This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A337247 #29 Dec 21 2024 16:46:27
%S A337247 25,809,23020,730325,27867142,1117643720,42658771456,1558395721085,
%T A337247 57260792702050,2179584653311070,84835851591609400,
%U A337247 3292250198848240760,126379831667243976400,4841030410501144484000,186842197443136622824960,7269291788529191112814925,283472902036823148786161530
%N A337247 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 A337247 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 A337247 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 A337247 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 A337247 (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 A337247 (ii) If p == 5,7 (mod 8), then S(p) == 0 (mod p^2).
%H A337247 Zhi-Wei Sun, <a href="/A337247/b337247.txt">Table of n, a(n) for n = 2..100</a>
%H A337247 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 A337247 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 A337247 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 A337247 a(2) = (160 - (4 + 1)*C(2,1)*(-20 + C(3,2)*C(2,1)))/(2*C(4,2)) = 300/12 = 25.
%t A337247 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 A337247 Table[a[n],{n,2,18}]
%Y A337247 Cf. A000796, A000984, A336981, A336982.
%K A337247 nonn
%O A337247 2,1
%A A337247 _Zhi-Wei Sun_, Aug 20 2020