A254882 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.

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, 3980, 2273, 800, 170, 20, 1, 0, 14400, 65760, 129076, 143700, 100805, 46710, 14523, 3000, 395, 30, 1, 0, 518400, 2540160, 5450256, 6787872, 5482456, 3034920, 1184153

Offset: 0

Author

Peter Luschny, Feb 10 2015

Keywords

• nonn
• tabf

Examples

			[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, 3980, 2273, 800, 170, 20, 1]
		

Crossrefs

Cf. A246117, A254881, A129256.

Programs

Maple

a := n -> (x^n*pochhammer(1+1/x,n))^2:
c := (n,k) -> coeff(expand(a(n)),x,n-k):
for n from 0 to 5 do: `if`(n=0,[1],[seq(c(n-1,k),k=-n..n-1)]) od;
# Second program, a special case of the recurrence given in A246117:
t := proc(n,k) option remember; if n=0 and k=0 then 1 elif
k <= 0 or k>n then 0 else iquo(n,2)*t(n-1,k)+t(n-1,k-1) fi end:
A254882 := (n,k) -> `if`(n=0,1,t(2*n-1,k)):
seq(print(seq(A254882(n,k), k=0..max(0,2*n-1))), n=0..5);

Mathematica

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 *)

Sage

def A254882(n,k):
    if n == 0: return 1
    return sum(stirling_number1(n,j+1)*stirling_number1(n,k-j) for j in range(k))
for n in range (5): [A254882(n,k) for k in (0..max(0,2*n-1))]

Formula

T(n+1, n+1) = A129256(n) for n>=0.