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I have found the hypergeometric identity $$\sum_{k=0}^\infty(22k^2-92k+11)\frac{\binom{4k}k}{16^k}=-5. \tag{1}$$ As the series converges fast, one can easily check $(1)$ numerically by Mathematica. Actually I have many other conjectural identities similar to $(1)$.

Question. How to prove the identity $(1)$? Can one deduce it via the generating function method or the WZ method?

Your comments are welcome!

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    $\begingroup$ Here is an idea that should lead to a solution but the calculations get out of hand. We need to figure out what the function $\sum_i {4n \choose n} x^n$ is. I will sketch a way to figure the function $\sum_i \frac{1}{3n+1}{4n \choose n} x^n$ which would lead to what the original function should be. We need to use Lagrange–Bürmann formula when $f(z) = \frac{z}{(z+1)^4}$. Then the inverse of this should have the generating function corresponding to $\frac{1}{3n+1}{4n \choose n}$ minus 1. $\endgroup$ Commented Feb 18 at 3:25
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    $\begingroup$ I also conjecture that for any odd prime $p$ we have the related congruence $$\sum_{k=0}^{p-1}(22k^2-92k+11)\frac{\binom{4k}k}{16^k}=-\frac52+27p\pmod{p^2}.$$ $\endgroup$ Commented Feb 18 at 4:29
  • $\begingroup$ A telescoping sum is $\sum_{k=0}^\infty(22 k^3 - 102 k^2 - 10 k- 3){4k\choose k}/16^k=0$, is it a coincidence that the leading term 22 is the same? $\endgroup$ Commented Feb 20 at 5:08

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Per my old answer in AoPS, the generating function $f(x) := \sum_{k\geq0} \binom{4k}k x^k$ satisfies the algebraic identity: $$(f(x) - 1)(3f(x) + 1)^3 = x ( 4 f(x) )^4.$$ From where by differentiating, we have can also express $f'(x)$ and $f''(x)$ in terms of $f(x)$.

We we need to compute $$\sum_{k\geq 0} (22k^2 - 92k + 11) \frac{\binom{4k}{k}}{16^k} = \frac{11}{128}f''(1/16) - \frac{35}8f'(1/16) + 11f(1/16).$$ As this Sage code shows, eliminating the derivatives from the sum value turns it into $$\frac{-55 f(1/16)^3 + 55 f(1/16)^2 + 35 f(1/16) - 15}4.$$ It remains to note that the algebraic identity for $x=1/16$ factors as $$(f(1/16) + 1) (11f(1/16)^3 - 11f(1/16)^2 - 7f(1/16) - 1)=0,$$ where the first factor is clearly nonzero. Reducing the the sum value modulo the second factor, we get $\frac{-20}{4}=-5$. QED

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