login
A228333
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(7).
6
1, 132, 4260, 120400, 3017700, 69776784, 1524611088, 31951782720, 648578888100, 12837530477200, 248966505964176, 4747739344525632, 89267646282614800, 1658349027407016000, 30489930211792680000, 555544747397829254400, 10042477557290424843300, 180267292319119226298000, 3215718323211443887530000
OFFSET
0,2
LINKS
Pedro J. Miana, Natalia Romero, Moments of combinatorial and Catalan numbers, Journal of Number Theory, Volume 130, Issue 8, August 2010, Pages 1876-1887. See Omega7. Remark 3 p. 1882.
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.
FORMULA
Conjecture: n^2*(304*n-411)*a(n) + 4*(-1814*n^3+2554*n^2-4776*n+7567)*a(n-1) + 32*(2*n-5)*(2*n-1)*(299*n-176)*a(n-2) = 0. - R. J. Mathar, Dec 04 2013
Recurrence: n^2*(6*n^3 - 12*n^2 + 6*n - 1)*a(n) = 4*(2*n-3)*(2*n+1)*(6*n^3 + 6*n^2 - 1)*a(n-1). - Vaclav Kotesovec, Dec 08 2013
a(n) = binomial(2*n,n)^2 * (2*n+1)*(6*n^3+6*n^2-1)/(2*n-1). - Vaclav Kotesovec, Dec 08 2013
G.f.: ((256*x+3)*hypergeom([1/2, 5/2],[1],16*x)+80*(38*x+1)*x*hypergeom([3/2, 7/2],[2],16*x))/3. - Mark van Hoeij, Apr 12 2014
MATHEMATICA
Table[Sum[(k+1)^7*(Binomial[2n+1, n-k]*2*(k+1)/(n+k+2))^2, {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Dec 08 2013 *)
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, Aug 26 2013
STATUS
approved