 $.

So, $ (6 a_2 - 5 a_1)^2 = 2016 a_1 $.

Therefore, $ 2016 a_1 $ must be a perfect square. Since $ a_1 $ is a positive integer, and 2016 is given, we can factor 2016 to see what conditions this imposes on $ a_1 $.

First, let me factor 2016:

2016 ÷ 2 = 1008

1008 ÷ 2 = 504

504 ÷ 2 = 252

252 ÷ 2 = 126

126 ÷ 2 = 63

63 ÷ 3 = 21

21 ÷ 3 = 7

7 ÷ 7 = 1

So, prime factors of 2016 are $ 2^5 \times 3^2 \times 7^1 $.

Therefore, 2016 = $ 2^5 \times 3^2 \times 7 $.

So, $ 2016 a_1 = 2^5 \times 3^2 \times 7 \times a_1 $ must be a perfect square.

For a number to be a perfect square, all exponents in its prime factorization must be even. Therefore, the exponents of primes in $ 2016 a_1 $ must be even.

Let me denote the prime factors of $ a_1 $. Let me suppose that $ a_1 $ has prime factors that can make the exponents of 2, 3, 7, and any other primes in $ a_1 $ even.

Let me write $ a_1 $ as $ 2^{x} \times 3^{y} \times 7^{z} \times \text{other primes} $.

Then, the exponents in $ 2016 a_1 $ would be:

- For prime 2: 5 + x
- For prime 3: 2 + y
- For prime 7: 1 + z
- For other primes: whatever they are in $ a_1 $, but since $ a_1 $ is multiplied by 2016, which doesn't have other primes, so other primes must be present in $ a_1 $ with even exponents.

Therefore, for $ 2016 a_1 $ to be a perfect square:

- 5 + x must be even ⇒ x is odd (since 5 is odd)
- 2 + y must be even ⇒ y is even
- 1 + z must be even ⇒ z is odd
- Any other primes in $ a_1 $ must have even exponents.

Therefore, $ a_1 $ must be of the form $ 2^{odd} \times 3^{even} \times 7^{odd} \times (\text{square}) $.

But since $ a_1 $ is a positive integer, and the sequence is strictly increasing, we need to find such $ a_1 $ and $ a_2 $ that satisfy the equations and the sequence is strictly increasing.

But we also have another equation from the previous relation: $ (6 a_2 - 5 a_1)^2 = 2016 a_1 $.

Let me denote $ 6 a_2 - 5 a_1 = \sqrt{2016 a_1} $. However, since $ a_1 $ and $ a_2 $ are positive integers, $ 6 a_2 - 5 a_1 $ must also be a positive integer. Therefore, $ 6 a_2 - 5 a_1 > 0 $, so $ a_2 > \frac{5}{6} a_1 $. Since $ a_2 > a_1 $, this is automatically satisfied because $ a_2 > a_1 $, so $ a_2 > a_1 > \frac{5}{6} a_1 $.

But let me think about how to relate $ a_2 $ and $ a_1 $. Let me denote $ 6 a_2 - 5 a_1 = d $, where $ d $ is a positive integer. Then, $ d^2 = 2016 a_1 $, so $ a_1 = \frac{d^2}{2016} $.

Since $ a_1 $ must be a positive integer, $ 2016 $ must divide $ d^2 $. Therefore, $ d $ must be a multiple of the square root of the square-free part of 2016.

First, let me find the square-free part of 2016. The prime factorization is $ 2^5 \times 3^2 \times 7 $. The square-free part is the product of primes with odd exponents. Here, 2 has exponent 5 (odd), 3 has exponent 2 (even), 7 has exponent 1 (odd). Therefore, square-free part is $ 2 \times 7 = 14 $. Therefore, $ d $ must be a multiple of 14. Let me check:

If $ d = 14 k $, then $ d^2 = 196 k^2 $, so $ a_1 = \frac{196 k^2}{2016} = \frac{49 k^2}{504} = \frac{7 k^2}{72} $. Wait, but this must be integer. Therefore, 72 divides 7 k^2. Since 7 and 72 are coprime, 72 divides k^2. Therefore, k must be a multiple of the square root of 72's square-free part. Wait, 72 factors into $ 2^3 \times 3^2 $. The square-free part is 2. Therefore, k must be a multiple of 2. Let me write k = 2 m. Then:

$ a_1 = \frac{7 (2m)^2 }{72} = \frac{7 \times 4 m^2 }{72} = \frac{28 m^2}{72} = \frac{7 m^2}{18} $.

Again, 18 divides 7 m^2. Since 7 and 18 are coprime, 18 divides m^2. Therefore, m must be a multiple of the square root of the square-free part of 18. 18 factors into $ 2 \times 3^2 $, square-free part is 2. Therefore, m must be a multiple of 2. Let me set m = 2 n. Then:

$ a_1 = \frac{7 (2n)^2 }{18} = \frac{7 \times 4 n^2 }{18} = \frac{28 n^2 }{18} = \frac{14 n^2 }{9} $.

Now, 9 divides 14 n^2. Since 14 and 9 are coprime, 9 divides n^2. Therefore, n must be a multiple of 3. Let me set n = 3 p. Then:

$ a_1 = \frac{14 (3p)^2 }{9} = \frac{14 \times 9 p^2 }{9} = 14 p^2 $.

Therefore, $ a_1 = 14 p^2 $, where p is a positive integer. Therefore, the minimal value of $ a_1 $ is 14 when p = 1, then 14*4=56 when p=2, etc.

But we need to find the value of $ a_1 $, so we need more constraints. Let me recall that $ d = 14 k = 14 \times 2 m = 14 \times 2 \times 2 n = 14 \times 4 n = 14 \times 4 \times 3 p = 14 \times 12 p $. Wait, let me retrace:

Earlier steps:

We had $ d = 14 k $, then k = 2 m, then m = 2 n, then n = 3 p. Therefore:

k = 2 m = 2*(2n) = 4n

Then, d = 14 * 4n = 56n

But then, since n = 3p, d = 56 * 3p = 168p.

But also, from $ a_1 = 14 p^2 $, and $ d = 6 a_2 - 5 a_1 $, so:

$ d = 6 a_2 - 5 a_1 \Rightarrow 6 a_2 = d + 5 a_1 \Rightarrow a_2 = \frac{d + 5 a_1}{6} $.

Substituting $ a_1 = 14 p^2 $ and $ d = 168 p $:

$ a_2 = \frac{168 p + 5 \times 14 p^2 }{6} = \frac{168 p + 70 p^2 }{6} = \frac{70 p^2 + 168 p }{6} $.

Simplify:

Divide numerator and denominator by 2:

$ \frac{35 p^2 + 84 p }{3} $.

Therefore, $ a_2 = \frac{35 p^2 + 84 p }{3} $.

Since $ a_2 $ must be an integer, the numerator must be divisible by 3. Let me check:

$ 35 p^2 + 84 p = 35 p^2 + 84 p $.

Note that 84 is divisible by 3 (84 ÷ 3 = 28), and 35 ≡ 2 mod 3. Therefore:

$ 35 p^2 + 84 p ≡ 2 p^2 + 0 ≡ 2 p^2 mod 3 $.

Therefore, for this to be divisible by 3, we need $ 2 p^2 ≡ 0 mod 3 $, which implies $ p^2 ≡ 0 mod 3 $, so $ p ≡ 0 mod 3 $. Therefore, p must be a multiple of 3. Let me set p = 3 q, where q is a positive integer.

Therefore, substituting p = 3 q:

$ a_1 = 14 p^2 = 14 (9 q^2 ) = 126 q^2 $.

$ a_2 = \frac{35 (9 q^2 ) + 84 (3 q ) }{3} = \frac{315 q^2 + 252 q }{3} = 105 q^2 + 84 q $.

Therefore, $ a_2 = 105 q^2 + 84 q $.

Now, since the sequence is strictly increasing, we need to ensure that $ a_2 > a_1 $. Let me check:

$ a_2 - a_1 = 105 q^2 + 84 q - 126 q^2 = -21 q^2 + 84 q = -21 q (q - 4) $.

For this to be positive (since $ a_2 > a_1 $):

$ -21 q (q - 4) > 0 $.

Since q is a positive integer, q ≥ 1.

Let me analyze the sign:

- For q = 1: -21(1)(1 - 4) = -21(-3) = 63 > 0. Okay.
- For q = 2: -21(2)(2 - 4) = -21(2)(-2) = 84 > 0.
- For q = 3: -21(3)(3 - 4) = -21(3)(-1) = 63 > 0.
- For q = 4: -21(4)(4 - 4) = 0. Not positive.
- For q ≥ 5: -21 q (q - 4) < 0, since q - 4 ≥ 1, so negative times negative? Wait:

Wait, let me re-express:

The expression is -21 q (q - 4). Let me think of it as:

If q - 4 is positive (i.e., q ≥ 5), then the entire expression is -21 q (positive) = negative. If q - 4 is negative (i.e., q ≤ 3), then the entire expression is -21 q (negative) = positive. If q = 4, it's zero.

Therefore, for q = 1, 2, 3, the difference is positive, so $ a_2 > a_1 $. For q ≥ 4, the difference is non-positive, which would violate the strictly increasing condition. Therefore, q can only be 1, 2, or 3.

Therefore, possible values of q are 1, 2, 3. Let me check each case.

First, let me note that we also need to ensure that all terms in the sequence are positive integers and that the sequence is strictly increasing. However, since we derived $ a_1 $ and $ a_2 $ based on the condition that $ a_{13} = 2016 $, and the sequence is strictly increasing, we need to check if for these values of q, the rest of the terms are integers and the sequence remains strictly increasing.

But since we have already ensured that $ a_1 $ and $ a_2 $ are integers and $ a_2 > a_1 $, and given the recursive relations, perhaps the rest of the terms will automatically be integers and strictly increasing? Let me check for q = 1, 2, 3.

But first, let me note that $ a_1 = 126 q^2 $, and $ a_{13} = 2016 $. Let me check for each q:

First, let me recall that $ a_{13} = \frac{(6 a_2 - 5 a_1)^2}{a_1} = 2016 $, which we already used to derive the relations. Therefore, for each q, we can compute $ a_1 $, $ a_2 $, and check if the rest of the terms are integers and the sequence is strictly increasing. However, since the problem states that the sequence is strictly increasing and consists of positive integers, and we have already derived $ a_1 $ and $ a_2 $ based on the given $ a_{13} $, perhaps there is only one valid q that makes all terms integers and the sequence strictly increasing. However, since we have three possible q's, we need to check which one(s) are valid.

Alternatively, maybe there is only one possible value of q that leads to the sequence being strictly increasing beyond the first few terms. Let me check for each q.

First, let me compute for q = 1:

Case 1: q = 1

Then:

$ a_1 = 126 * 1^2 = 126 $

$ a_2 = 105 * 1^2 + 84 * 1 = 105 + 84 = 189 $

Check if $ a_2 > a_1 $: 189 > 126, yes.

Now, let me compute the rest of the terms up to $ a_{13} $ to check if they are integers and strictly increasing. However, since we already have $ a_{13} = 2016 $, maybe we can check if with these values, $ a_{13} $ is indeed 2016.

But we already derived that $ a_{13} = 2016 $, so that's already satisfied. However, we need to ensure that all previous terms are integers and the sequence is strictly increasing. Let me check some terms.

First, compute $ a_3 $:

From earlier, $ a_3 = \frac{a_2^2}{a_1} = \frac{189^2}{126} $.

Calculate 189^2: 189*189. Let me compute:

First, 200^2 = 40000, subtract 11*200 + 11^2? Wait, maybe better to compute directly:

189 * 189:

= (200 - 11)^2 = 200^2 - 2*200*11 + 11^2 = 40000 - 4400 + 121 = 40000 - 4400 = 35600; 35600 + 121 = 35721.

Therefore, $ a_3 = 35721 / 126 $.

Let me divide 35721 by 126:

First, divide numerator and denominator by 3: 35721 ÷ 3 = 11907; 126 ÷ 3 = 42.

So 11907 / 42. Divide numerator and denominator by 3 again: 11907 ÷ 3 = 3969; 42 ÷ 3 = 14.

So 3969 / 14. Let me divide 3969 by 14:

14 * 283 = 14 * 280 + 14 * 3 = 3920 + 42 = 3962. Then 3969 - 3962 = 7. Therefore, 3969 / 14 = 283 + 7/14 = 283.5. Wait, but this is not an integer. But this contradicts our earlier assumption that all terms are integers. Therefore, there must be a mistake here.

Wait, but according to our previous derivation, $ a_3 = \frac{a_2^2}{a_1} $, and since $ a_1 = 126 $, $ a_2 = 189 $, which is 189^2 / 126. However, 189 is divisible by 3, 126 is divisible by 3, but maybe there's more?

