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).

1, 132, 4260, 120400, 3017700, 69776784, 1524611088, 31951782720, 648578888100, 12837530477200, 248966505964176, 4747739344525632, 89267646282614800, 1658349027407016000, 30489930211792680000, 555544747397829254400, 10042477557290424843300, 180267292319119226298000, 3215718323211443887530000

Offset

0,2

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..200

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 *)

Crossrefs

Cf. A000108, A039598, A024492, A000894, A228329, A000515, A228330, A228331, A228332.

Sequence in context: A168180 A236260 A264179 * A035837 A335857 A270411

Adjacent sequences: A228330 A228331 A228332 * A228334 A228335 A228336

Keyword

nonn

Author

N. J. A. Sloane, Aug 26 2013

Status

approved