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In Question 488044, I asked for a proof of the hypergeometric identity $$\sum_{k=0}^\infty (22k^2-92k+11)\frac{\binom{4k}k}{16^k}=-5.\tag{1}$$ Here I propose some further identities motivated by $(1)$.

Recall that the harmonic numbers are given by $$H_n:=\sum_{0<k\le n}\frac{1}k\quad \ (n=0,1,2,\ldots).$$ Based on my computation, I have formulated the following conjecture inspired by $(1)$.

Conjecture. Let $P(k)=22k^2-92k+11$. Then we have \begin{align}\sum_{k=1}^\infty\frac{\binom{4k}k}{16^k}\left(P(k)H_k-54k+108-\frac{10}{3k}\right)&=-\frac{20}3\log2,\tag{2} \\\sum_{k=1}^\infty\frac{\binom{4k}k}{16^k}\left(P(k)H_{2k}+287k-115-\frac{25}{6k}\right)&=214-\frac{40}3\log2,\tag{3} \\\sum_{k=1}^\infty\frac{\binom{4k}k}{16^k}\left(P(k)H_{3k}-296k+178-\frac{25}{3k}\right)&=-196-\frac{80}3\log2,\tag{4} \\\sum_{k=1}^\infty\frac{\binom{4k}k}{16^k}\left(P(k)H_{4k}-\frac{449k-275}2-\frac{85}{12k}\right)&=-151-\frac{80}3\log2.\tag{5} \end{align}

I have checked the identities $(2)$-$(5)$ numerically via Mathematica.

QUESTION. Any ideas to prove the conjectural identities $(2)$-$(5)$?

Your comments are welcome!

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  • $\begingroup$ For $(2)$-$(5)$, I also have corresponding conjectural congruences. For example, I conjecture that for any prime $p>3$ we have the congruence \begin{align}&\ \sum_{k=1}^{p-1}\frac{\binom{4k}k}{16^k}\left(P(k)H_k-54k+108-\frac{10}{3k}\right)\\\equiv&\ -10q_p(2)+5p\,q_p(2)^2-\frac{10}3p^2q_p(2)^3\pmod{p^3},\end{align} where $q_p(2)=(2^{p-1}-1)/p$. $\endgroup$ Commented Feb 20 at 12:28

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