Wait, 189 = 3^3 * 7, and 126 = 2 * 3^2 * 7. Therefore, 189^2 / 126 = (3^6 * 7^2) / (2 * 3^2 * 7) ) = 3^4 * 7 / 2 = 81 * 7 / 2 = 567 / 2 = 283.5. Which is not an integer. But this contradicts the requirement that all terms are integers. Therefore, there's a problem here. Which suggests that my earlier reasoning is flawed.

Wait, but how? Earlier, I thought that $ a_1 = 14 p^2 $, and with p = 3 q, so for q = 1, p = 3, so a_1 = 14 * 9 = 126. But according to the formula for $ a_3 $, which is $ a_2^2 / a_1 $, this is not an integer. Therefore, there must be an error in my derivation.

Wait, let me check back. Earlier, I had:

From $ (6 a_2 - 5 a_1)^2 = 2016 a_1 $, and then I set $ d = 6 a_2 - 5 a_1 $, so $ d^2 = 2016 a_1 $, hence $ a_1 = d^2 / 2016 $. Then I tried to find d such that a_1 is integer. Then I considered the prime factors and found that d must be multiple of 14, etc., leading to a_1 = 14 p^2. However, in this case, with q = 1, p = 3, a_1 = 126, but then a_3 is not integer. Therefore, there's a mistake here.

Wait, but why? Because even though $ a_1 $ is integer, the term $ a_3 = a_2^2 / a_1 $ may not be integer. Therefore, my previous assumption that $ a_1 = 14 p^2 $ is not sufficient? Because even though $ d^2 = 2016 a_1 $, which ensures that $ a_1 $ divides $ d^2 $, but in addition, $ a_1 $ must divide $ a_2^2 $, since $ a_3 = a_2^2 / a_1 $. Therefore, there's another condition here. Therefore, my mistake was that I only considered the condition for $ a_{13} $, but not for the previous terms. Therefore, my previous approach is incomplete.

Therefore, I need to ensure that not only $ a_1 $ divides $ d^2 $, but also that $ a_1 $ divides $ a_2^2 $, $ a_3 $ divides $ a_4^2 $, etc. However, since the sequence is defined recursively, maybe there's a way to ensure that all terms are integers. However, in my case, with q = 1, a_3 is not integer, which suggests that my earlier approach is missing some constraints.

Therefore, I need to revisit my earlier steps. Let me think again.

We had:

From the geometric sequence $ a_1, a_2, a_3 $, so $ a_3 = a_2^2 / a_1 $. Therefore, $ a_1 $ must divide $ a_2^2 $. Similarly, from the arithmetic sequence $ a_2, a_3, a_4 $, we have $ a_4 = 2a_3 - a_2 $. Since $ a_3 = a_2^2 / a_1 $, then $ a_4 = 2(a_2^2 / a_1) - a_2 $. Therefore, $ a_4 $ is integer only if $ a_1 $ divides $ 2a_2^2 - a_1 a_2 $, which is $ a_1 $ divides $ a_2(2a_2 - a_1) $. Since $ a_1 $ divides $ a_2^2 $, let me denote $ a_2 = k a_1 $, where k is a rational number. Wait, but since $ a_1 $ divides $ a_2^2 $, $ a_2 $ must be a multiple of some factors of $ a_1 $. However, since $ a_2 > a_1 $, and the sequence is strictly increasing, k must be greater than 1.

Alternatively, perhaps I need to find $ a_1 $ and $ a_2 $ such that $ a_1 $ divides $ a_2^2 $, and $ a_1 $ divides $ a_2(2a_2 - a_1) $, etc. But this seems complex.

Alternatively, maybe there's a relation between $ a_1 $ and $ a_2 $ that can be derived from the recursive relations. Let me think.

From earlier, we had:

For $ a_3 = \frac{a_2^2}{a_1} $, and $ a_4 = 2a_3 - a_2 = \frac{2a_2^2}{a_1} - a_2 = \frac{2a_2^2 - a_1 a_2}{a_1} = \frac{a_2(2a_2 - a_1)}{a_1} $.

Similarly, for $ a_5 = \frac{a_4^2}{a_3} = \frac{ [a_2(2a_2 - a_1)/a_1]^2 }{a_2^2 / a_1} } = \frac{a_2^2 (2a_2 - a_1)^2 / a_1^2 }{a_2^2 / a_1} } = \frac{(2a_2 - a_1)^2}{a_1} $.

Similarly, $ a_6 = 2a_5 - a_4 = 2*(2a_2 - a_1)^2 / a_1 - a_2(2a_2 - a_1)/a_1 = [2(2a_2 - a_1)^2 - a_2(2a_2 - a_1)] / a_1 = (2a_2 - a_1)[2(2a_2 - a_1) - a_2] / a_1 = (2a_2 - a_1)(3a_2 - 2a_1)/a_1 $.

Similarly, $ a_7 = \frac{a_6^2}{a_5} = \frac{[(2a_2 - a_1)(3a_2 - 2a_1)/a_1]^2}{(2a_2 - a_1)^2 / a_1} } = \frac{(3a_2 - 2a_1)^2}{a_1} $.

Continuing this way, we can see that each term is defined in terms of $ a_1 $ and $ a_2 $, and for all these terms to be integers, certain divisibility conditions must hold.

In particular, for $ a_3 $ to be integer, $ a_1 $ divides $ a_2^2 $.

For $ a_4 $ to be integer, $ a_1 $ divides $ a_2(2a_2 - a_1) $.

For $ a_5 $ to be integer, $ a_1 $ divides $ (2a_2 - a_1)^2 $.

For $ a_6 $ to be integer, $ a_1 $ divides $ (2a_2 - a_1)(3a_2 - 2a_1) $.

And so on.

Therefore, there are multiple divisibility conditions. However, since we already have $ a_1 $ and $ a_2 $ related through $ a_{13} = 2016 $, perhaps we can find relations that connect these divisibility conditions.

Alternatively, maybe there's a pattern in the terms that can be expressed as linear combinations of $ a_1 $ and $ a_2 $, leading to a relation between $ a_1 $ and $ a_2 $.

Alternatively, perhaps there's a way to express $ a_n $ in terms of $ a_1 $ and $ a_2 $, and then find a relation for $ a_{13} $.

But given that earlier, with q = 1, we get a non-integer $ a_3 $, which suggests that my previous approach is missing something. Let me check where I went wrong.

Earlier, I derived that $ a_1 = 14 p^2 $, with p = 3 q, leading to q = 1, 2, 3. However, this led to a non-integer $ a_3 $. Therefore, there must be additional constraints on $ a_1 $ and $ a_2 $.

Let me think again about the equation $ (6 a_2 - 5 a_1)^2 = 2016 a_1 $. Let me denote $ x = a_1 $, $ y = a_2 $. Then:

$ (6y - 5x)^2 = 2016x $.

Let me expand the left side:

$ 36y^2 - 60xy + 25x^2 = 2016x $.

Rearranged:

$ 36y^2 - 60xy + 25x^2 - 2016x = 0 $.

This is a quadratic equation in terms of y:

$ 36y^2 - 60x y + (25x^2 - 2016x) = 0 $.

Let me solve for y:

Using quadratic formula:

$ y = [60x ± sqrt( (60x)^2 - 4 * 36 * (25x^2 - 2016x) ) ] / (2 * 36) $.

Compute discriminant D:

D = (60x)^2 - 4 * 36 * (25x^2 - 2016x)

= 3600x^2 - 144*(25x^2 - 2016x)

= 3600x^2 - 3600x^2 + 144*2016x

= 0 + 144*2016x

= 144 * 2016 * x.

Therefore, sqrt(D) = sqrt(144 * 2016 * x) = 12 * sqrt(2016x).

Therefore, y = [60x ± 12√(2016x)] / 72 = [5x ± √(2016x)] / 6.

Since y must be positive, we take the positive root:

y = [5x + √(2016x)] / 6.

But since y must be rational (as it's an integer), √(2016x) must be rational. Since 2016x is a perfect square, as we had earlier. Let me denote √(2016x) = k, where k is a positive integer. Then:

k^2 = 2016x ⇒ x = k^2 / 2016.

Since x must be a positive integer, 2016 divides k^2. As before, this implies that k must be a multiple of the square-free part of 2016, which is 14, so k = 14m, where m is a positive integer. Then:

x = (14m)^2 / 2016 = 196m^2 / 2016 = 49m^2 / 504 = 7m^2 / 72.

Therefore, x = 7m^2 / 72 must be integer. Hence, 72 divides 7m^2. Since 7 and 72 are coprime, 72 divides m^2. Therefore, m must be a multiple of the square root of the square-free part of 72. The prime factorization of 72 is 2^3 * 3^2. The square-free part is 2. Therefore, m must be a multiple of 2. Let me set m = 2n, where n is a positive integer. Then:

x = 7*(2n)^2 / 72 = 7*4n^2 / 72 = 28n^2 / 72 = 7n^2 / 18.

Therefore, x = 7n^2 / 18 must be integer. Hence, 18 divides 7n^2. Since 7 and 18 are coprime, 18 divides n^2. Therefore, n must be a multiple of the square root of the square-free part of 18. The square-free part of 18 is 2, so n must be a multiple of 2. Let me set n = 2p, where p is a positive integer. Then:

x = 7*(2p)^2 / 18 = 7*4p^2 / 18 = 28p^2 / 18 = 14p^2 / 9.

Therefore, x = 14p^2 / 9 must be integer. Hence, 9 divides 14p^2. Since 14 and 9 are coprime, 9 divides p^2. Therefore, p must be a multiple of 3. Let me set p = 3q, where q is a positive integer. Then:

x = 14*(3q)^2 / 9 = 14*9q^2 / 9 = 14q^2.

Therefore, x = a_1 = 14q^2, and k = 14m = 14*(2n) = 14*(2*(2p)) = 14*4p = 14*4*3q = 168q.

Wait, let me retrace:

Earlier, we had:

k = 14m, m = 2n, n = 2p, p = 3q.

Wait, no:

Wait, initially, k = 14m, where m = 2n, then n = 2p, then p = 3q. So:

m = 2n = 2*(2p) = 4p

n = 2p = 2*(3q) = 6q

p = 3q

Therefore, k = 14m = 14*4p = 14*4*3q = 168q.

Then, x = a_1 = 14q^2.

Then, y = [5x + k]/6 = [5*(14q^2) + 168q]/6 = [70q^2 + 168q]/6 = [70q^2]/6 + 168q/6 = (35q^2)/3 + 28q.

Wait, but this must be integer. Therefore, 35q^2 must be divisible by 3. Since 35 and 3 are coprime, this implies that q^2 must be divisible by 3, hence q must be divisible by 3. Let me set q = 3r, where r is a positive integer. Then:

a_1 = 14q^2 = 14*(9r^2) = 126r^2.

y = [5x + k]/6 = [5*14q^2 + 168q]/6 = [70q^2 + 168q]/6 = [70*(9r^2) + 168*(3r)] / 6 = [630r^2 + 504r]/6 = 105r^2 + 84r.

Therefore, a_2 = 105r^2 + 84r.

Now, this is the same as before, with q = r. Earlier, I had set q = 1, 2, 3, but now with this substitution, q = r, and we need to ensure that a_3 is integer.

Let me check for r = 1:

a_1 = 126*1^2 = 126

a_2 = 105*1^2 + 84*1 = 105 + 84 = 189

Then, a_3 = a_2^2 / a_1 = 189^2 / 126 = 35721 / 126 = 283.5, which is not integer. So this is invalid.

Wait, but earlier, we derived that a_1 = 14q^2 and a_2 = 105q^2 + 84q, but this leads to a non-integer a_3. What's wrong here?

Wait, but earlier, we had that k = 168q, and x = 14q^2, and y = [5x + k]/6 = [70q^2 + 168q]/6 = 35q^2/3 + 28q. For this to be integer, 35q^2 must be divisible by 3, hence q must be divisible by 3. So we set q = 3r, leading to a_1 = 126r^2 and a_2 = 105*(9r^2) + 84*(3r)? Wait, no:

Wait, no, when we set q = 3r, then:

Wait, earlier, we had:

After setting q = 3r, then:

a_1 = 14q^2 = 14*(3r)^2 = 14*9r^2 = 126r^2.

a_2 = 105q^2 + 84q = 105*(3r)^2 + 84*(3r) = 105*9r^2 + 252r = 945r^2 + 252r.

Wait, previously, I had written a_2 = 105r^2 + 84r, but that was incorrect. Let me correct that.

