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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.
9
1, 8, 127, 2624, 61501, 1552760, 41186755, 1131614720, 31923047665, 919243356008, 26908963456783, 798379043762624, 23954974906866901, 725620080605773592, 22159617936375571627, 681528994326392115200, 21090805673899997148025, 656256696917886135153800
OFFSET
1,2
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
LINKS
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
EXAMPLE
For n=3 we have a(3)=(A_0+3A_1+5A_2)/3=(1+3*5+5*73)/3=127.
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 *)
PROG
(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
(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
(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
(GAP) List([1..30], n-> Sum([0..n], j-> Binomial(n-1, k)*Binomial(n+k, k) *Binomial(n+k, 2*k+1)* Binomial(2*k, k) )) # G. C. Greubel, Jan 24 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Jun 14 2010
STATUS
approved