%I #30 Sep 08 2022 08:46:05
%S 1,20,362,6504,114686,1992536,34231540,583027920,9862508790,
%T 165918037560,2778642667020,46358257249200,770951008563372,
%U 12785838603285104,211540243555702376,3492587812271418784,57557091526140668070,946970607665938615032,15557339429900195819164,255246113991506558429936
%N 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(4).
%H G. C. Greubel, <a href="/A228330/b228330.txt">Table of n, a(n) for n = 0..825</a>
%H Pedro J. Miana, 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. Omega4(n) = a(n-1).
%H Yidong Sun and Fei Ma, <a href="http://arxiv.org/abs/1305.2017">Four transformations on the Catalan triangle</a>, arXiv:1305.2017 [math.CO], 2013.
%H Yidong Sun and Fei Ma, <a href="http://www.combinatorics.org/ojs/index.php/eljc/article/view/v21i1p33">Some new binomial sums related to the Catalan triangle</a>, Electronic Journal of Combinatorics 21(1) (2014), #P1.33.
%F Conjecture: n*(2*n+1)*(3467*n-4029)*a(n) + 8*(-36721*n^3 + 109040*n^2 - 137926*n + 69822)*a(n-1) + 4*(4*n-9)*(45706*n-7907)*(4*n-7)*a(n-2) = 0. - _R. J. Mathar_, Sep 08 2013
%F Recurrence: n*(2*n+1)*(15*n^3 - 30*n^2 + 16*n - 2)*a(n) = 2*(4*n-5)*(4*n-3)*(15*n^3 + 15*n^2 + n - 1)*a(n-1). - _Vaclav Kotesovec_, Dec 08 2013
%F From _Vaclav Kotesovec_, Dec 08 2013: (Start)
%F a(n) = binomial(4*n,2*n) * (15*n^3+15*n^2+n-1)/((2*n+1)*(4*n-1)).
%F a(n) = 4*Sum_{k=0..n} (k+1)^6*(binomial(2*n+1, n-k)/(n+k+2))^2. (End)
%t Table[4*Sum[(k+1)^6*(Binomial[2n+1, n-k]/(n+k+2))^2, {k,0,n}], {n,0,20}] (* _Vaclav Kotesovec_, Dec 08 2013 *)
%o (PARI) vector(20, n, n--; binomial(4*n,2*n)*(15*n^3+15*n^2+n-1)/((2*n+1)*(4*n-1))) \\ _G. C. Greubel_, Mar 02 2019
%o (Magma) [Binomial(4*n,2*n)*(15*n^3+15*n^2+n-1)/((2*n+1)*(4*n-1)): n in [0..20]]; // _G. C. Greubel_, Mar 02 2019
%o (Sage) [binomial(4*n,2*n)*(15*n^3+15*n^2+n-1)/((2*n+1)*(4*n-1)) for n in (0..20)] # _G. C. Greubel_, Mar 02 2019
%o (GAP) List([0..20], n-> Binomial(4*n,2*n)*(15*n^3+15*n^2+n-1)/((2*n+1)*(4*n-1))) # _G. C. Greubel_, Mar 02 2019
%Y Cf. A000108, A039598, A024492 (h(0)), A000894 (h(1)), A228329 (h(2)), A000515 (h(3)), this sequence (h(4)), A228331 (h(5)), A228332 (h(6)), A228333 (h(7)).
%K nonn
%O 0,2
%A _N. J. A. Sloane_, Aug 26 2013