cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A112007 Coefficient triangle for polynomials used for o.g.f.s for unsigned Stirling1 diagonals.

Original entry on oeis.org

1, 2, 1, 6, 8, 1, 24, 58, 22, 1, 120, 444, 328, 52, 1, 720, 3708, 4400, 1452, 114, 1, 5040, 33984, 58140, 32120, 5610, 240, 1, 40320, 341136, 785304, 644020, 195800, 19950, 494, 1, 362880, 3733920, 11026296, 12440064, 5765500, 1062500, 67260, 1004, 1
Offset: 0

Views

Author

Wolfdieter Lang, Sep 12 2005

Keywords

Comments

This is the row reversed second-order Eulerian triangle A008517(k+1,k+1-m). For references see A008517.
The o.g.f. for the k-th diagonal, k >= 1, of the unsigned Stirling1 triangle |A008275| is G1(1,x)=1/(1-x) if k=1 and G1(k,x) = g1(k-2,x)/(1-x)^(2*k-1), if k >= 2, with the row polynomials g1(k;x):=Sum_{m=0..k} a(k,m)*x^m.
The recurrence eq. for the row polynomials is g1(k,x)=((k+1)+k*x)*g1(k-1,x) + x*(1-x)*(d/dx)g1(k-1,x), k >= 1, with input g1(0,x):=1.
The column sequences start with A000142 (factorials), A002538, A002539, A112008, A112485.
This o.g.f. computation was inspired by Bender et al. article where the Stirling polynomials have been rediscussed.
The A163936 triangle is identical to the triangle given above except for an extra right hand column [1, 0, 0, 0, ... ]. The A163936 triangle is related to the higher order exponential integrals E(x,m,n), see A163931 and A163932. - Johannes W. Meijer, Oct 16 2009

Examples

			Triangle begins:
    1;
    2,   1;
    6,   8,   1;
   24,  58,  22,   1;
  120, 444, 328,  52,   1;
  ...
G.f. for k=3 sequence A000914(n-1), [2,11,35,85,175,322,546,...], is G1(3,x)= g1(1,x)/(1-x)^5= (2+x)/(1-x)^5.
		

Crossrefs

Row sums give A001147(k+1) = (2*k+1)!!, k>=0.

Programs

  • Maple
    a:= proc(k,m) option remember; if m >= 0 and k >= 0 then (k+m+1)*procname(k-1,m)+(k-m+1)*procname(k-1,m-1) else 0 fi end proc:
    a(0,0):= 1:
    seq(seq(a(k,m),m=0..k),k=0..10); # Robert Israel, Jul 20 2017
  • Mathematica
    a[k_, m_] = Sum[(-1)^(k + n + 1)*Binomial[2k + 3, n]*StirlingS1[m + k - n + 2, m + 1 - n], {n, 0, m}]; Flatten[Table[a[k, m], {k, 0, 8}, {m, 0, k}]][[1 ;; 45]] (* Jean-François Alcover, Jun 01 2011, after Johannes W. Meijer *)
  • PARI
    a(k, m)=sum(n=0, m, (-1)^(k + n + 1)*binomial(2*k + 3, n)*stirling(m + k - n + 2, m + 1 - n, 1));
    for(k=0, 10, for(m=0, k, print1(a(k, m),", "))) \\ Indranil Ghosh, Jul 21 2017

Formula

a(k, m) = (k+m+1)*a(k-1, m) + (k-m+1)*a(k-1, m-1), if k >= m >= 0, a(0, 0)=1; a(k, -1):=0, otherwise 0.
a(k,m) = Sum_{n=0..m} (-1)^(k+n+1)*C(2*k+3,n)*Stirling1(m+k-n+2,m+1-n). - Johannes W. Meijer, Oct 16 2009
The compositional inverse (with respect to x) of y = y(t,x) = (x+t*log(1-x)) is x = x(t,y) = 1/(1-t)*y + t/(1-t)^3*y^2/2! + (2*t+t^2)/(1-t)^5*y^3/3! + (6*t+8*t^2+t^3)/(1-t)^7*y^4/4! + .... The numerator polynomials of the rational functions in t are the row polynomials of this triangle. As observed above, the rational functions in t are the generating functions for the diagonals of |A008275|. See the Bala link for a proof. Cf. A008517. - Peter Bala, Dec 02 2011