A370516 Triangle of numbers read by rows: T(n,k) = Sum_{i=0..n-k} binomial(n+1,n-k-i)*Stirling2(i+3,i+1)*(-1)^i for n >= 0, 0 <= k <= n.

1, -5, 1, 7, -4, 1, -3, 3, -3, 1, 0, 0, 0, -2, 1, 0, 0, 0, -2, -1, 1, 0, 0, 0, -2, -3, 0, 1, 0, 0, 0, -2, -5, -3, 1, 1, 0, 0, 0, -2, -7, -8, -2, 2, 1, 0, 0, 0, -2, -9, -15, -10, 0, 3, 1, 0, 0, 0, -2, -11, -24, -25, -10, 3, 4

Offset

0,2

Comments

Generalized binomial coefficients of the second order.

Links

Table of n, a(n) for n=0..64.

Igor Victorovich Statsenko, On the ordinal numbers of triangles of generalized special numbers, Innovation science No 2-2, State Ufa, Aeterna Publishing House, 2024, pp. 15-19. In Russian.

Formula

T(n,k) = Sum_{i=0..n-k} binomial(n+1,n-k-i)*Stirling2(i+m+1,i+1)*(-1)^i where m = 2 for n >= 0, 0 <= k <= n.

Example

n\k   0    1    2    3    4    5    6
0:    1
1:   -5    1
2:    7   -4    1
3:   -3    3   -3    1
4:    0    0    0   -2    1
5:    0    0    0   -2   -1    1
6:    0    0    0   -2   -3    0    1

Maple

C:=(n, k)->n!/(k!*(n-k)!) : T:=(m, n, k)->sum(C(n+1, n-k-r)*Stirling2(r+m+1, r+1)*((-1)^r), r=0..n-k) : m:=2 : seq(seq T(m, n, k), k=0..n), n=0..10);

Crossrefs

For m=0 the formula gives the sequence A007318; for m=1 the formula gives the sequence A159854. In this case, we assume that A007318 consists of generalized binomial coefficients of order zero and A159854 consists of generalized binomial coefficients of order one.

Sequence in context: A195429 A158552 A322050 * A021663 A099218 A198129

Adjacent sequences: A370513 A370514 A370515 * A370517 A370518 A370519

Keyword

tabl,sign

Author

Igor Victorovich Statsenko, Feb 21 2024

Status

approved