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I am trying to obtain an explicit formula for $(a_k)_{k \in \mathbb{N}}$, where $a_1=A>1$ and $$(a_k)^{k^2}=\prod_{i=1}^{k}a_i, \; k=2, 3,...$$ What makes this recursive relation interesting, among other things, is the cumulative dependence of a given term on all terms of the sequence preceding that term. From this recursive formula, after some iterative factoring, I found that $$a_2=A^{\frac{1}{3}}, a_3=A^{(1+\frac{1}{3})\frac{1}{8}}, a_4=A^{(1+\frac{1}{3})(1+\frac{1}{8})\frac{1}{15}}, a_5=A^{(1+\frac{1}{3})(1+\frac{1}{8})(1+\frac{1}{15})\frac{1}{24}}.$$

So long as I did not make any computational error, I see the pattern is

$$a_k=A^{\frac{1}{k^2-1}\prod_{i=2}^{k-1}{1+\frac{1}{i^2-1}}}, \; k \ge 3 \qquad (1).$$

This formula is not in a particularly satisfying form; I would like to obtain a closed form expression of (1) without the product notation, and seek to prove that the resultant explicit formula corresponds to the recursive relation. I would also be interested in other more direct, more elegant ways to obtain an explicit formula besides obtaining a closed form expression from (1).

Emily
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  • The obvious suspicion is that your formula is as good as it gets. Most sums (and products) simply do not have closed forms. See this discussion and its references. – Rus May Oct 04 '16 at 19:41

1 Answers1

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Despite my comment above, Maple gives a closed-form expression for your product in terms of the Gamma function. $$ \prod_{i=1}^{k-1}\left(1-\frac1{i^2-1}\right)= 2\,{\frac {\Gamma \left( k-\sqrt {2} \right) \Gamma \left( k+\sqrt { 2} \right) }{\Gamma \left( k-1 \right) \Gamma \left( k+1 \right) \Gamma \left( 2-\sqrt {2} \right) \Gamma \left( 2+\sqrt {2} \right) }}. $$

Rus May
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