A105793 Expansion of e.g.f. (1 + y)^(1 + x).

1, 1, 1, 0, 1, 1, 0, -1, 0, 1, 0, 2, -1, -2, 1, 0, -6, 5, 5, -5, 1, 0, 24, -26, -15, 25, -9, 1, 0, -120, 154, 49, -140, 70, -14, 1, 0, 720, -1044, -140, 889, -560, 154, -20, 1, 0, -5040, 8028, -64, -6363, 4809, -1638, 294, -27, 1, 0, 40320, -69264, 8540, 50840, -44835, 17913, -3990, 510, -35, 1

Offset: 0

Author

Paul Barry, Apr 20 2005

Keywords

• easy
• sign
• tabl

Comments

Generalized Stirling number triangle of first kind. Row sums are (1,2,2,0,0,0,...) = 2C(2,n) - 2C(1,n) + C(0,n). Inverse is A105794.

Triangle T(n,k), 0 <= k <= n, read by rows, given by [1, -1, 0, -2, -1, -3, -2, -4, -3, -5, -4, -6, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Aug 23 2006

Examples

			From _Wolfdieter Lang_, Jun 19 2017: (Start)
The triangle T(n, k) starts
  n\k  0     1      2    3     4      5     6     7   8   9 10 ...
  0:   1
  1:   1     1
  2:   0     1      1
  3:   0    -1      0    1
  4:   0     2     -1   -2     1
  5:   0    -6      5    5    -5      1
  6:   0    24    -26  -15    25     -9     1
  7:   0  -120    154   49  -140     70   -14     1
  8:   0   720  -1044 -140   889   -560   154   -20   1
  9:   0 -5040   8028  -64 -6363   4809 -1638   294 -27   1
  10:  0 40320 -69264 8540 50840 -44835 17913 -3990 510 -35  1
  ... reformatted
Recurrence from a- and z-sequence (see above): T(1, 0) = T(0, 0) = 1; T(1, 1) = (1/1)*(1*T(0, 0)) = 1, T(2, 0) = 2*(T(1, 0) - T(1, 1)) = 0, T(2, 1) = (2/1)*(T(1,0) + (-1/2)*T(1, 1)) = 1. T(3, 1) = (3/1)*(0 + (-1/2)*T(2, 1) + (1/6)*T(2, 2)) = -1. (End)
		

Crossrefs

Cf. A007318, A027641/A027642, A033999, A048994, A094645, A269953.

Programs

Mathematica

t[0, 0] = 1; t[n_, k_] := StirlingS1[n, k] + n*StirlingS1[n-1, k]; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Dec 04 2013, after Giuliano Cabrele *)

Formula

E.g.f.: (1+y)^(1+x); rows have g.f. k!*binomial(x+1, k); Columns have g.f. (1+x)*log(1+x)^k.

If we define f(n,i,a) = Sum_{k=0..n-i} binomial(n,k)*Stirling1(n-k,i)*Product_{j=0..k-1} (-a-j), then T(n,i) = f(n,i,-1), for n=1,2,...; i=0...n. - Milan Janjic, Dec 21 2008

So T(n,k) = Stirling1(n,k) + n*Stirling1(n-1,k), Stirling1 being the (signed) Stirling numbers of first kind A048994. In terms of lower triangular matrices, 0<= k <= n, T is also the product [Stirling1] * [Pascal] = [A048994] * [A007318], i.e., T(n,k) = Sum_{j=0..n} Stirling1(n,j) * binomial(j,k). - Giuliano Cabrele, Jan 19 2009

This is the triangle of connection constants for expressing the basis of falling factorial polynomials x_(k) := x*(x-1)*...*(x-k+1) in terms of the polynomial sequence (x-1)^n, that is, x_(n) = Sum_{k = 0..n} T(n,k)*(x-1)^k. - Peter Bala, Jul 10 2013

From Wolfdieter Lang, Jun 19 2017: (Start)

Triangle T is the (infinite) matrix product of A048994 (Stirling1) and A007318 (Pascal): T(n,k) = Sum_{m=k..n} Stirling1(n, m)*Pascal(m, k), n >= k >= 0, and 0 for n < k. Note that the Pascal matrix is Sheffer (e^t, t) of the Appell type.

T is the Sheffer (aka exponential Riordan) matrix (1+t, log(1+t)).

E.g.f. column k: (1+x)*(log(1+x))^k/k!, k >= 0.

The a-sequence for T is A027641/A027642 (Bernoulli), and the z-sequence is A033999 (repeat(1,-1)) (see a W. Lang link under A006232 for a- and z-sequences for Sheffer matrices, also for references).

Therefore the combined recurrence is: T(n, 0) = n*Sum_{j=0..n-1} (-1)^j*T(n-1, j), n >= 1, T(0, 0) = 1, and T(n, m) = (n/m)*Sum_{j=0..n-m} binomial(m-1+j, m-1)*Bernoulli(j)*T(n-1, m-1+j), n >= m >= 1. (End)