Wait, no, earlier, when we set q = 3r, then:

Wait, no, let me go back. Initially, we had:

After setting q = 3r, then:

Wait, no, let me retrace:

Initially, we had:

After setting p = 3q, then:

Wait, this is getting too convoluted. Let me start over with the substitutions.

We had:

From the equation $ (6y - 5x)^2 = 2016x $, we derived that x = 14q^2, y = 105q^2 + 84q, but this led to a non-integer a_3. However, we also had the condition that 35q^2 must be divisible by 3, hence q must be divisible by 3. Let me set q = 3r, then:

x = 14q^2 = 14*(9r^2) = 126r^2

y = [5x + k]/6, where k = 168q = 168*3r = 504r

So y = [5*126r^2 + 504r]/6 = [630r^2 + 504r]/6 = 105r^2 + 84r.

Therefore, a_2 = 105r^2 + 84r.

Now, let's compute a_3 = y^2 / x = (105r^2 + 84r)^2 / (126r^2).

Let me factor numerator and denominator:

Numerator: (105r^2 + 84r)^2 = [21r(5r + 4)]^2 = 441r^2(5r + 4)^2

Denominator: 126r^2 = 126r^2

Therefore, a_3 = 441r^2(5r + 4)^2 / 126r^2 = 441 / 126 * (5r + 4)^2 = 3.5 * (5r + 4)^2.

Wait, 441 / 126 = 3.5? Let me compute 441 ÷ 126:

126 * 3 = 378, 441 - 378 = 63, so 3 + 63/126 = 3.5. Yes. So a_3 = 3.5 * (5r + 4)^2. But this is not an integer unless (5r + 4)^2 is even, which would require 5r + 4 to be even. Let me check:

5r + 4 is even ⇨ 5r is even - 4 ⇨ 5r is even ⇨ r is even, since 5 is odd. Therefore, r must be even. Let me set r = 2s, where s is a positive integer.

Then, a_1 = 126r^2 = 126*(4s^2) = 504s^2

a_2 = 105r^2 + 84r = 105*(4s^2) + 84*(2s) = 420s^2 + 168s

Now, compute a_3 = 3.5 * (5r + 4)^2 = 3.5 * (5*(2s) + 4)^2 = 3.5 * (10s + 4)^2 = 3.5 * [2(5s + 2)]^2 = 3.5 * 4*(5s + 2)^2 = 14*(5s + 2)^2.

Which is an integer, since 14 is integer and (5s + 2)^2 is integer. Great! So now, a_3 is integer.

Now, let's check if this works. Let me take s = 1:

s = 1:

a_1 = 504*1 = 504

a_2 = 420*1 + 168*1 = 420 + 168 = 588

Check if a_2 > a_1: 588 > 504, yes.

Compute a_3 = 14*(5*1 + 2)^2 = 14*(7)^2 = 14*49 = 686.

Check if a_3 is integer: yes.

Now, check if a_4 is integer. From earlier, a_4 = 2a_3 - a_2 = 2*686 - 588 = 1372 - 588 = 784.

Which is integer. Good.

Now, compute a_5 = a_4^2 / a_3 = 784^2 / 686.

Calculate 784^2: 784 * 784. Let me compute:

784 = 28^2, so 784^2 = (28^2)^2 = 28^4. But maybe better to compute directly:

784 * 784: Let me note that 784 = 700 + 84.

But maybe better to note that 784 / 686 = 784 / 686 = 1.142857... Wait, but let me compute 784^2 / 686:

First, simplify the fraction:

784 / 686 = 784 ÷ 14 = 56; 686 ÷ 14 = 49. So 784 / 686 = 56 / 49 = 8 / 7.

Therefore, 784^2 / 686 = (784 / 686) * 784 = (8/7) * 784 = 8 * 112 = 896.

Yes, because 784 ÷ 7 = 112, so 8 * 112 = 896. So a_5 = 896, which is integer.

Good. Now, a_6 = 2a_5 - a_4 = 2*896 - 784 = 1792 - 784 = 1008.

Integer. Good.

a_7 = a_6^2 / a_5 = 1008^2 / 896.

Simplify: 1008 / 896 = 1.125 = 9/8. So 1008^2 / 896 = (1008 / 896) * 1008 = (9/8) * 1008 = 9 * 126 = 1134.

Yes, since 1008 ÷ 8 = 126. So a_7 = 1134.

Good. a_8 = 2a_7 - a_6 = 2*1134 - 1008 = 2268 - 1008 = 1260.

Integer. Good.

a_9 = a_8^2 / a_7 = 1260^2 / 1134.

Simplify: 1260 / 1134 = 1.111... = 10/9. So 1260^2 / 1134 = (1260 / 1134) * 1260 = (10/9) * 1260 = 10 * 140 = 1400.

Yes, since 1260 ÷ 9 = 140. So a_9 = 1400.

Good. a_10 = 2a_9 - a_8 = 2*1400 - 1260 = 2800 - 1260 = 1540.

Integer. Good.

a_11 = a_10^2 / a_9 = 1540^2 / 1400.

Simplify: 1540 / 1400 = 1.1 = 11/10. So 1540^2 / 1400 = (1540 / 1400) * 1540 = (11/10) * 1540 = 11 * 154 = 1694.

Yes, since 1540 ÷ 10 = 154. So a_11 = 1694.

Good. a_12 = 2a_11 - a_10 = 2*1694 - 1540 = 3388 - 1540 = 1848.

Integer. Good.

a_13 = a_12^2 / a_11 = 1848^2 / 1694.

Let me compute this. First, simplify 1848 / 1694.

Divide numerator and denominator by 2: 924 / 847.

Check if they have common factors. Let me see:

847 ÷ 7 = 121, since 7*121 = 847. 924 ÷ 7 = 132. So 924 / 847 = 132 / 121.

Therefore, 1848^2 / 1694 = (1848 / 1694) * 1848 = (132 / 121) * 1848.

Compute 1848 ÷ 121: 121 * 15 = 1815, 1848 - 1815 = 33, so 15 + 33/121 = 15.2727...

But let me compute it as fractions:

(132 / 121) * 1848 = 132 * 1848 / 121.

Note that 1848 = 121 * 15 + 33 = 121*15 + 33, but maybe better to factor:

1848 = 121 * 15 + 33 = 121*15 + 3*11. Not sure. Alternatively, note that 1848 = 121 * 15.2727... but perhaps there's a better way.

Wait, 132 and 121: 132 = 12 * 11, 121 = 11^2. So:

132 / 121 = (12 * 11) / 11^2 = 12 / 11.

Therefore, 132 / 121 * 1848 = (12 / 11) * 1848 = 12 * (1848 / 11).

Compute 1848 ÷ 11: 11*168 = 1848, since 11*160 = 1760, 11*8 = 88, so 1760 + 88 = 1848. Therefore, 1848 / 11 = 168.

Therefore, a_13 = 12 * 168 = 2016. Which matches the given value! Great!

So with s = 1, we have a_1 = 504, and a_13 = 2016. Therefore, this is a valid solution.

Now, let me check if there are other possible values of s. Earlier, we had to set r = 2s, and s is a positive integer. Let me check for s = 2:

s = 2:

a_1 = 504 * 4 = 2016

a_2 = 420*(4) + 168*(2) = 1680 + 336 = 2016

Wait, but a_2 = 2016, and a_1 = 2016. But the sequence is strictly increasing, so a_2 must be greater than a_1. Here, a_2 = a_1, which is invalid. Therefore, s = 2 is invalid.

Wait, let me check the calculation:

Wait, a_1 = 504s^2, for s = 2, a_1 = 504 * 4 = 2016.

a_2 = 420s^2 + 168s = 420*(4) + 168*(2) = 1680 + 336 = 2016. Yes, so a_2 = a_1, which violates the strictly increasing condition. Therefore, s = 2 is invalid.

Similarly, for s ≥ 2, a_2 = 420s^2 + 168s. Let me check for s = 2, a_2 = 2016, which equals a_1 = 2016. For s = 3:

a_1 = 504 * 9 = 4536

a_2 = 420*9 + 168*3 = 3780 + 504 = 4284

Check if a_2 > a_1: 4284 < 4536? No, 4284 is less than 4536. Wait, this is a problem. Wait, this suggests that for s = 3, a_2 < a_1, which violates the strictly increasing condition. Wait, what's happening here?

Wait, let me recalculate a_2 for s = 3:

a_2 = 420s^2 + 168s = 420*(9) + 168*3 = 3780 + 504 = 4284.

a_1 = 504s^2 = 504*9 = 4536.

So 4284 < 4536, which means a_2 < a_1, which is invalid. Therefore, s = 3 is invalid.

Wait, this suggests that only s = 1 is valid? Let me check for s = 1, a_1 = 504, a_2 = 588 > 504, which is good. For s = 2, a_1 = 2016, a_2 = 2016, invalid. For s = 3, a_2 < a_1, invalid. For s = 0, not positive. Therefore, only s = 1 is valid.

Therefore, the only valid solution is s = 1, leading to a_1 = 504.

But let me check if there are other possible values of r or q that could lead to valid sequences. Earlier, we had to set r = 2s, but maybe there are other constraints?

Wait, let me think. When we set q = 3r, then r had to be even to make a_3 integer, leading to r = 2s. But perhaps there are other ways to satisfy the divisibility conditions? Or maybe not. Since we derived that for a_3 to be integer, r must be even, and then with s = 1 being the only valid value that keeps the sequence strictly increasing.

Therefore, the only possible value is a_1 = 504.

But let me check if there are other possibilities. For example, could there be other values of r that make a_2 > a_1?

Wait, for s = 1, a_1 = 504, a_2 = 588.

For s = 1/2? No, s must be integer.

Alternatively, maybe there are other solutions where the earlier terms are integers, but with different relations? But given the way we derived the relations, and the constraints from the divisibility, it seems that s = 1 is the only valid solution.

Therefore, the answer is a_1 = 504.

But let me verify with the sequence:

We had:

a_1 = 504

a_2 = 588

a_3 = 686

a_4 = 784

a_5 = 896

a_6 = 1008

a_7 = 1134

a_8 = 1260

a_9 = 1400

a_10 = 1540

a_11 = 1694

a_12 = 1848

a_13 = 2016

Yes, each term is strictly increasing, and all terms are integers. Also, the given condition a_13 = 2016 is satisfied. Therefore, this must be the correct value.

Therefore, the answer is 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: Understanding the Structure

Let’s analyze the structure of the sequence based on the given properties.

- The subsequence $ a_{2k-1}, a_{2k}, a_{2k+1} $ is geometric. So:
  $$
  a_{2k}^2 = a_{2k-1} \cdot a_{2k+1}
  $$
- The subsequence $ a_{2k}, a_{2k+1}, a_{2k+2} $ is arithmetic. So:
  $$
  2a_{2k+1} = a_{2k} + a_{2k+2}
  $$

These recursive relationships allow us to express terms in terms of earlier ones. For example, from the geometric condition:
$$
a_{2k+1} = \frac{a_{2k}^2}{a_{2k-1}}
$$
From the arithmetic condition:
$$
a_{2k+2} = 2a_{2k+1} - a_{2k}
$$

This suggests that the sequence can be built recursively from $ a_1 $ and $ a_2 $, and all terms must be positive integers.

---

### Step 2: Expressing $ a_{13} $ in Terms of $ a_1 $ and $ a_2 $

We are told that $ a_{13} = 2016 $. Since 13 is odd, it corresponds to the term $ a_{2k-1} $, where $ k = 7 $. Using the earlier derived formula for odd-indexed terms:

$$
a_{2k-1} = \frac{( (k-1)a_2 - (k-2)a_1 )^2}{a_1}
$$

For $ k = 7 $, this becomes:
$$
a_{13} = \frac{(6a_2 - 5a_1)^2}{a_1}
$$

Given $ a_{13} = 2016 $, we have:
$$
\frac{(6a_2 - 5a_1)^2}{a_1} = 2016
$$

Multiplying both sides by $ a_1 $:
$$
(6a_2 - 5a_1)^2 = 2016a_1
$$

Let’s denote $ x = a_1 $, $ y = a_2 $. Then:
$$
(6y - 5x)^2 = 2016x
$$

---

### Step 3: Solving the Equation

We solve the equation:
$$
(6y - 5x)^2 = 2016x
$$

Let’s denote $ d = 6y - 5x $, so:
$$
d^2 = 2016x \Rightarrow x = \frac{d^2}{2016}
$$

Since $ x $ must be a positive integer, $ d^2 $ must be divisible by 2016. The prime factorization of 2016 is:
$$
2016 = 2^5 \cdot 3^2 \cdot 7
$$

So $ d $ must be a multiple of the square root of the square-free part of 2016. The square-free part is $ 2 \cdot 7 = 14 $, so $ d = 14k $ for some integer $ k $.

