%I #16 Feb 10 2015 17:19:51
%S 1,0,1,0,1,2,1,0,4,12,13,6,1,0,36,132,193,144,58,12,1,0,576,2400,4180,
%T 3980,2273,800,170,20,1,0,14400,65760,129076,143700,100805,46710,
%U 14523,3000,395,30,1,0,518400,2540160,5450256,6787872,5482456,3034920,1184153
%N Triangle read by rows, T(n,k) = Sum_{j=0..k-1} S(n,j+1)*S(n,k-j) where S denotes the Stirling cycle numbers A132393, T(0,0)=1, n>=0, 0<=k<=2n-1.
%F T(n+1, n+1) = A129256(n) for n>=0.
%e [1]
%e [0, 1]
%e [0, 1, 2, 1]
%e [0, 4, 12, 13, 6, 1]
%e [0, 36, 132, 193, 144, 58, 12, 1]
%e [0, 576, 2400, 4180, 3980, 2273, 800, 170, 20, 1]
%p a := n -> (x^n*pochhammer(1+1/x,n))^2:
%p c := (n,k) -> coeff(expand(a(n)),x,n-k):
%p for n from 0 to 5 do: `if`(n=0,[1],[seq(c(n-1,k),k=-n..n-1)]) od;
%p # Second program, a special case of the recurrence given in A246117:
%p t := proc(n,k) option remember; if n=0 and k=0 then 1 elif
%p k <= 0 or k>n then 0 else iquo(n,2)*t(n-1,k)+t(n-1,k-1) fi end:
%p A254882 := (n,k) -> `if`(n=0,1,t(2*n-1,k)):
%p seq(print(seq(A254882(n,k), k=0..max(0,2*n-1))), n=0..5);
%t Flatten[{1,Table[Table[Sum[Abs[StirlingS1[n,j+1]] * Abs[StirlingS1[n,k-j]],{j,0,k-1}],{k,0,2*n-1}],{n,1,10}]}] (* _Vaclav Kotesovec_, Feb 10 2015 *)
%o (Sage)
%o def A254882(n,k):
%o if n == 0: return 1
%o return sum(stirling_number1(n,j+1)*stirling_number1(n,k-j) for j in range(k))
%o for n in range (5): [A254882(n,k) for k in (0..max(0,2*n-1))]
%Y Cf. A246117, A254881, A129256.
%K nonn,tabf
%O 0,6
%A _Peter Luschny_, Feb 10 2015