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%I #46 Mar 02 2019 12:07:40
%S 1,8,98,1320,18590,268736,3952228,58837680,883941750,13373883600,
%T 203487733020,3110407163760,47726453450988,734694122886080,
%U 11341161925265480,175489379096245984,2721169178975361702,42273090191785999728,657788911222324942060,10250564041646388681200
%N a(n) = Sum_{k=0..n} (k+1)^2*T(n,k)^2 where T(n,k) is the Catalan triangle A039598.
%C 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(2).
%H G. C. Greubel, <a href="/A228329/b228329.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. Omega2(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)*a(n) + 2*(-26*n^2+25*n-11)*a(n-1) + 20*(4*n-5)*(4*n-7)*a(n-2) = 0. - _R. J. Mathar_, Sep 08 2013
%F a(n) = ((4n)!*(3n+1))/((2n)!^2*(2n+1)) = binomial(4n,2n)*(3n+1)/(2n+1). - _Philippe Deléham_, Nov 25 2013
%F From _Peter Luschny_, Nov 26 2013: (Start)
%F a(n) = 16^n*(3*n+1)*gamma(2*n+1/2)/(sqrt(Pi)*gamma(2*n+2)).
%F a(n) = a(n-1)*(6*n+2)*(4*n-3)*(4*n-1)/(n*(2*n+1)*(3*n-2)) if n > 0 else 1.
%F a(n) = [x^n] I*HeunG(8/5,0,-1/4,1/4,3/2,1/2,16*x)/sqrt(16*x-1) where [x^n] f(x) is the coefficient of x^n in f(x) and HeunG is the Heun general function. (End)
%p B:=(n,k)->binomial(2*n, n-k) - binomial(2*n, n-k-2); #A039598
%p Omega:=(m,n)->add((k+1)^m*B(n,k)^2,k=0..n);
%p h:=m->[seq(Omega(m,n),n=0..20)];
%p h(2);
%p # Second solution:
%p h := n -> I*HeunG(8/5,0,-1/4,1/4,3/2,1/2,16*x)/sqrt(16*x-1);
%p seq(coeff(series(h(x),x,n+2),x,n),n=0..19); # _Peter Luschny_, Nov 26 2013
%t a[n_] := Binomial[4n, 2n] (3n+1)/(2n+1);
%t Table[a[n], {n, 0, 19}] (* _Jean-François Alcover_, Jul 30 2018, after _Philippe Deléham_ *)
%o (Sage)
%o @CachedFunction
%o def A228329(n):
%o return A228329(n-1)*(6*n+2)*(4*n-3)*(4*n-1)/(n*(2*n+1)*(3*n-2)) if n>0 else 1
%o [A228329(n) for n in (0..19)] # _Peter Luschny_, Nov 26 2013
%Y Cf. A039598, A000108, A024492 (h(0)), A000894 (h(1)), A000515 (h(3)), A228330 (h(4)), A228331 (h(5)) - A228333 (h(7)).
%Y Cf. A000142, A007318.
%K nonn
%O 0,2
%A _N. J. A. Sloane_, Aug 26 2013
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