Then:
$$
x = \frac{(14k)^2}{2016} = \frac{196k^2}{2016} = \frac{49k^2}{504}
$$

To make $ x $ an integer, $ 504 $ must divide $ 49k^2 $. Since $ \gcd(49, 504) = 7 $, we divide both sides by 7:
$$
x = \frac{7k^2}{72}
$$

So $ 72 $ must divide $ 7k^2 $. Since $ \gcd(7, 72) = 1 $, $ 72 $ must divide $ k^2 $. The square-free part of 72 is 2, so $ k $ must be a multiple of 2. Let $ k = 2m $, then:
$$
x = \frac{7(2m)^2}{72} = \frac{28m^2}{72} = \frac{7m^2}{18}
$$

Now $ 18 $ must divide $ 7m^2 $. Since $ \gcd(7, 18) = 1 $, $ 18 $ must divide $ m^2 $. The square-free part of 18 is 2, so $ m $ must be a multiple of 2. Let $ m = 2n $, then:
$$
x = \frac{7(2n)^2}{18} = \frac{28n^2}{18} = \frac{14n^2}{9}
$$

Now $ 9 $ must divide $ 14n^2 $. Since $ \gcd(14, 9) = 1 $, $ 9 $ must divide $ n^2 $. The square-free part of 9 is 1, so $ n $ must be a multiple of 3. Let $ n = 3p $, then:
$$
x = \frac{14(3p)^2}{9} = \frac{14 \cdot 9p^2}{9} = 14p^2
$$

So:
$$
a_1 = 14p^2
$$

Now, recall that $ d = 14k = 14 \cdot 2m = 14 \cdot 2 \cdot 2n = 14 \cdot 4n = 14 \cdot 4 \cdot 3p = 168p $

Then:
$$
a_2 = \frac{d + 5a_1}{6} = \frac{168p + 5 \cdot 14p^2}{6} = \frac{168p + 70p^2}{6}
$$

Simplify:
$$
a_2 = \frac{70p^2 + 168p}{6} = \frac{35p^2 + 84p}{3}
$$

For $ a_2 $ to be an integer, $ 35p^2 + 84p $ must be divisible by 3. Since $ 35 \equiv 2 \mod 3 $, and $ 84 \equiv 0 \mod 3 $, we have:
$$
35p^2 + 84p \equiv 2p^2 \mod 3
$$

So $ 2p^2 \equiv 0 \mod 3 \Rightarrow p^2 \equiv 0 \mod 3 \Rightarrow p \equiv 0 \mod 3 $

Let $ p = 3q $, then:
$$
a_1 = 14p^2 = 14(9q^2) = 126q^2
$$
$$
a_2 = \frac{35p^2 + 84p}{3} = \frac{35(9q^2) + 84(3q)}{3} = \frac{315q^2 + 252q}{3} = 105q^2 + 84q
$$

Now, we check for the smallest positive integer $ q $ such that the sequence is strictly increasing and all terms are integers.

Try $ q = 1 $:
- $ a_1 = 126 $
- $ a_2 = 105 + 84 = 189 $
- $ a_3 = \frac{a_2^2}{a_1} = \frac{189^2}{126} = \frac{35721}{126} = 283.5 $ → **Not an integer**

Try $ q = 2 $:
- $ a_1 = 126 \cdot 4 = 504 $
- $ a_2 = 105 \cdot 4 + 84 \cdot 2 = 420 + 168 = 588 $
- $ a_3 = \frac{588^2}{504} = \frac{345744}{504} = 686 $ → **Integer**
- $ a_4 = 2a_3 - a_2 = 2 \cdot 686 - 588 = 1372 - 588 = 784 $
- $ a_5 = \frac{a_4^2}{a_3} = \frac{784^2}{686} = \frac{614656}{686} = 896 $
- $ a_6 = 2a_5 - a_4 = 2 \cdot 896 - 784 = 1792 - 784 = 1008 $
- $ a_7 = \frac{a_6^2}{a_5} = \frac{1008^2}{896} = \frac{1016064}{896} = 1134 $
- $ a_8 = 2a_7 - a_6 = 2 \cdot 1134 - 1008 = 2268 - 1008 = 1260 $
- $ a_9 = \frac{a_8^2}{a_7} = \frac{1260^2}{1134} = \frac{1587600}{1134} = 1400 $
- $ a_{10} = 2a_9 - a_8 = 2 \cdot 1400 - 1260 = 2800 - 1260 = 1540 $
- $ a_{11} = \frac{a_{10}^2}{a_9} = \frac{1540^2}{1400} = \frac{2371600}{1400} = 1694 $
- $ a_{12} = 2a_{11} - a_{10} = 2 \cdot 1694 - 1540 = 3388 - 1540 = 1848 $
- $ a_{13} = \frac{a_{12}^2}{a_{11}} = \frac{1848^2}{1694} = \frac{3415104}{1694} = 2016 $

All terms are integers, and the sequence is strictly increasing. Thus, this is a valid solution.

---

### 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 to make sure I understand it correctly.

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 $. The goal is 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 $ a $, $ b $, $ c $ are geometric, then $ b^2 = a \cdot c $. An arithmetic sequence has a common difference, so if three terms $ a $, $ b $, $ c $ are arithmetic, then $ 2b = a + c $.

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 $.

Now, the problem mentions that for every positive integer $ k $, the subsequence $ a_{2k-1} $, $ a_{2k} $, $ a_{2k+1} $ is geometric. Let me note that for each $ k $, this gives us a geometric progression starting at position $ 2k - 1 $, then $ 2k $, then $ 2k + 1 $. Similarly, for each $ k $, the subsequence $ a_{2k} $, $ a_{2k+1} $, $ a_{2k+2} $ is arithmetic.

So, for example, when $ k = 1 $, the first geometric subsequence is $ a_1, a_2, a_3 $, and the first arithmetic subsequence is $ a_2, a_3, a_4 $. Similarly, for $ k = 2 $, the geometric subsequence is $ a_3, a_4, a_5 $, and the arithmetic subsequence is $ a_4, a_5, a_6 $, and so on.

Therefore, each odd-indexed term is part of a geometric progression with the next two terms, and each even-indexed term is part of an arithmetic progression with the next two terms. This seems like there's a recursive relationship here.

Let me try to write down equations based on these properties.

Starting with $ k = 1 $:

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

$ a_2^2 = a_1 \cdot a_3 $. (1)

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

$ 2a_3 = a_2 + a_4 $. (2)

Similarly, for $ k = 2 $:

Geometric sequence $ a_3, a_4, a_5 $:

$ a_4^2 = a_3 \cdot a_5 $. (3)

Arithmetic sequence $ a_4, a_5, a_6 $:

$ 2a_5 = a_4 + a_6 $. (4)

For $ k = 3 $:

Geometric sequence $ a_5, a_6, a_7 $:

$ a_6^2 = a_5 \cdot a_7 $. (5)

Arithmetic sequence $ a_6, a_7, a_8 $:

$ 2a_7 = a_6 + a_8 $. (6)

And so on. It seems like there's a pattern here. Each odd term is part of a geometric progression with the next two terms, and each even term is part of an arithmetic progression with the next two terms.

Since the sequence is strictly increasing, all the terms are positive integers, and each term is larger than the previous. Therefore, the common ratio in the geometric sequences must be greater than 1, and the common difference in the arithmetic sequences must be positive.

Let me try to find a recursive relationship between terms. Let me denote the terms as $ a_1, a_2, a_3, \dots $.

From equation (1): $ a_2^2 = a_1 a_3 $. Therefore, $ a_3 = \frac{a_2^2}{a_1} $. Since $ a_3 $ must be an integer, $ a_1 $ must divide $ a_2^2 $.

From equation (2): $ 2a_3 = a_2 + a_4 $. Therefore, $ a_4 = 2a_3 - a_2 $. Substituting $ a_3 $ from above, we get:

$ a_4 = 2 \cdot \frac{a_2^2}{a_1} - a_2 $.

Similarly, from equation (3): $ a_4^2 = a_3 a_5 $. Therefore, $ a_5 = \frac{a_4^2}{a_3} $.

From equation (4): $ 2a_5 = a_4 + a_6 $, so $ a_6 = 2a_5 - a_4 $.

Continuing this way, perhaps we can find a pattern or recurrence relation.

But since the problem gives us $ a_{13} = 2016 $, maybe we can find expressions for terms up to $ a_{13} $ in terms of $ a_1 $ and $ a_2 $, then set up an equation and solve for $ a_1 $.

But this might get complicated. Let me see if there's a pattern or if the sequence can be expressed in terms of a common ratio or something similar.

Alternatively, maybe there's a way to express each term in terms of the previous ones, leading to a recursive formula.

Alternatively, perhaps there's a relation between consecutive terms. Let me try to find relations between terms.

Let me consider the first few terms:

From equation (1): $ a_3 = \frac{a_2^2}{a_1} $.

From equation (2): $ a_4 = 2a_3 - a_2 = 2 \cdot \frac{a_2^2}{a_1} - a_2 $.

From equation (3): $ a_5 = \frac{a_4^2}{a_3} $.

Let me compute $ a_5 $:

First, compute $ a_4 $:

$ a_4 = \frac{2a_2^2}{a_1} - a_2 = \frac{2a_2^2 - a_1 a_2}{a_1} = \frac{a_2(2a_2 - a_1)}{a_1} $.

Then $ a_5 = \frac{a_4^2}{a_3} = \frac{ \left( \frac{a_2(2a_2 - a_1)}{a_1} \right)^2 }{ \frac{a_2^2}{a_1} } = \frac{ a_2^2 (2a_2 - a_1)^2 }{ a_1^2 } \cdot \frac{a_1}{a_2^2} } = \frac{(2a_2 - a_1)^2}{a_1} $.

So $ a_5 = \frac{(2a_2 - a_1)^2}{a_1} $.

Similarly, from equation (4): $ a_6 = 2a_5 - a_4 $.

Let me compute $ a_6 $:

First, $ 2a_5 = 2 \cdot \frac{(2a_2 - a_1)^2}{a_1} $.

Then subtract $ a_4 $:

$ a_6 = \frac{2(2a_2 - a_1)^2}{a_1} - \frac{a_2(2a_2 - a_1)}{a_1} = \frac{2(2a_2 - a_1)^2 - a_2(2a_2 - a_1)}{a_1} $.

Factor out $ (2a_2 - a_1) $:

$ a_6 = \frac{(2a_2 - a_1)[2(2a_2 - a_1) - a_2]}{a_1} $.

Simplify the bracket:

$ 2(2a_2 - a_1) - a_2 = 4a_2 - 2a_1 - a_2 = 3a_2 - 2a_1 $.

Therefore, $ a_6 = \frac{(2a_2 - a_1)(3a_2 - 2a_1)}{a_1} $.

Now, moving on to equation (5): $ a_6^2 = a_5 a_7 $, so $ a_7 = \frac{a_6^2}{a_5} $.

Let me compute $ a_7 $:

First, compute $ a_6^2 $:

$ \left( \frac{(2a_2 - a_1)(3a_2 - 2a_1)}{a_1} \right)^2 $.

Then divide by $ a_5 = \frac{(2a_2 - a_1)^2}{a_1} $:

So $ a_7 = \frac{ \left( (2a_2 - a_1)^2 (3a_2 - 2a_1)^2 \right) / a_1^2 }{ (2a_2 - a_1)^2 / a_1 } } = \frac{(3a_2 - 2a_1)^2}{a_1} $.

Therefore, $ a_7 = \frac{(3a_2 - 2a_1)^2}{a_1} $.

Similarly, from equation (6): $ 2a_7 = a_6 + a_8 $, so $ a_8 = 2a_7 - a_6 $.

Let me compute $ a_8 $:

First, $ 2a_7 = 2 \cdot \frac{(3a_2 - 2a_1)^2}{a_1} $.

Subtract $ a_6 $:

$ a_8 = \frac{2(3a_2 - 2a_1)^2}{a_1} - \frac{(2a_2 - a_1)(3a_2 - 2a_1)}{a_1} $.

Factor out $ \frac{(3a_2 - 2a_1)}{a_1} $:

$ a_8 = \frac{(3a_2 - 2a_1)}{a_1} [ 2(3a_2 - 2a_1) - (2a_2 - a_1) ] $.

Compute the bracket:

$ 2(3a_2 - 2a_1) - (2a_2 - a_1) = 6a_2 - 4a_1 - 2a_2 + a_1 = 4a_2 - 3a_1 $.

Therefore, $ a_8 = \frac{(3a_2 - 2a_1)(4a_2 - 3a_1)}{a_1} $.

Hmm, interesting. So there seems to be a pattern here. Let me try to see if I can find a pattern in terms of coefficients.

