A228332 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(6).
1, 68, 1778, 43080, 958430, 20119736, 405350788, 7921691280, 151231519350, 2834134359000, 52320693313020, 953960351550960, 17212782834351468, 307826474156801840, 5462948893700675720, 96303960593503261984, 1687752152779483045542, 29424712141610821296408, 510621541414656188646220
Offset: 0
Keywords
Links
- 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 Remark 3 p. 1882. Omega6(n) = a(n-1).
- Yidong Sun and Fei Ma, Four transformations on the Catalan triangle, arXiv preprint arXiv:1305.2017 [math.CO], 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)^6*(Binomial[2n+1, n-k]*2*(k+1)/(n+k+2))^2,{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Dec 08 2013 *)
Formula
Recurrence: n*(2*n+1)*(105*n^5 - 420*n^4 + 588*n^3 - 356*n^2 + 96*n - 10)*a(n) = 2*(4*n-7)*(4*n-5)*(105*n^5 + 105*n^4 - 42*n^3 - 62*n^2 - 7*n + 3)*a(n-1). - Vaclav Kotesovec, Dec 08 2013
a(n) = binomial(4*n,2*n) * (105*n^5 + 105*n^4 - 42*n^3 - 62*n^2 - 7*n + 3) / ((2*n+1)*(4*n-3)*(4*n-1)). - Vaclav Kotesovec, Dec 08 2013