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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).

%I A228332 #26 Feb 17 2025 01:33:11
%S A228332 1,68,1778,43080,958430,20119736,405350788,7921691280,151231519350,
%T A228332 2834134359000,52320693313020,953960351550960,17212782834351468,
%U A228332 307826474156801840,5462948893700675720,96303960593503261984,1687752152779483045542,29424712141610821296408,510621541414656188646220
%N 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).
%H A228332 Pedro J. Miana and Natalia Romero, <a href="https://doi.org/10.1016/j.jnt.2010.01.018">Moments of combinatorial and Catalan numbers</a>, Journal of Number Theory, Volume 130, Issue 8, August 2010, Pages 1876-1887. See Remark 3 p. 1882. Omega6(n) = a(n-1).
%H A228332 Yidong Sun and Fei Ma, <a href="http://arxiv.org/abs/1305.2017">Four transformations on the Catalan triangle</a>, arXiv preprint arXiv:1305.2017 [math.CO], 2013.
%H A228332 Yidong Sun and Fei Ma, <a href="https://doi.org/10.37236/3701">Some new binomial sums related to the Catalan triangle</a>, Electronic Journal of Combinatorics 21(1) (2014), #P1.33.
%F A228332 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
%F A228332 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
%t A228332 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 *)
%Y A228332 Cf. A000108, A039598, A024492, A000894, A228329, A000515, A228330, A228331, A228333.
%K A228332 nonn
%O A228332 0,2
%A A228332 _N. J. A. Sloane_, Aug 26 2013