Looking at the expressions for $ a_3, a_5, a_7 $, they are of the form $ \frac{(m a_2 - n a_1)^2}{a_1} $, where m and n seem to follow some pattern.

Similarly, the even terms $ a_4, a_6, a_8 $ are products of terms like $ (m a_2 - n a_1) $ divided by $ a_1 $.

Let me try to see if there's a recursive pattern here. Let me note the coefficients:

For $ a_3 $: numerator is $ a_2^2 $, denominator $ a_1 $.

For $ a_5 $: numerator is $ (2a_2 - a_1)^2 $, denominator $ a_1 $.

For $ a_7 $: numerator is $ (3a_2 - 2a_1)^2 $, denominator $ a_1 $.

Similarly, for $ a_4 $: numerator is $ a_2(2a_2 - a_1) $, denominator $ a_1 $.

For $ a_6 $: numerator is $ (2a_2 - a_1)(3a_2 - 2a_1) $, denominator $ a_1 $.

For $ a_8 $: numerator is $ (3a_2 - 2a_1)(4a_2 - 3a_1) $, denominator $ a_1 $.

So, if I denote $ b_k = k a_2 - (k - 1)a_1 $, then:

- $ a_3 = \frac{(1 a_2 - 0 a_1)^2}{a_1} = \frac{a_2^2}{a_1} $
- $ a_5 = \frac{(2 a_2 - 1 a_1)^2}{a_1} $
- $ a_7 = \frac{(3 a_2 - 2 a_1)^2}{a_1} $
- Similarly, $ a_4 = \frac{a_2 (2a_2 - a_1)}{a_1} $
- $ a_6 = \frac{(2a_2 - a_1)(3a_2 - 2a_1)}{a_1} $
- $ a_8 = \frac{(3a_2 - 2a_1)(4a_2 - 3a_1)}{a_1} $

So, it seems like for odd indices $ a_{2k+1} $, the term is $ \frac{(k a_2 - (k - 1)a_1)^2}{a_1} $, and for even indices $ a_{2k} $, the term is $ \frac{( (k - 1)a_2 - (k - 2)a_1 ) (k a_2 - (k - 1)a_1 ) }{a_1} $.

Wait, let me check:

For $ a_4 $, which is $ k = 2 $, since $ a_{2k} $, so $ k = 2 $, then:

$ \frac{( (2 - 1)a_2 - (2 - 2)a_1 ) (2 a_2 - (2 - 1)a_1 ) }{a_1} = \frac{(a_2 - 0 a_1)(2a_2 - a_1)}{a_1} = \frac{a_2(2a_2 - a_1)}{a_1} $, which matches.

Similarly, for $ a_6 $, $ k = 3 $:

$ \frac{( (3 - 1)a_2 - (3 - 2)a_1 ) (3 a_2 - (3 - 1)a_1 ) }{a_1} = \frac{(2a_2 - a_1)(3a_2 - 2a_1)}{a_1} $, which matches.

Therefore, generalizing:

For each $ k \geq 1 $:

- $ a_{2k - 1} = \frac{( (k - 1)a_2 - (k - 2)a_1 )^2 }{a_1} $ ?

Wait, no. Wait for $ a_3 $, which is $ k = 2 $, since $ a_{2k - 1} = a_3 $ when $ k = 2 $. Let me check:

If $ a_{2k - 1} = \frac{( (k - 1)a_2 - (k - 2)a_1 )^2 }{a_1} $, then for $ k = 2 $, it would be $ \frac{(1 a_2 - 0 a_1)^2}{a_1} = \frac{a_2^2}{a_1} $, which is correct. For $ k = 3 $, $ a_5 = \frac{(2a_2 - 1a_1)^2}{a_1} $, which is correct. Similarly, for $ k = 4 $, $ a_7 = \frac{(3a_2 - 2a_1)^2}{a_1} $, which is correct. So yes, the general formula for $ a_{2k - 1} $ is $ \frac{( (k - 1)a_2 - (k - 2)a_1 )^2 }{a_1} $.

Similarly, for even terms $ a_{2k} $, the formula is $ \frac{ ( (k - 1)a_2 - (k - 2)a_1 ) (k a_2 - (k - 1)a_1 ) }{a_1} $.

Therefore, if I can find a general formula for $ a_n $, maybe I can express $ a_{13} $ in terms of $ a_1 $ and $ a_2 $, then solve for $ a_1 $.

But first, let me check if this pattern continues. Let me compute $ a_9 $, for example.

From equation (7): For $ k = 4 $, the geometric sequence $ a_7, a_8, a_9 $:

So $ a_8^2 = a_7 a_9 $, so $ a_9 = \frac{a_8^2}{a_7} $.

From previous expressions:

$ a_7 = \frac{(3a_2 - 2a_1)^2}{a_1} $

$ a_8 = \frac{(3a_2 - 2a_1)(4a_2 - 3a_1)}{a_1} $

Therefore, $ a_9 = \frac{ [ (3a_2 - 2a_1)(4a_2 - 3a_1) / a_1 ]^2 }{ (3a_2 - 2a_1)^2 / a_1 } } = \frac{ (4a_2 - 3a_1)^2 }{a_1 } $.

Which fits the pattern: for $ a_{2k - 1} $, when $ k = 5 $, $ a_9 = \frac{(4a_2 - 3a_1)^2}{a_1} $, which is consistent with the formula $ \frac{( (k - 1)a_2 - (k - 2)a_1 )^2 }{a_1} $, since $ k = 5 $, $ (5 - 1)a_2 - (5 - 2)a_1 = 4a_2 - 3a_1 $.

Similarly, $ a_{10} $ would be $ 2a_9 - a_8 $, which would be:

$ 2 \cdot \frac{(4a_2 - 3a_1)^2}{a_1} - \frac{(3a_2 - 2a_1)(4a_2 - 3a_1)}{a_1} = \frac{2(4a_2 - 3a_1)^2 - (3a_2 - 2a_1)(4a_2 - 3a_1)}{a_1} $.

Factor out $ (4a_2 - 3a_1) $:

$ \frac{(4a_2 - 3a_1)[2(4a_2 - 3a_1) - (3a_2 - 2a_1)]}{a_1} $.

Compute the bracket:

$ 2(4a_2 - 3a_1) - (3a_2 - 2a_1) = 8a_2 - 6a_1 - 3a_2 + 2a_1 = 5a_2 - 4a_1 $.

Therefore, $ a_{10} = \frac{(4a_2 - 3a_1)(5a_2 - 4a_1)}{a_1} $, which fits the pattern for even terms.

So, the pattern seems consistent. Therefore, we can generalize:

For each $ n $, if $ n $ is odd, say $ n = 2k - 1 $, then:

$ a_n = \frac{( (k - 1)a_2 - (k - 2)a_1 )^2 }{a_1} $.

If $ n $ is even, say $ n = 2k $, then:

$ a_n = \frac{ ( (k - 1)a_2 - (k - 2)a_1 ) (k a_2 - (k - 1)a_1 ) }{a_1} $.

Alternatively, if we let $ m = k - 1 $, then for odd indices $ n = 2m + 1 $:

$ a_{2m + 1} = \frac{(m a_2 - (m - 1)a_1)^2}{a_1} $.

Similarly, for even indices $ n = 2m $:

$ a_{2m} = \frac{( (m - 1)a_2 - (m - 2)a_1 ) (m a_2 - (m - 1)a_1 ) }{a_1} $.

Therefore, if we can express $ a_{13} $ in terms of $ a_1 $ and $ a_2 $, then we can set up an equation.

Since 13 is odd, $ n = 13 = 2m + 1 $, so $ m = 6 $. Therefore:

$ a_{13} = \frac{(6 a_2 - 5 a_1)^2}{a_1} $.

Given that $ a_{13} = 2016 $, we have:

$ \frac{(6 a_2 - 5 a_1)^2}{a_1} = 2016 $.

So, $ (6 a_2 - 5 a_1)^2 = 2016 a_1 $.

Therefore, $ 2016 a_1 $ must be a perfect square. Since $ a_1 $ is a positive integer, and 2016 is given, we can factor 2016 to see what conditions this imposes on $ a_1 $.

First, let me factor 2016:

2016 ÷ 2 = 1008

1008 ÷ 2 = 504

504 ÷ 2 = 252

252 ÷ 2 = 126

126 ÷ 2 = 63

63 ÷ 3 = 21

21 ÷ 3 = 7

7 ÷ 7 = 1

So, prime factors of 2016 are $ 2^5 \times 3^2 \times 7^1 $.

Therefore, 2016 = $ 2^5 \times 3^2 \times 7 $.

So, $ 2016 a_1 = 2^5 \times 3^2 \times 7 \times a_1 $ must be a perfect square.

For a number to be a perfect square, all exponents in its prime factorization must be even. Therefore, the exponents of primes in $ 2016 a_1 $ must be even.

Let me denote the prime factors of $ a_1 $. Let me suppose that $ a_1 $ has prime factors that can make the exponents of 2, 3, 7, and any other primes in $ a_1 $ even.

Let me write $ a_1 $ as $ 2^{x} \times 3^{y} \times 7^{z} \times \text{other primes} $.

Then, the exponents in $ 2016 a_1 $ would be:

- For prime 2: 5 + x
- For prime 3: 2 + y
- For prime 7: 1 + z
- For other primes: whatever they are in $ a_1 $, but since $ a_1 $ is multiplied by 2016, which doesn't have other primes, so other primes must be present in $ a_1 $ with even exponents.

Therefore, for $ 2016 a_1 $ to be a perfect square:

- 5 + x must be even ⇒ x is odd (since 5 is odd)
- 2 + y must be even ⇒ y is even
- 1 + z must be even ⇒ z is odd
- Any other primes in $ a_1 $ must have even exponents.

Therefore, $ a_1 $ must be of the form $ 2^{odd} \times 3^{even} \times 7^{odd} \times (\text{square}) $.

But since $ a_1 $ is a positive integer, and the sequence is strictly increasing, we need to find such $ a_1 $ and $ a_2 $ that satisfy the equations and the sequence is strictly increasing.

But we also have another equation from the previous relation: $ (6 a_2 - 5 a_1)^2 = 2016 a_1 $.

Let me denote $ 6 a_2 - 5 a_1 = \sqrt{2016 a_1} $. However, since $ a_1 $ and $ a_2 $ are positive integers, $ 6 a_2 - 5 a_1 $ must also be a positive integer. Therefore, $ 6 a_2 - 5 a_1 > 0 $, so $ a_2 > \frac{5}{6} a_1 $. Since $ a_2 > a_1 $, this is automatically satisfied because $ a_2 > a_1 $, so $ a_2 > a_1 > \frac{5}{6} a_1 $.

But let me think about how to relate $ a_2 $ and $ a_1 $. Let me denote $ 6 a_2 - 5 a_1 = d $, where $ d $ is a positive integer. Then, $ d^2 = 2016 a_1 $, so $ a_1 = \frac{d^2}{2016} $.

Since $ a_1 $ must be a positive integer, $ 2016 $ must divide $ d^2 $. Therefore, $ d $ must be a multiple of the square root of the square-free part of 2016.

First, let me find the square-free part of 2016. The prime factorization is $ 2^5 \times 3^2 \times 7 $. The square-free part is the product of primes with odd exponents. Here, 2 has exponent 5 (odd), 3 has exponent 2 (even), 7 has exponent 1 (odd). Therefore, square-free part is $ 2 \times 7 = 14 $. Therefore, $ d $ must be a multiple of 14. Let me check:

If $ d = 14 k $, then $ d^2 = 196 k^2 $, so $ a_1 = \frac{196 k^2}{2016} = \frac{49 k^2}{504} = \frac{7 k^2}{72} $. Wait, but this must be integer. Therefore, 72 divides 7 k^2. Since 7 and 72 are coprime, 72 divides k^2. Therefore, k must be a multiple of the square root of 72's square-free part. Wait, 72 factors into $ 2^3 \times 3^2 $. The square-free part is 2. Therefore, k must be a multiple of 2. Let me write k = 2 m. Then:

$ a_1 = \frac{7 (2m)^2 }{72} = \frac{7 \times 4 m^2 }{72} = \frac{28 m^2}{72} = \frac{7 m^2}{18} $.

Again, 18 divides 7 m^2. Since 7 and 18 are coprime, 18 divides m^2. Therefore, m must be a multiple of the square root of the square-free part of 18. 18 factors into $ 2 \times 3^2 $, square-free part is 2. Therefore, m must be a multiple of 2. Let me set m = 2 n. Then:

$ a_1 = \frac{7 (2n)^2 }{18} = \frac{7 \times 4 n^2 }{18} = \frac{28 n^2 }{18} = \frac{14 n^2 }{9} $.

