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%I A228333 #39 Oct 16 2025 12:00:27
%S A228333 1,132,4260,120400,3017700,69776784,1524611088,31951782720,
%T A228333 648578888100,12837530477200,248966505964176,4747739344525632,
%U A228333 89267646282614800,1658349027407016000,30489930211792680000,555544747397829254400,10042477557290424843300,180267292319119226298000,3215718323211443887530000
%N 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).
%H A228333 Vincenzo Librandi, Table of n, a(n) for n = 0..200
%H A228333 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 Omega7. Remark 3 p. 1882.
%H A228333 Yidong Sun and Fei Ma, Four transformations on the Catalan triangle, arXiv preprint arXiv:1305.2017 [math.CO], 2013.
%H A228333 Yidong Sun and Fei Ma, Some new binomial sums related to the Catalan triangle, Electronic Journal of Combinatorics 21(1) (2014), #P1.33.
%F A228333 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
%F A228333 From _Vaclav Kotesovec_, Dec 08 2013: (Start)
%F A228333 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).
%F A228333 a(n) = binomial(2*n,n)^2 * (2*n+1)*(6*n^3+6*n^2-1)/(2*n-1). (End)
%F A228333 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
%F A228333 a(n) ~ 3 * 2^(4*n+1) * n^2 / Pi. - _Amiram Eldar_, Oct 16 2025
%t A228333 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 *)
%Y A228333 Cf. A000108, A039598, A024492, A000894, A228329, A000515, A228330, A228331, A228332.
%K A228333 nonn,easy
%O A228333 0,2
%A A228333 _N. J. A. Sloane_, Aug 26 2013
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