A370518 Triangle of numbers read by rows: T(n,k) = Sum_{i=0..n} binomial(n,i)*(n-i)!*Stirling1(i,k)*TC(n,i) where TC(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, 14, -9, 1, -18, 29, -12, 1, 0, -22, 35, -14, 1, 0, -26, 15, 25, -15, 1, 0, -60, 4, 75, -5, -15, 1, 0, -204, -56, 259, 70, -56, -14, 1, 0, -912, -484, 1092, 609, -168, -126, -12, 1, 0, -5040, -3708, 5480, 4599, -231, -882, -210, -9, 1, 0, -33120, -30024, 31820, 36350, 3675, -6027, -2370, -300, -5, 1

Offset: 0

Author

Igor Victorovich Statsenko, Feb 21 2024

Keywords

• tabl
• sign

Comments

Generalized Stirling numbers of the first kind of the second order.

Examples

			n\k     0     1     2     3     4     5     6
0:      1
1:     -5     1
2:     14    -9     1
3:    -18    29   -12     1
4:      0   -22    35   -14     1
5:      0   -26    15    25   -15     1
6:      0   -60     4    75    -5   -15     1
		

Links

• 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.

Crossrefs

For m=0 the formula gives the sequence A130534; for m=1 the formula gives the sequence A094645. In this case, we assume that A130534 consists of generalized Stirling numbers of the first kind of zero order, and A094645 consists of generalized Stirling numbers of the first kind of the first order.

Programs

Maple

C:=(n,k)->n!/(k!*(n-k)!) : T0:=(m,n,k)->sum(C(n+1,n-k-p)*Stirling2(p+m+1,p+1)*((-1)^p), p=0..n-k) : T:=(m,n,k)->sum(C(n,r)*(n-r)!*Stirling1(r,k)*T0(m,n,r), r=0..n)  m:=2 : seq(seq T(m,n,k), k=0..n), n=0..10);

Formula

T(n,k) = Sum_{i=0..n} binomial(n,i)*(n-i)!*Stirling1(i,k)*TC(m,n,i) where TC(m,n,k) = Sum_{i=0..n-k} binomial(n+1,n-k-i)*Stirling2(i+m+1,i+1)*(-1)^(i),m = 2 for n >= 0.