Now, 9 divides 14 n^2. Since 14 and 9 are coprime, 9 divides n^2. Therefore, n must be a multiple of 3. Let me set n = 3 p. Then:

$ a_1 = \frac{14 (3p)^2 }{9} = \frac{14 \times 9 p^2 }{9} = 14 p^2 $.

Therefore, $ a_1 = 14 p^2 $, where p is a positive integer. Therefore, the minimal value of $ a_1 $ is 14 when p = 1, then 14*4=56 when p=2, etc.

But we need to find the value of $ a_1 $, so we need more constraints. Let me recall that $ d = 14 k = 14 \times 2 m = 14 \times 2 \times 2 n = 14 \times 4 n = 14 \times 4 \times 3 p = 14 \times 12 p $. Wait, let me retrace:

Earlier steps:

We had $ d = 14 k $, then k = 2 m, then m = 2 n, then n = 3 p. Therefore:

k = 2 m = 2*(2n) = 4n

Then, d = 14 * 4n = 56n

But then, since n = 3p, d = 56 * 3p = 168p.

But also, from $ a_1 = 14 p^2 $, and $ d = 6 a_2 - 5 a_1 $, so:

$ d = 6 a_2 - 5 a_1 \Rightarrow 6 a_2 = d + 5 a_1 \Rightarrow a_2 = \frac{d + 5 a_1}{6} $.

Substituting $ a_1 = 14 p^2 $ and $ d = 168 p $:

$ a_2 = \frac{168 p + 5 \times 14 p^2 }{6} = \frac{168 p + 70 p^2 }{6} = \frac{70 p^2 + 168 p }{6} $.

Simplify:

Divide numerator and denominator by 2:

$ \frac{35 p^2 + 84 p }{3} $.

Therefore, $ a_2 = \frac{35 p^2 + 84 p }{3} $.

Since $ a_2 $ must be an integer, the numerator must be divisible by 3. Let me check:

$ 35 p^2 + 84 p = 35 p^2 + 84 p $.

Note that 84 is divisible by 3 (84 ÷ 3 = 28), and 35 ≡ 2 mod 3. Therefore:

$ 35 p^2 + 84 p ≡ 2 p^2 + 0 ≡ 2 p^2 mod 3 $.

Therefore, for this to be divisible by 3, we need $ 2 p^2 ≡ 0 mod 3 $, which implies $ p^2 ≡ 0 mod 3 $, so $ p ≡ 0 mod 3 $. Therefore, p must be a multiple of 3. Let me set p = 3 q, where q is a positive integer.

Therefore, substituting p = 3 q:

$ a_1 = 14 p^2 = 14 (9 q^2 ) = 126 q^2 $.

$ a_2 = \frac{35 (9 q^2 ) + 84 (3 q ) }{3} = \frac{315 q^2 + 252 q }{3} = 105 q^2 + 84 q $.

Therefore, $ a_2 = 105 q^2 + 84 q $.

Now, since the sequence is strictly increasing, we need to ensure that $ a_2 > a_1 $. Let me check:

$ a_2 - a_1 = 105 q^2 + 84 q - 126 q^2 = -21 q^2 + 84 q = -21 q (q - 4) $.

For this to be positive (since $ a_2 > a_1 $):

$ -21 q (q - 4) > 0 $.

Since q is a positive integer, q ≥ 1.

Let me analyze the sign:

- For q = 1: -21(1)(1 - 4) = -21(-3) = 63 > 0. Okay.
- For q = 2: -21(2)(2 - 4) = -21(2)(-2) = 84 > 0.
- For q = 3: -21(3)(3 - 4) = -21(3)(-1) = 63 > 0.
- For q = 4: -21(4)(4 - 4) = 0. Not positive.
- For q ≥ 5: -21 q (q - 4) < 0, since q - 4 ≥ 1, so negative times negative? Wait:

Wait, let me re-express:

The expression is -21 q (q - 4). Let me think of it as:

If q - 4 is positive (i.e., q ≥ 5), then the entire expression is -21 q (positive) = negative. If q - 4 is negative (i.e., q ≤ 3), then the entire expression is -21 q (negative) = positive. If q = 4, it's zero.

Therefore, for q = 1, 2, 3, the difference is positive, so $ a_2 > a_1 $. For q ≥ 4, the difference is non-positive, which would violate the strictly increasing condition. Therefore, q can only be 1, 2, or 3.

Therefore, possible values of q are 1, 2, 3. Let me check each case.

First, let me note that we also need to ensure that all terms in the sequence are positive integers and that the sequence is strictly increasing. However, since we derived $ a_1 $ and $ a_2 $ based on the condition that $ a_{13} = 2016 $, and the sequence is strictly increasing, we need to check if for these values of q, the rest of the terms are integers and the sequence remains strictly increasing.

But since we have already ensured that $ a_1 $ and $ a_2 $ are integers and $ a_2 > a_1 $, and given the recursive relations, perhaps the rest of the terms will automatically be integers and strictly increasing? Let me check for q = 1, 2, 3.

But first, let me note that $ a_1 = 126 q^2 $, and $ a_{13} = 2016 $. Let me check for each q:

First, let me recall that $ a_{13} = \frac{(6 a_2 - 5 a_1)^2}{a_1} = 2016 $, which we already used to derive the relations. Therefore, for each q, we can compute $ a_1 $, $ a_2 $, and check if the rest of the terms are integers and the sequence is strictly increasing. However, since the problem states that the sequence is strictly increasing and consists of positive integers, and we have already derived $ a_1 $ and $ a_2 $ based on the given $ a_{13} $, perhaps there is only one valid q that makes all terms integers and the sequence strictly increasing. However, since we have three possible q's, we need to check which one(s) are valid.

Alternatively, maybe there is only one possible value of q that leads to the sequence being strictly increasing beyond the first few terms. Let me check for each q.

First, let me compute for q = 1:

Case 1: q = 1

Then:

$ a_1 = 126 * 1^2 = 126 $

$ a_2 = 105 * 1^2 + 84 * 1 = 105 + 84 = 189 $

Check if $ a_2 > a_1 $: 189 > 126, yes.

Now, let me compute the rest of the terms up to $ a_{13} $ to check if they are integers and strictly increasing. However, since we already have $ a_{13} = 2016 $, maybe we can check if with these values, $ a_{13} $ is indeed 2016.

But we already derived that $ a_{13} = 2016 $, so that's already satisfied. However, we need to ensure that all previous terms are integers and the sequence is strictly increasing. Let me check some terms.

First, compute $ a_3 $:

From earlier, $ a_3 = \frac{a_2^2}{a_1} = \frac{189^2}{126} $.

Calculate 189^2: 189*189. Let me compute:

First, 200^2 = 40000, subtract 11*200 + 11^2? Wait, maybe better to compute directly:

189 * 189:

= (200 - 11)^2 = 200^2 - 2*200*11 + 11^2 = 40000 - 4400 + 121 = 40000 - 4400 = 35600; 35600 + 121 = 35721.

Therefore, $ a_3 = 35721 / 126 $.

Let me divide 35721 by 126:

First, divide numerator and denominator by 3: 35721 ÷ 3 = 11907; 126 ÷ 3 = 42.

So 11907 / 42. Divide numerator and denominator by 3 again: 11907 ÷ 3 = 3969; 42 ÷ 3 = 14.

So 3969 / 14. Let me divide 3969 by 14:

14 * 283 = 14 * 280 + 14 * 3 = 3920 + 42 = 3962. Then 3969 - 3962 = 7. Therefore, 3969 / 14 = 283 + 7/14 = 283.5. Wait, but this is not an integer. But this contradicts our earlier assumption that all terms are integers. Therefore, there must be a mistake here.

Wait, but according to our previous derivation, $ a_3 = \frac{a_2^2}{a_1} $, and since $ a_1 = 126 $, $ a_2 = 189 $, which is 189^2 / 126. However, 189 is divisible by 3, 126 is divisible by 3, but maybe there's more?

Wait, 189 = 3^3 * 7, and 126 = 2 * 3^2 * 7. Therefore, 189^2 / 126 = (3^6 * 7^2) / (2 * 3^2 * 7) ) = 3^4 * 7 / 2 = 81 * 7 / 2 = 567 / 2 = 283.5. Which is not an integer. But this contradicts the requirement that all terms are integers. Therefore, there's a problem here. Which suggests that my earlier reasoning is flawed.

Wait, but how? Earlier, I thought that $ a_1 = 14 p^2 $, and with p = 3 q, so for q = 1, p = 3, so a_1 = 14 * 9 = 126. But according to the formula for $ a_3 $, which is $ a_2^2 / a_1 $, this is not an integer. Therefore, there must be an error in my derivation.

Wait, let me check back. Earlier, I had:

From $ (6 a_2 - 5 a_1)^2 = 2016 a_1 $, and then I set $ d = 6 a_2 - 5 a_1 $, so $ d^2 = 2016 a_1 $, hence $ a_1 = d^2 / 2016 $. Then I tried to find d such that a_1 is integer. Then I considered the prime factors and found that d must be multiple of 14, etc., leading to a_1 = 14 p^2. However, in this case, with q = 1, p = 3, a_1 = 126, but then a_3 is not integer. Therefore, there's a mistake here.

Wait, but why? Because even though $ a_1 $ is integer, the term $ a_3 = a_2^2 / a_1 $ may not be integer. Therefore, my previous assumption that $ a_1 = 14 p^2 $ is not sufficient? Because even though $ d^2 = 2016 a_1 $, which ensures that $ a_1 $ divides $ d^2 $, but in addition, $ a_1 $ must divide $ a_2^2 $, since $ a_3 = a_2^2 / a_1 $. Therefore, there's another condition here. Therefore, my mistake was that I only considered the condition for $ a_{13} $, but not for the previous terms. Therefore, my previous approach is incomplete.

Therefore, I need to ensure that not only $ a_1 $ divides $ d^2 $, but also that $ a_1 $ divides $ a_2^2 $, $ a_3 $ divides $ a_4^2 $, etc. However, since the sequence is defined recursively, maybe there's a way to ensure that all terms are integers. However, in my case, with q = 1, a_3 is not integer, which suggests that my earlier approach is missing some constraints.

Therefore, I need to revisit my earlier steps. Let me think again.

We had:

From the geometric sequence $ a_1, a_2, a_3 $, so $ a_3 = a_2^2 / a_1 $. Therefore, $ a_1 $ must divide $ a_2^2 $. Similarly, from the arithmetic sequence $ a_2, a_3, a_4 $, we have $ a_4 = 2a_3 - a_2 $. Since $ a_3 = a_2^2 / a_1 $, then $ a_4 = 2(a_2^2 / a_1) - a_2 $. Therefore, $ a_4 $ is integer only if $ a_1 $ divides $ 2a_2^2 - a_1 a_2 $, which is $ a_1 $ divides $ a_2(2a_2 - a_1) $. Since $ a_1 $ divides $ a_2^2 $, let me denote $ a_2 = k a_1 $, where k is a rational number. Wait, but since $ a_1 $ divides $ a_2^2 $, $ a_2 $ must be a multiple of some factors of $ a_1 $. However, since $ a_2 > a_1 $, and the sequence is strictly increasing, k must be greater than 1.

Alternatively, perhaps I need to find $ a_1 $ and $ a_2 $ such that $ a_1 $ divides $ a_2^2 $, and $ a_1 $ divides $ a_2(2a_2 - a_1) $, etc. But this seems complex.

Alternatively, maybe there's a relation between $ a_1 $ and $ a_2 $ that can be derived from the recursive relations. Let me think.

From earlier, we had:

For $ a_3 = \frac{a_2^2}{a_1} $, and $ a_4 = 2a_3 - a_2 = \frac{2a_2^2}{a_1} - a_2 = \frac{2a_2^2 - a_1 a_2}{a_1} = \frac{a_2(2a_2 - a_1)}{a_1} $.

Similarly, for $ a_5 = \frac{a_4^2}{a_3} = \frac{ [a_2(2a_2 - a_1)/a_1]^2 }{a_2^2 / a_1} } = \frac{a_2^2 (2a_2 - a_1)^2 / a_1^2 }{a_2^2 / a_1} } = \frac{(2a_2 - a_1)^2}{a_1} $.

Similarly, $ a_6 = 2a_5 - a_4 = 2*(2a_2 - a_1)^2 / a_1 - a_2(2a_2 - a_1)/a_1 = [2(2a_2 - a_1)^2 - a_2(2a_2 - a_1)] / a_1 = (2a_2 - a_1)[2(2a_2 - a_1) - a_2] / a_1 = (2a_2 - a_1)(3a_2 - 2a_1)/a_1 $.

