A178790 The arithmetic mean of (2*k+1)*A_k (k=0,...,n-1), where A_0,A_1,... are Apery numbers given by A005259.

1, 8, 127, 2624, 61501, 1552760, 41186755, 1131614720, 31923047665, 919243356008, 26908963456783, 798379043762624, 23954974906866901, 725620080605773592, 22159617936375571627, 681528994326392115200, 21090805673899997148025, 656256696917886135153800

Offset: 1

Author

Zhi-Wei Sun, Jun 14 2010

Keywords

• nonn

Comments

Conjecture: the number a(n) = n^(-1)*Sum_{k=0..n-1}(2*k+1)*A_k is always an integer.

We can prove that for any prime p>3 we have a(p)=p (mod p^4).

Conjecture: If p=5,7 (mod 8) is a prime then sum_{k=0}^{p-1}A_k=0 (mod p^2); if p=1,3 (mod 8) is a prime greater than 3 and p=x^2+2y^2 with x,y integers then sum_{k=0}^{p-1}A_k=4x^2-2p (mod p^2).

a(n) is always an integer. The detailed proof can be found in the latest version of arXiv:1006.2776 . - Zhi-Wei Sun, Jun 17 2010

Examples

			For n=3 we have a(3)=(A_0+3A_1+5A_2)/3=(1+3*5+5*73)/3=127.
		

Links

• G. C. Greubel, Table of n, a(n) for n = 1..500
• Zhi-Wei Sun, Arithmetic properties of Apery numbers and central Delannoy numbers, arXiv:1006.2776 [math.NT], 2011.

Crossrefs

Cf. A005259, A173774, A189766.

Programs

GAP

List([1..30], n-> Sum([0..n], k -> Binomial(n-1,k)*Binomial(n+k,k) *Binomial(n+k, 2*k+1)* Binomial(2*k,k) )); # G. C. Greubel, Jan 24 2019

Magma

[(&+[Binomial(n-1,k)*Binomial(n+k,k)*Binomial(n+k, 2*k+1)* Binomial(2*k,k): k in [0..n-1]]): n in [1..30]]; // G. C. Greubel, Jan 24 2019

Maple

G := (-1/2)*(3*x-3+(x^2-34*x+1)^(1/2))*(x+1)^(-2)*hypergeom([1/3,2/3],[1],(-1/2)*(x^2-7*x+1)*(x+1)^(-3)*(x^2-34*x+1)^(1/2)+(1/2)*(x^3+30*x^2-24*x+1)*(x+1)^(-3))^2;
ogf := 2*x*G/(1-x)-Int((x+1)*G/(x-1)^2,x);
series(ogf,x=0,25); # Mark van Hoeij, May 07 2013

Mathematica

Apery[n_]:= Sum[Binomial[n+k,k]^2 Binomial[n,k]^2,{k,0,n}]; AA[n_]:= Sum[(2k+1)*Apery[k],{k,0,n-1}]/n; Table[AA[n],{n,1,25}]
Table[Sum[(Binomial[n-1,k]*Binomial[n+k,k]*Binomial[n+k,2*k+1]* Binomial[2*k,k]), {k,0,n-1}], {n,1,30}] (* G. C. Greubel, Jan 24 2019 *)

PARI

{a(n) = sum(k=0,n-1, binomial(n-1,k)*binomial(n+k,k)*binomial(n+k, 2*k+1)*binomial(2*k,k))}; \\ G. C. Greubel, Jan 24 2019

Sage

[sum(binomial(n-1,k)*binomial(n+k,k)*binomial(n+k, 2*k+1)* binomial(2*k,k) for k in (0..n-1)) for n in (1..30)] # G. C. Greubel, Jan 24 2019

Formula

Recursion : (n+2)^3 *(n+3) *(2n+1) *a(n+3) = (n+2) *(2n+1) *(35*n^3+193*n^2+345*n+203) *a(n+2) -(n+1) *(2n+5) *(35*n^3+122*n^2+132*n+40) * a(n+1) +n *(n+1)^3 *(2n+5) *a(n).

a(n) = Sum(k=0..n-1, (binomial(n-1,k)* binomial(n+k,k)* binomial(n+k,2*k+1)*binomial(2*k,k)) ). - Zhi-Wei Sun, Jun 17 2010

a(n) = A189766(n) / n = trace( HilbertMatrix(n)^(-1) )/n. - Richard Penner, Jun 04 2011

a(n) = (1/n)*Sum_{k=0..n-1} (2*k+1)*binomial(n+k,2*k+1)^2*binomial(2*k, k)^2. - Richard Penner, Jun 04 2011

G.f.: 2*x*G/(1-x)-Int((x+1)*G/(x-1)^2,x) where G is the generating function of A005259. - Mark van Hoeij, May 07 2013

a(n) ~ 2^(1/4) * (1 + sqrt(2))^(4*n) / (16*(Pi*n)^(3/2)). - Vaclav Kotesovec, Jan 24 2019