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.

2, 181, 23488, 3625081, 619898336, 113451041232, 21790823094272, 4339409873332321, 888730714063587232, 186141207745025911376, 39707252850926474171392, 8600444322930062324576656

Offset

1,1

Comments

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.

Links

Table of n, a(n) for n=1..12.

Kasper Andersen, Re: A somewhat surprising conjecture

Jesús Guillera, About a new kind of Ramanujan-type series, Experiment. Math. 12(2003), 507-510.

Zhi-Wei Sun, Open Conjectures on Congruences, preprint, arXiv:0911.5665.

Zhi-Wei Sun, A somewhat surprising conjecture

Zhi-Wei Sun, Re: A somewhat surprising conjecture

Formula

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).

Example

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.

Mathematica

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}]

Crossrefs

Cf. A176285, A173774, A000984, A001700.

Sequence in context: A094221 A324442 A032593 * A094483 A258426 A163794

Adjacent sequences: A176474 A176475 A176476 * A176478 A176479 A176480

Keyword

nonn

Author

Zhi-Wei Sun, Apr 18 2010

Status

approved