Similarly, $ a_7 = \frac{a_6^2}{a_5} = \frac{[(2a_2 - a_1)(3a_2 - 2a_1)/a_1]^2}{(2a_2 - a_1)^2 / a_1} } = \frac{(3a_2 - 2a_1)^2}{a_1} $.

Continuing this way, we can see that each term is defined in terms of $ a_1 $ and $ a_2 $, and for all these terms to be integers, certain divisibility conditions must hold.

In particular, for $ a_3 $ to be integer, $ a_1 $ divides $ a_2^2 $.

For $ a_4 $ to be integer, $ a_1 $ divides $ a_2(2a_2 - a_1) $.

For $ a_5 $ to be integer, $ a_1 $ divides $ (2a_2 - a_1)^2 $.

For $ a_6 $ to be integer, $ a_1 $ divides $ (2a_2 - a_1)(3a_2 - 2a_1) $.

And so on.

Therefore, there are multiple divisibility conditions. However, since we already have $ a_1 $ and $ a_2 $ related through $ a_{13} = 2016 $, perhaps we can find relations that connect these divisibility conditions.

Alternatively, maybe there's a pattern in the terms that can be expressed as linear combinations of $ a_1 $ and $ a_2 $, leading to a relation between $ a_1 $ and $ a_2 $.

Alternatively, perhaps there's a way to express $ a_n $ in terms of $ a_1 $ and $ a_2 $, and then find a relation for $ a_{13} $.

But given that earlier, with q = 1, we get a non-integer $ a_3 $, which suggests that my previous approach is missing something. Let me check where I went wrong.

Earlier, I derived that $ a_1 = 14 p^2 $, with p = 3 q, leading to q = 1, 2, 3. However, this led to a non-integer $ a_3 $. Therefore, there must be additional constraints on $ a_1 $ and $ a_2 $.

Let me think again about the equation $ (6 a_2 - 5 a_1)^2 = 2016 a_1 $. Let me denote $ x = a_1 $, $ y = a_2 $. Then:

$ (6y - 5x)^2 = 2016x $.

Let me expand the left side:

$ 36y^2 - 60xy + 25x^2 = 2016x $.

Rearranged:

$ 36y^2 - 60xy + 25x^2 - 2016x = 0 $.

This is a quadratic equation in terms of y:

$ 36y^2 - 60x y + (25x^2 - 2016x) = 0 $.

Let me solve for y:

Using quadratic formula:

$ y = [60x ± sqrt( (60x)^2 - 4 * 36 * (25x^2 - 2016x) ) ] / (2 * 36) $.

Compute discriminant D:

D = (60x)^2 - 4 * 36 * (25x^2 - 2016x)

= 3600x^2 - 144*(25x^2 - 2016x)

= 3600x^2 - 3600x^2 + 144*2016x

= 0 + 144*2016x

= 144 * 2016 * x.

Therefore, sqrt(D) = sqrt(144 * 2016 * x) = 12 * sqrt(2016x).

Therefore, y = [60x ± 12√(2016x)] / 72 = [5x ± √(2016x)] / 6.

Since y must be positive, we take the positive root:

y = [5x + √(2016x)] / 6.

But since y must be rational (as it's an integer), √(2016x) must be rational. Since 2016x is a perfect square, as we had earlier. Let me denote √(2016x) = k, where k is a positive integer. Then:

k^2 = 2016x ⇒ x = k^2 / 2016.

Since x must be a positive integer, 2016 divides k^2. As before, this implies that k must be a multiple of the square-free part of 2016, which is 14, so k = 14m, where m is a positive integer. Then:

x = (14m)^2 / 2016 = 196m^2 / 2016 = 49m^2 / 504 = 7m^2 / 72.

Therefore, x = 7m^2 / 72 must be integer. Hence, 72 divides 7m^2. Since 7 and 72 are coprime, 72 divides m^2. Therefore, m must be a multiple of the square root of the square-free part of 72. The prime factorization of 72 is 2^3 * 3^2. The square-free part is 2. Therefore, m must be a multiple of 2. Let me set m = 2n, where n is a positive integer. Then:

x = 7*(2n)^2 / 72 = 7*4n^2 / 72 = 28n^2 / 72 = 7n^2 / 18.

Therefore, x = 7n^2 / 18 must be integer. Hence, 18 divides 7n^2. Since 7 and 18 are coprime, 18 divides n^2. Therefore, n must be a multiple of the square root of the square-free part of 18. The square-free part of 18 is 2, so n must be a multiple of 2. Let me set n = 2p, where p is a positive integer. Then:

x = 7*(2p)^2 / 18 = 7*4p^2 / 18 = 28p^2 / 18 = 14p^2 / 9.

Therefore, x = 14p^2 / 9 must be integer. Hence, 9 divides 14p^2. Since 14 and 9 are coprime, 9 divides p^2. Therefore, p must be a multiple of 3. Let me set p = 3q, where q is a positive integer. Then:

x = 14*(3q)^2 / 9 = 14*9q^2 / 9 = 14q^2.

Therefore, x = a_1 = 14q^2, and k = 14m = 14*(2n) = 14*(2*(2p)) = 14*4p = 14*4*3q = 168q.

Wait, let me retrace:

Earlier, we had:

k = 14m, m = 2n, n = 2p, p = 3q.

Wait, no:

Wait, initially, k = 14m, where m = 2n, then n = 2p, then p = 3q. So:

m = 2n = 2*(2p) = 4p

n = 2p = 2*(3q) = 6q

p = 3q

Therefore, k = 14m = 14*4p = 14*4*3q = 168q.

Then, x = a_1 = 14q^2.

Then, y = [5x + k]/6 = [5*(14q^2) + 168q]/6 = [70q^2 + 168q]/6 = [70q^2]/6 + 168q/6 = (35q^2)/3 + 28q.

Wait, but this must be integer. Therefore, 35q^2 must be divisible by 3. Since 35 and 3 are coprime, this implies that q^2 must be divisible by 3, hence q must be divisible by 3. Let me set q = 3r, where r is a positive integer. Then:

a_1 = 14q^2 = 14*(9r^2) = 126r^2.

y = [5x + k]/6 = [5*14q^2 + 168q]/6 = [70q^2 + 168q]/6 = [70*(9r^2) + 168*(3r)] / 6 = [630r^2 + 504r]/6 = 105r^2 + 84r.

Therefore, a_2 = 105r^2 + 84r.

Now, this is the same as before, with q = r. Earlier, I had set q = 1, 2, 3, but now with this substitution, q = r, and we need to ensure that a_3 is integer.

Let me check for r = 1:

a_1 = 126*1^2 = 126

a_2 = 105*1^2 + 84*1 = 105 + 84 = 189

Then, a_3 = a_2^2 / a_1 = 189^2 / 126 = 35721 / 126 = 283.5, which is not integer. So this is invalid.

Wait, but earlier, we derived that a_1 = 14q^2 and a_2 = 105q^2 + 84q, but this leads to a non-integer a_3. What's wrong here?

Wait, but earlier, we had that k = 168q, and x = 14q^2, and y = [5x + k]/6 = [70q^2 + 168q]/6 = 35q^2/3 + 28q. For this to be integer, 35q^2 must be divisible by 3, hence q must be divisible by 3. So we set q = 3r, leading to a_1 = 126r^2 and a_2 = 105*(9r^2) + 84*(3r)? Wait, no:

Wait, no, when we set q = 3r, then:

Wait, earlier, we had:

After setting q = 3r, then:

a_1 = 14q^2 = 14*(3r)^2 = 14*9r^2 = 126r^2.

a_2 = 105q^2 + 84q = 105*(3r)^2 + 84*(3r) = 105*9r^2 + 252r = 945r^2 + 252r.

Wait, previously, I had written a_2 = 105r^2 + 84r, but that was incorrect. Let me correct that.

Wait, no, earlier, when we set q = 3r, then:

Wait, no, let me go back. Initially, we had:

After setting q = 3r, then:

Wait, no, let me retrace:

Initially, we had:

After setting p = 3q, then:

Wait, this is getting too convoluted. Let me start over with the substitutions.

We had:

From the equation $ (6y - 5x)^2 = 2016x $, we derived that x = 14q^2, y = 105q^2 + 84q, but this led to a non-integer a_3. However, we also had the condition that 35q^2 must be divisible by 3, hence q must be divisible by 3. Let me set q = 3r, then:

x = 14q^2 = 14*(9r^2) = 126r^2

y = [5x + k]/6, where k = 168q = 168*3r = 504r

So y = [5*126r^2 + 504r]/6 = [630r^2 + 504r]/6 = 105r^2 + 84r.

Therefore, a_2 = 105r^2 + 84r.

Now, let's compute a_3 = y^2 / x = (105r^2 + 84r)^2 / (126r^2).

Let me factor numerator and denominator:

Numerator: (105r^2 + 84r)^2 = [21r(5r + 4)]^2 = 441r^2(5r + 4)^2

Denominator: 126r^2 = 126r^2

Therefore, a_3 = 441r^2(5r + 4)^2 / 126r^2 = 441 / 126 * (5r + 4)^2 = 3.5 * (5r + 4)^2.

Wait, 441 / 126 = 3.5? Let me compute 441 ÷ 126:

126 * 3 = 378, 441 - 378 = 63, so 3 + 63/126 = 3.5. Yes. So a_3 = 3.5 * (5r + 4)^2. But this is not an integer unless (5r + 4)^2 is even, which would require 5r + 4 to be even. Let me check:

5r + 4 is even ⇨ 5r is even - 4 ⇨ 5r is even ⇨ r is even, since 5 is odd. Therefore, r must be even. Let me set r = 2s, where s is a positive integer.

Then, a_1 = 126r^2 = 126*(4s^2) = 504s^2

a_2 = 105r^2 + 84r = 105*(4s^2) + 84*(2s) = 420s^2 + 168s

Now, compute a_3 = 3.5 * (5r + 4)^2 = 3.5 * (5*(2s) + 4)^2 = 3.5 * (10s + 4)^2 = 3.5 * [2(5s + 2)]^2 = 3.5 * 4*(5s + 2)^2 = 14*(5s + 2)^2.

Which is an integer, since 14 is integer and (5s + 2)^2 is integer. Great! So now, a_3 is integer.

Now, let's check if this works. Let me take s = 1:

s = 1:

a_1 = 504*1 = 504

a_2 = 420*1 + 168*1 = 420 + 168 = 588

Check if a_2 > a_1: 588 > 504, yes.

Compute a_3 = 14*(5*1 + 2)^2 = 14*(7)^2 = 14*49 = 686.

Check if a_3 is integer: yes.

Now, check if a_4 is integer. From earlier, a_4 = 2a_3 - a_2 = 2*686 - 588 = 1372 - 588 = 784.

Which is integer. Good.

Now, compute a_5 = a_4^2 / a_3 = 784^2 / 686.

Calculate 784^2: 784 * 784. Let me compute:

784 = 28^2, so 784^2 = (28^2)^2 = 28^4. But maybe better to compute directly:

784 * 784: Let me note that 784 = 700 + 84.

But maybe better to note that 784 / 686 = 784 / 686 = 1.142857... Wait, but let me compute 784^2 / 686:

First, simplify the fraction:

784 / 686 = 784 ÷ 14 = 56; 686 ÷ 14 = 49. So 784 / 686 = 56 / 49 = 8 / 7.

Therefore, 784^2 / 686 = (784 / 686) * 784 = (8/7) * 784 = 8 * 112 = 896.

Yes, because 784 ÷ 7 = 112, so 8 * 112 = 896. So a_5 = 896, which is integer.

Good. Now, a_6 = 2a_5 - a_4 = 2*896 - 784 = 1792 - 784 = 1008.

Integer. Good.

a_7 = a_6^2 / a_5 = 1008^2 / 896.

Simplify: 1008 / 896 = 1.125 = 9/8. So 1008^2 / 896 = (1008 / 896) * 1008 = (9/8) * 1008 = 9 * 126 = 1134.

Yes, since 1008 ÷ 8 = 126. So a_7 = 1134.

Good. a_8 = 2a_7 - a_6 = 2*1134 - 1008 = 2268 - 1008 = 1260.

Integer. Good.

a_9 = a_8^2 / a_7 = 1260^2 / 1134.

Simplify: 1260 / 1134 = 1.111... = 10/9. So 1260^2 / 1134 = (1260 / 1134) * 1260 = (10/9) * 1260 = 10 * 140 = 1400.

Yes, since 1260 ÷ 9 = 140. So a_9 = 1400.

Good. a_10 = 2a_9 - a_8 = 2*1400 - 1260 = 2800 - 1260 = 1540.

Integer. Good.

a_11 = a_10^2 / a_9 = 1540^2 / 1400.

Simplify: 1540 / 1400 = 1.1 = 11/10. So 1540^2 / 1400 = (1540 / 1400) * 1540 = (11/10) * 1540 = 11 * 154 = 1694.

