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A176477 a(1)=2; for n >= 2, (2n+1)^3*a(n) = 32n^3*a(n-1) + (21n^3 + 22n^2 + 8n + 1)*binomial(2n-1,n)^4.

%I A176477 #15 Sep 22 2021 06:01:01
%S A176477 2,181,23488,3625081,619898336,113451041232,21790823094272,
%T A176477 4339409873332321,888730714063587232,186141207745025911376,
%U A176477 39707252850926474171392,8600444322930062324576656
%N A176477 a(1)=2; for n >= 2, (2n+1)^3*a(n) = 32n^3*a(n-1) + (21n^3 + 22n^2 + 8n + 1)*binomial(2n-1,n)^4.
%C A176477 On Apr 06 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 = 2, 2^2, 2^3, .... It is easy to see that 16*(2n+1)^3*binomial(2n,n)^3*a(n) equals Sum_{k=0..n} (21k^3 + 22k^2 + 8k + 1)*256^(n-k)*binomial(2k,k)^7. Sun also conjectured that for any prime p > 5 we have Sum_{k=0..p-1} (21k^3 + 22k^2 + 8k + 1)*binomial(2k,k)^7/256^k == p^3 (mod p^8). It is also remarkable that Sum_{n>=1} 256^n*(21n^3 - 22n^2 + 8n - 1)/(n^7*binomial(2n,n)^7) = Pi^4/8 as conjectured by J. Guillera.
%H A176477 Kasper Andersen, <a href="http://listserv.nodak.edu/cgi-bin/wa.exe?A2=ind1002&amp;L=nmbrthry&amp;T=0&amp;P=1395">Re: A somewhat surprising conjecture</a>
%H A176477 Jesús Guillera, <a href="http://emis.icm.edu.pl/journals/EM/expmath/volumes/12/12.4/Guillera.pdf">About a new kind of Ramanujan-type series</a>, Experiment. Math. 12(2003), 507-510.
%H A176477 Zhi-Wei Sun, <a href="http://arxiv.org/abs/0911.5665">Open Conjectures on Congruences</a>, preprint, arXiv:0911.5665.
%H A176477 Zhi-Wei Sun, <a href="http://listserv.nodak.edu/cgi-bin/wa.exe?A2=ind1002&amp;L=nmbrthry&amp;T=0&amp;P=956">A somewhat surprising conjecture</a>
%H A176477 Zhi-Wei Sun, <a href="http://listserv.nodak.edu/cgi-bin/wa.exe?A2=ind1002&amp;L=nmbrthry&amp;T=0&amp;P=1289">Re: A somewhat surprising conjecture</a>
%F A176477 a(n) = (Sum_{k=0..n} (21k^3 + 22k^2 + 8k + 1)*256^(n-k)*binomial(2k,k)^7) / (16(2n+1)^3*binomial(2n,n)^3).
%e A176477 For n=2 we have a(2) = (32*2^3*a(1) + (21*2^3 + 22*2^2 + 8*2 + 1)*binomial(2*2-1,2)^4)/(2*2 + 1)^3 = 181.
%t A176477 u[n_]:=u[n]=((21n^3+22n^2+8n+1)Binomial[2n-1,n]^4+32*n^3*u[n-1])/((2n+1)^3) u[1]=2 Table[u[n],{n,1,50}]
%Y A176477 Cf. A176285, A173774, A000984, A001700.
%K A176477 nonn
%O A176477 1,1
%A A176477 _Zhi-Wei Sun_, Apr 18 2010