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%I #18 Dec 01 2018 08:02:40
%S 1,11,316,12011,522376,24593348,1219951188,62798884331,3323228619736,
%T 179665076698136,9880531254032176,550994628527745476,
%U 31084678988906064016,1770908612898043660556,101738260887234550287316
%N a(1) = 1, and then 4*(2*n + 1)^2*a(n+1) + n^2*a(n) = (205*n^2 + 160*n + 32)*binomial(2n-1,n)^3 (n = 1, 2, 3, ...).
%C On Apr 04 2010, _Zhi-Wei Sun_ introduced this sequence and conjectured that each term a(n) is a positive integer. He also guessed that a(n) is odd if and only if n is a power of two. It is easy to see that 8*n^2*binomial(2*n,n)^2*a(n) equals Sum_{k=0..n-1}(205*k^2 + 160*k + 32)*(-1)^{n - 1 - k}*binomial(2*k,k)^5. Sun also conjectured that for any prime p > 5 we have Sum_{k=0..p-1}(205*k^2 + 160*k + 32)(-1)^k*binomial(2*k,k)^5 = 32*p^2 + 64*p^3*Sum_{k=1..p-1}1/k (mod p^7) and Sum_{k=0..(p-1)/2}(205*k^2 + 160*k + 32)(-1)^k*binomial(2*k,k)^5 = 32*p^2 + 896/3*p^5*B_{p-3} (mod p^6), where B_0, B_1, B_2, ... are Bernoulli numbers. It is also remarkable that Sum_{n>0}(-1)^n(205*n^2 - 160*n + 32)/(n^5*binomial(2*n,n)^5) = -2*zeta(3) as proved by T. Amdeberhan and D. Zeilberger via the WZ method.
%H T. Amdeberhan and D. Zeilberger, <a href="http://www.combinatorics.org/ojs/index.php/eljc/article/view/v4i2r3">Hypergeometric series acceleration via the WZ method</a>, Electron. J. Combin. 4(1997), no.2, #R3.
%H Kasper Andersen, <a href="https://listserv.nodak.edu/cgi-bin/wa.exe?A2=NMBRTHRY;72aeca0c.1002">Re: A somewhat surprising conjecture</a>
%H J. Guillera, <a href="https://arxiv.org/abs/1104.0396">Hypergeometric identities for 10 extended Ramanujan-type series</a>, arXiv:1104.0396 [math.NT], 2011; Ramanujan J. 15(2008), 219-234.
%H Z. W. Sun, <a href="http://arxiv.org/abs/0911.5665">Open Conjectures on Congruences</a>, preprint, arXiv:0911.5665 [math.NT], 2009-2011.
%H Zhi-Wei Sun, <a href="https://listserv.nodak.edu/cgi-bin/wa.exe?A2=NMBRTHRY;a39bf149.1002">A somewhat surprising conjecture</a>
%H Zhi-Wei Sun, <a href="https://listserv.nodak.edu/cgi-bin/wa.exe?A2=NMBRTHRY;4d3d48b9.1002">Re: A somewhat surprising conjecture</a>
%F a(n) = (Sum_{k=0..n-1} (205*k^2 + 160*k + 32)(-1)^{n - 1 - k}*binomial(2*k,k)^5)/(8*n^2*binomial(2*n,n)^2).
%e For n = 2 we have a(2) = 11 since 4*(2*1 + 1)^2*a(2) = -1^2*a(1) + (205*1^2 + 160*1 + 32)*binomial(2*1 - 1,1)^3 = 396.
%t u[n_]:=u[n]=((205(n-1)^2+160(n-1)+32)Binomial[2n-3,n-1]^3-(n-1)^2*u[n-1])/(4(2n-1)^2) u[1]=1 Table[u[n],{n,1,50}]
%Y Cf. A173774, A000984, A001700.
%K nonn
%O 1,2
%A _Zhi-Wei Sun_, Apr 14 2010
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