Yes, since 1540 ÷ 10 = 154. So a_11 = 1694.

Good. a_12 = 2a_11 - a_10 = 2*1694 - 1540 = 3388 - 1540 = 1848.

Integer. Good.

a_13 = a_12^2 / a_11 = 1848^2 / 1694.

Let me compute this. First, simplify 1848 / 1694.

Divide numerator and denominator by 2: 924 / 847.

Check if they have common factors. Let me see:

847 ÷ 7 = 121, since 7*121 = 847. 924 ÷ 7 = 132. So 924 / 847 = 132 / 121.

Therefore, 1848^2 / 1694 = (1848 / 1694) * 1848 = (132 / 121) * 1848.

Compute 1848 ÷ 121: 121 * 15 = 1815, 1848 - 1815 = 33, so 15 + 33/121 = 15.2727...

But let me compute it as fractions:

(132 / 121) * 1848 = 132 * 1848 / 121.

Note that 1848 = 121 * 15 + 33 = 121*15 + 33, but maybe better to factor:

1848 = 121 * 15 + 33 = 121*15 + 3*11. Not sure. Alternatively, note that 1848 = 121 * 15.2727... but perhaps there's a better way.

Wait, 132 and 121: 132 = 12 * 11, 121 = 11^2. So:

132 / 121 = (12 * 11) / 11^2 = 12 / 11.

Therefore, 132 / 121 * 1848 = (12 / 11) * 1848 = 12 * (1848 / 11).

Compute 1848 ÷ 11: 11*168 = 1848, since 11*160 = 1760, 11*8 = 88, so 1760 + 88 = 1848. Therefore, 1848 / 11 = 168.

Therefore, a_13 = 12 * 168 = 2016. Which matches the given value! Great!

So with s = 1, we have a_1 = 504, and a_13 = 2016. Therefore, this is a valid solution.

Now, let me check if there are other possible values of s. Earlier, we had to set r = 2s, and s is a positive integer. Let me check for s = 2:

s = 2:

a_1 = 504 * 4 = 2016

a_2 = 420*(4) + 168*(2) = 1680 + 336 = 2016

Wait, but a_2 = 2016, and a_1 = 2016. But the sequence is strictly increasing, so a_2 must be greater than a_1. Here, a_2 = a_1, which is invalid. Therefore, s = 2 is invalid.

Wait, let me check the calculation:

Wait, a_1 = 504s^2, for s = 2, a_1 = 504 * 4 = 2016.

a_2 = 420s^2 + 168s = 420*(4) + 168*(2) = 1680 + 336 = 2016. Yes, so a_2 = a_1, which violates the strictly increasing condition. Therefore, s = 2 is invalid.

Similarly, for s ≥ 2, a_2 = 420s^2 + 168s. Let me check for s = 2, a_2 = 2016, which equals a_1 = 2016. For s = 3:

a_1 = 504 * 9 = 4536

a_2 = 420*9 + 168*3 = 3780 + 504 = 4284

Check if a_2 > a_1: 4284 < 4536? No, 4284 is less than 4536. Wait, this is a problem. Wait, this suggests that for s = 3, a_2 < a_1, which violates the strictly increasing condition. Wait, what's happening here?

Wait, let me recalculate a_2 for s = 3:

a_2 = 420s^2 + 168s = 420*(9) + 168*3 = 3780 + 504 = 4284.

a_1 = 504s^2 = 504*9 = 4536.

So 4284 < 4536, which means a_2 < a_1, which is invalid. Therefore, s = 3 is invalid.

Wait, this suggests that only s = 1 is valid? Let me check for s = 1, a_1 = 504, a_2 = 588 > 504, which is good. For s = 2, a_1 = 2016, a_2 = 2016, invalid. For s = 3, a_2 < a_1, invalid. For s = 0, not positive. Therefore, only s = 1 is valid.

Therefore, the only valid solution is s = 1, leading to a_1 = 504.

But let me check if there are other possible values of r or q that could lead to valid sequences. Earlier, we had to set r = 2s, but maybe there are other constraints?

Wait, let me think. When we set q = 3r, then r had to be even to make a_3 integer, leading to r = 2s. But perhaps there are other ways to satisfy the divisibility conditions? Or maybe not. Since we derived that for a_3 to be integer, r must be even, and then with s = 1 being the only valid value that keeps the sequence strictly increasing.

Therefore, the only possible value is a_1 = 504.

But let me check if there are other possibilities. For example, could there be other values of r that make a_2 > a_1?

Wait, for s = 1, a_1 = 504, a_2 = 588.

For s = 1/2? No, s must be integer.

Alternatively, maybe there are other solutions where the earlier terms are integers, but with different relations? But given the way we derived the relations, and the constraints from the divisibility, it seems that s = 1 is the only valid solution.

Therefore, the answer is a_1 = 504.

But let me verify with the sequence:

We had:

a_1 = 504

a_2 = 588

a_3 = 686

a_4 = 784

a_5 = 896

a_6 = 1008

a_7 = 1134

a_8 = 1260

a_9 = 1400

a_10 = 1540

a_11 = 1694

a_12 = 1848

a_13 = 2016

Yes, each term is strictly increasing, and all terms are integers. Also, the given condition a_13 = 2016 is satisfied. Therefore, this must be the correct value.

Therefore, the answer is 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: Understanding the Structure

Let’s analyze the structure of the sequence based on the given properties.

- The subsequence $ a_{2k-1}, a_{2k}, a_{2k+1} $ is geometric. So:
  $$
  a_{2k}^2 = a_{2k-1} \cdot a_{2k+1}
  $$
- The subsequence $ a_{2k}, a_{2k+1}, a_{2k+2} $ is arithmetic. So:
  $$
  2a_{2k+1} = a_{2k} + a_{2k+2}
  $$

These recursive relationships allow us to express terms in terms of earlier ones. For example, from the geometric condition:
$$
a_{2k+1} = \frac{a_{2k}^2}{a_{2k-1}}
$$
From the arithmetic condition:
$$
a_{2k+2} = 2a_{2k+1} - a_{2k}
$$

This suggests that the sequence can be built recursively from $ a_1 $ and $ a_2 $, and all terms must be positive integers.

---

### Step 2: Expressing $ a_{13} $ in Terms of $ a_1 $ and $ a_2 $

We are told that $ a_{13} = 2016 $. Since 13 is odd, it corresponds to the term $ a_{2k-1} $, where $ k = 7 $. Using the earlier derived formula for odd-indexed terms:

$$
a_{2k-1} = \frac{( (k-1)a_2 - (k-2)a_1 )^2}{a_1}
$$

For $ k = 7 $, this becomes:
$$
a_{13} = \frac{(6a_2 - 5a_1)^2}{a_1}
$$

Given $ a_{13} = 2016 $, we have:
$$
\frac{(6a_2 - 5a_1)^2}{a_1} = 2016
$$

Multiplying both sides by $ a_1 $:
$$
(6a_2 - 5a_1)^2 = 2016a_1
$$

Let’s denote $ x = a_1 $, $ y = a_2 $. Then:
$$
(6y - 5x)^2 = 2016x
$$

---

### Step 3: Solving the Equation

We solve the equation:
$$
(6y - 5x)^2 = 2016x
$$

Let’s denote $ d = 6y - 5x $, so:
$$
d^2 = 2016x \Rightarrow x = \frac{d^2}{2016}
$$

Since $ x $ must be a positive integer, $ d^2 $ must be divisible by 2016. The prime factorization of 2016 is:
$$
2016 = 2^5 \cdot 3^2 \cdot 7
$$

So $ d $ must be a multiple of the square root of the square-free part of 2016. The square-free part is $ 2 \cdot 7 = 14 $, so $ d = 14k $ for some integer $ k $.

Then:
$$
x = \frac{(14k)^2}{2016} = \frac{196k^2}{2016} = \frac{49k^2}{504}
$$

To make $ x $ an integer, $ 504 $ must divide $ 49k^2 $. Since $ \gcd(49, 504) = 7 $, we divide both sides by 7:
$$
x = \frac{7k^2}{72}
$$

So $ 72 $ must divide $ 7k^2 $. Since $ \gcd(7, 72) = 1 $, $ 72 $ must divide $ k^2 $. The square-free part of 72 is 2, so $ k $ must be a multiple of 2. Let $ k = 2m $, then:
$$
x = \frac{7(2m)^2}{72} = \frac{28m^2}{72} = \frac{7m^2}{18}
$$

Now $ 18 $ must divide $ 7m^2 $. Since $ \gcd(7, 18) = 1 $, $ 18 $ must divide $ m^2 $. The square-free part of 18 is 2, so $ m $ must be a multiple of 2. Let $ m = 2n $, then:
$$
x = \frac{7(2n)^2}{18} = \frac{28n^2}{18} = \frac{14n^2}{9}
$$

Now $ 9 $ must divide $ 14n^2 $. Since $ \gcd(14, 9) = 1 $, $ 9 $ must divide $ n^2 $. The square-free part of 9 is 1, so $ n $ must be a multiple of 3. Let $ n = 3p $, then:
$$
x = \frac{14(3p)^2}{9} = \frac{14 \cdot 9p^2}{9} = 14p^2
$$

So:
$$
a_1 = 14p^2
$$

Now, recall that $ d = 14k = 14 \cdot 2m = 14 \cdot 2 \cdot 2n = 14 \cdot 4n = 14 \cdot 4 \cdot 3p = 168p $

Then:
$$
a_2 = \frac{d + 5a_1}{6} = \frac{168p + 5 \cdot 14p^2}{6} = \frac{168p + 70p^2}{6}
$$

Simplify:
$$
a_2 = \frac{70p^2 + 168p}{6} = \frac{35p^2 + 84p}{3}
$$

For $ a_2 $ to be an integer, $ 35p^2 + 84p $ must be divisible by 3. Since $ 35 \equiv 2 \mod 3 $, and $ 84 \equiv 0 \mod 3 $, we have:
$$
35p^2 + 84p \equiv 2p^2 \mod 3
$$

So $ 2p^2 \equiv 0 \mod 3 \Rightarrow p^2 \equiv 0 \mod 3 \Rightarrow p \equiv 0 \mod 3 $

Let $ p = 3q $, then:
$$
a_1 = 14p^2 = 14(9q^2) = 126q^2
$$
$$
a_2 = \frac{35p^2 + 84p}{3} = \frac{35(9q^2) + 84(3q)}{3} = \frac{315q^2 + 252q}{3} = 105q^2 + 84q
$$

Now, we check for the smallest positive integer $ q $ such that the sequence is strictly increasing and all terms are integers.

Try $ q = 1 $:
- $ a_1 = 126 $
- $ a_2 = 105 + 84 = 189 $
- $ a_3 = \frac{a_2^2}{a_1} = \frac{189^2}{126} = \frac{35721}{126} = 283.5 $ → **Not an integer**

Try $ q = 2 $:
- $ a_1 = 126 \cdot 4 = 504 $
- $ a_2 = 105 \cdot 4 + 84 \cdot 2 = 420 + 168 = 588 $
- $ a_3 = \frac{588^2}{504} = \frac{345744}{504} = 686 $ → **Integer**
- $ a_4 = 2a_3 - a_2 = 2 \cdot 686 - 588 = 1372 - 588 = 784 $
- $ a_5 = \frac{a_4^2}{a_3} = \frac{784^2}{686} = \frac{614656}{686} = 896 $
- $ a_6 = 2a_5 - a_4 = 2 \cdot 896 - 784 = 1792 - 784 = 1008 $
- $ a_7 = \frac{a_6^2}{a_5} = \frac{1008^2}{896} = \frac{1016064}{896} = 1134 $
- $ a_8 = 2a_7 - a_6 = 2 \cdot 1134 - 1008 = 2268 - 1008 = 1260 $
- $ a_9 = \frac{a_8^2}{a_7} = \frac{1260^2}{1134} = \frac{1587600}{1134} = 1400 $
- $ a_{10} = 2a_9 - a_8 = 2 \cdot 1400 - 1260 = 2800 - 1260 = 1540 $
- $ a_{11} = \frac{a_{10}^2}{a_9} = \frac{1540^2}{1400} = \frac{2371600}{1400} = 1694 $
- $ a_{12} = 2a_{11} - a_{10} = 2 \cdot 1694 - 1540 = 3388 - 1540 = 1848 $
- $ a_{13} = \frac{a_{12}^2}{a_{11}} = \frac{1848^2}{1694} = \frac{3415104}{1694} = 2016 $

All terms are integers, and the sequence is strictly increasing. Thus, this is a valid solution.

---

### 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 $a_1 = 14 \cdot 6^2 = \boxed{504}.$