Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).
%I #29 Dec 21 2024 16:46:27
%S 25,809,23020,730325,27867142,1117643720,42658771456,1558395721085,
%T 57260792702050,2179584653311070,84835851591609400,
%U 3292250198848240760,126379831667243976400,4841030410501144484000,186842197443136622824960,7269291788529191112814925,283472902036823148786161530
%N a(n) = (Sum_{k=0..n-1} (-1)^k * (4k+1) * 160^(n-1-k) * C(2k,k) * Sum_{j=0..k} C(k,j) * C(k+2j,2j) * C(2j,j) * (-20)^(k-j)) / (n * C(2n,n)).
%C Conjecture 1: a(n) is a positive integer for each n > 1. Moreover, a(n) is odd if and only if n = 2^k + 1 for some nonnegative integer k.
%C Conjecture 2: The infinite series Sum_{k>=0} (4*k+1)/(-160)^k * C(2k,k) * Sum_{j=0..k} C(k,j) * C(k+2j,2j) * C(2j,j) * (-20)^(k-j) has the value sqrt(30)/(5*Pi)*(5+c^(1/3))/c^(1/6), where c = 145 + 30*sqrt(6).
%C Conjecture 3. Let p be an odd prime different from 5, and let S(p) denote the sum Sum_{k=0..p-1} C(2k,k)/(-160)^k * Sum_{j=0..k} C(k,j) * C(k+2j,2j) * C(2j,j) * (-20)^(k-j).
%C (i) If p == 1,3 (mod 8) and p = x^2 + 2*y^2 with x and y integers, then S(p) == (5/p)*(4x^2 - 2p) (mod p^2), where (5/p) is the Legendre symbol.
%C (ii) If p == 5,7 (mod 8), then S(p) == 0 (mod p^2).
%H Zhi-Wei Sun, <a href="/A337247/b337247.txt">Table of n, a(n) for n = 2..100</a>
%H Zhi-Wei Sun, <a href="http://mathoverflow.net/questions/369569">Two curious series for 1/Pi</a>, Question 369569 at MathOverflow, August 19-20, 2020.
%H Zhi-Wei Sun, <a href="http://dx.doi.org/10.3934/era.2020070">New series for powers of Pi and related congruences</a>, Electron. Res. Arch. 28(2020), no. 3, 1273-1342.
%H Zhi-Wei Sun, <a href="https://arxiv.org/abs/2009.04379">Some new series for 1/Pi motivated by congruences</a>, arXiv:2009.04379 [math.NT], 2020.
%e a(2) = (160 - (4 + 1)*C(2,1)*(-20 + C(3,2)*C(2,1)))/(2*C(4,2)) = 300/12 = 25.
%t a[n_]:=a[n]=Sum[(4k+1)(-1)^k*160^(n-1-k)*Binomial[2k,k]*Sum[Binomial[k,j]Binomial[k+2j,2j]Binomial[2j,j](-20)^(k-j),{j,0,k}],{k,0,n-1}]/(n*Binomial[2n,n])
%t Table[a[n],{n,2,18}]
%Y Cf. A000796, A000984, A336981, A336982.
%K nonn
%O 2,1
%A _Zhi-Wei Sun_, Aug 20 2020