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, ...).

1, 11, 316, 12011, 522376, 24593348, 1219951188, 62798884331, 3323228619736, 179665076698136, 9880531254032176, 550994628527745476, 31084678988906064016, 1770908612898043660556, 101738260887234550287316

Offset

1,2

Comments

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.

Links

Table of n, a(n) for n=1..15.

T. Amdeberhan and D. Zeilberger, Hypergeometric series acceleration via the WZ method, Electron. J. Combin. 4(1997), no.2, #R3.

Kasper Andersen, Re: A somewhat surprising conjecture

J. Guillera, Hypergeometric identities for 10 extended Ramanujan-type series, arXiv:1104.0396 [math.NT], 2011; Ramanujan J. 15(2008), 219-234.

Zhi-Wei Sun, Open Conjectures on Congruences, preprint, arXiv:0911.5665 [math.NT], 2009-2011.

Zhi-Wei Sun, A somewhat surprising conjecture

Zhi-Wei Sun, Re: A somewhat surprising conjecture

Formula

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).

From Peter Bala, Jul 11 2024: (Start)

a(n+1) = Sum_{k = 0..n} binomial(n, k)^2 * binomial(n+k, k) * binomial(2*n+k+1, n).

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).

a(n) = binomial(2*n-1, n-1)*hypergeom([-n+1, -n+1, n, 2*n], [1, 1, n+1], 1).

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.

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)

a(n) ~ 2^(6*n-3) / (5*Pi^(3/2)*n^(3/2)). - Vaclav Kotesovec, Jul 17 2024

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

Example

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.

Maple

a := proc(n) option remember; if n = 1 then 1 elif n = 2 then 11 else
 ( (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:
seq(a(n), n = 1..20); # Peter Bala, Jul 12 2024

Mathematica

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}]

Crossrefs

Cf. A000984, A001700, A002897, A173774.

Sequence in context: A266029 A106827 A276977 * A266484 A219979 A115609

Adjacent sequences: A176282 A176283 A176284 * A176286 A176287 A176288

Keyword

nonn,easy

Author

Zhi-Wei Sun, Apr 14 2010

Status

approved