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
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
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Apr 18 2010
STATUS
approved