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

0,6

Links

Table of n, a(n) for n=0..50.

Formula

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

Example

[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]

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 *)

Prog

(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))]

Crossrefs

Cf. A246117, A254881, A129256.

Sequence in context: A077929 A178039 A185411 * A086095 A322119 A363731

Adjacent sequences: A254879 A254880 A254881 * A254883 A254884 A254885

Keyword

nonn,tabf

Author

Peter Luschny, Feb 10 2015

Status

approved