A228331 Let h(m) denote the sequence whose n-th term is Sum__{k=0..n} (k+1)^m*T(n,k)^2, where T(n,k) is the Catalan triangle A039598. This is h(5).
1, 36, 780, 16240, 321300, 6131664, 114017904, 2079380160, 37356642180, 663144710800, 11657925495216, 203295462691776, 3521108298744400, 60632838691387200, 1038859802556120000, 17721669103065158400, 301147406355880764900, 5099997408534884394000, 86106549929771707182000
Offset: 0
Keywords
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..200
- Pedro J. Miana and Natalia Romero, Moments of combinatorial and Catalan numbers, Journal of Number Theory, Volume 130, Issue 8, August 2010, Pages 1876-1887. See Omega5. Remark 3 p. 1882.
- Yidong Sun and Fei Ma, Four transformations on the Catalan triangle, arXiv preprint arXiv:1305.2017, 2013
- Yidong Sun and Fei Ma, Some new binomial sums related to the Catalan triangle, Electronic Journal of Combinatorics 21(1) (2014), #P1.33
Programs
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Mathematica
Table[Sum[(k+1)^5*(Binomial[2n+1, n-k]*2*(k+1)/(n+k+2))^2,{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Dec 08 2013 *)
Formula
Conjecture: n^2*a(n) +4*(2*n^2-22*n+11)*a(n-1) +16*(-7*n^2-54*n+166)*a(n-2) -1088*(2*n-3)*(2*n-7)*a(n-3)=0. - R. J. Mathar, Sep 08 2013
Recurrence: n^2*(3*n^2 - 5*n + 1)*a(n) = 4*(2*n-3)*(2*n+1)*(3*n^2 + n - 1)*a(n-1). - Vaclav Kotesovec, Dec 08 2013
a(n) = binomial(2*n,n)^2 * (2*n+1)*(3*n^2+n-1)/(2*n-1). - Vaclav Kotesovec, Dec 08 2013
G.f.: ((28*x+3)*hypergeom([1/2, 5/2],[1],16*x)+20*x*(1-16*x)*hypergeom([3/2, 7/2],[2],16*x))/3. - Mark van Hoeij, Apr 12 2014