%I #23 Jun 05 2021 01:37:48
%S 1,1,2,7,31,162,968,6481,47893,386098,3364562,31460324,313743665,
%T 3320211313,37124987124,436985496790,5397178181290,69748452377058,
%U 940762812167126,13213888481979449,192891251215160017
%N Eigensequence of triangle A039599: a(n) = Sum_{k=0..n-1} A039599(n-1,k)*a(k) for n > 0 with a(0) = 1.
%C Starting with offset 1, these are the row sums of triangle A147294. - _Gary W. Adamson_, Nov 05 2008
%H Vaclav Kotesovec, <a href="/A125275/b125275.txt">Table of n, a(n) for n = 0..500</a>
%H Guo-Niu Han, <a href="/A196265/a196265.pdf">Enumeration of Standard Puzzles</a>, 2011. [Cached copy]
%H Guo-Niu Han, <a href="https://arxiv.org/abs/2006.14070">Enumeration of Standard Puzzles</a>, arXiv:2006.14070 [math.CO], 2020.
%F a(n) = Sum_{k=0..n-1} a(k) * C(2*n-1, n-k-1) * (2*k + 1)/(2*n - 1) for n > 0 with a(0) = 1.
%e a(3) = 2*(1) + 3*(1) + 1*(2) = 7;
%e a(4) = 5*(1) + 9*(1) + 5*(2) + 1*(7) = 31;
%e a(5) = 14*(1) + 28*(1) + 20*(2) + 7*(7) + 1*(31) = 162.
%e Triangle A039599(n,k) = C(2*n+1, n-k)*(2*k+1)/(2*n+1) (with rows n >= 0 and columns k = 0..n) begins:
%e 1;
%e 1, 1;
%e 2, 3, 1;
%e 5, 9, 5, 1;
%e 14, 28, 20, 7, 1;
%e 42, 90, 75, 35, 9, 1;
%e ...
%e where the g.f. of column k is G000108(x)^(2*k+1)
%e and G000108(x) = (1 - sqrt(1 - 4*x))/(2*x) is the Catalan g.f. function.
%t A125275=ConstantArray[0,20]; A125275[[1]]=1; Do[A125275[[n]]=Binomial[2*n-1,n-1]/(2*n-1)+Sum[A125275[[k]]*Binomial[2*n-1,n-k-1]*(2*k+1)/(2*n-1),{k,1,n-1}];,{n,2,20}]; Flatten[{1,A125275}] (* _Vaclav Kotesovec_, Dec 09 2013 *)
%o (PARI) a(n)=if(n==0,1,sum(k=0,n-1, a(k)*binomial(2*n-1, n-k-1)*(2*k+1)/(2*n-1)))
%Y Cf. A000108, A039599, A125276 (variant), A147294.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Nov 26 2006