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A176285 a(1) = 1, and then 4*(2*n + 1)^2*a(n+1) + n^2*a(n) = (205*n^2 + 160*n + 32)*binomial(2*n-1, n)^3 (n = 1, 2, 3, ...).

%I A176285 #34 Oct 18 2024 11:42:37
%S A176285 1,11,316,12011,522376,24593348,1219951188,62798884331,3323228619736,
%T A176285 179665076698136,9880531254032176,550994628527745476,
%U A176285 31084678988906064016,1770908612898043660556,101738260887234550287316
%N A176285 a(1) = 1, and then 4*(2*n + 1)^2*a(n+1) + n^2*a(n) = (205*n^2 + 160*n + 32)*binomial(2*n-1, n)^3 (n = 1, 2, 3, ...).
%C A176285 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 A176285 T. Amdeberhan and D. Zeilberger, <a href="https://doi.org/10.37236/1318">Hypergeometric series acceleration via the WZ method</a>, Electron. J. Combin. 4(1997), no.2, #R3.
%H A176285 Kasper Andersen, <a href="https://listserv.nodak.edu/cgi-bin/wa.exe?A2=NMBRTHRY;72aeca0c.1002">Re: A somewhat surprising conjecture</a>
%H A176285 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 A176285 Zhi-Wei Sun, <a href="http://arxiv.org/abs/0911.5665">Open Conjectures on Congruences</a>, preprint, arXiv:0911.5665 [math.NT], 2009-2011.
%H A176285 Zhi-Wei Sun, <a href="https://listserv.nodak.edu/cgi-bin/wa.exe?A2=NMBRTHRY;a39bf149.1002">A somewhat surprising conjecture</a>
%H A176285 Zhi-Wei Sun, <a href="https://listserv.nodak.edu/cgi-bin/wa.exe?A2=NMBRTHRY;4d3d48b9.1002">Re: A somewhat surprising conjecture</a>
%F A176285 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).
%F A176285 From _Peter Bala_, Jul 11 2024: (Start)
%F A176285 a(n+1) = Sum_{k = 0..n} binomial(n, k)^2 * binomial(n+k, k) * binomial(2*n+k+1, n).
%F A176285 Compare with the identity Sum_{k = 0..n} binomial(n, k)^2 * binomial(n+k, k) * binomial(2*n+k, n) = binomial(2*n, n)^3 = A002897(n).
%F A176285 a(n) = binomial(2*n-1, n-1)*hypergeom([-n+1, -n+1, n, 2*n], [1, 1, n+1], 1).
%F A176285 P-recursive: 4*(205*n^2 - 250*n + 77)*(2*n + 1)^2*n^3*a(n+1) = (209715*n^7 - 360710*n^6 + 147891*n^5 + 65280*n^4 - 57280*n^3 + 3680*n^2 + 5120*n - 1024)*a(n) + 8*(205*n^2 + 160*n + 32)*(n - 1)^2*(2*n - 1)^3*a(n-1) with a(1) = 1 and a(2) = 11.
%F A176285 Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and all positive integers n and r. (End)
%F A176285 a(n) ~ 2^(6*n-3) / (5*Pi^(3/2)*n^(3/2)). - _Vaclav Kotesovec_, Jul 17 2024
%F A176285 a(n) = Sum_{k = 0..n-1} (-1)^(n+k+1) * binomial(n-1, k) * binomial(2*n+k-1, k) * A108625(n-1, k). Cf. A002897. - _Peter Bala_, Oct 16 2024
%e A176285 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.
%p A176285 a := proc(n) option remember; if n = 1 then 1 elif n = 2 then 11 else
%p A176285  ( (209715*n^7 - 1828715*n^6 + 6716166*n^5 - 13424850*n^4 + 15714735*n^3 - 10726375*n^2 + 3936520*n - 598220)*a(n-1) + 8*(n - 2)^2*(2*n - 3)^3*(205*n^2 - 250*n + 77)*a(n-2) )/( 4*(n - 1)^3*(2*n - 1)^2*(205*n^2 - 660*n + 532) ) end if; end:
%p A176285 seq(a(n), n = 1..20); # _Peter Bala_, Jul 12 2024
%t A176285 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 A176285 Cf. A000984, A001700, A002897, A173774.
%K A176285 nonn,easy
%O A176285 1,2
%A A176285 _Zhi-Wei Sun_, Apr 14 2010