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A112007 Coefficient triangle for polynomials used for o.g.f.s for unsigned Stirling1 diagonals. 29
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 (list; table; graph; refs; listen; history; text; internal format)
OFFSET
0,2
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
LINKS
Robert Israel, Table of n, a(n) for n = 0..10010 (rows 0 to 140, flattened)
C. M. Bender, D. C. Brody and B. K. Meister, Bernoulli-like polynomials associated with Stirling Numbers, arXiv:math-ph/0509008 [math-ph], 2005.
Wolfdieter Lang, First 10 rows.
Wolfdieter Lang, On Generating functions of Diagonals Sequences of Sheffer and Riordan Number Triangles, arXiv:1708.01421 [math.NT], August 2017.
R. Paris, A uniform asymptotic expansion for the incomplete gamma function, Journal of Computational and Applied Mathematics, 148 (2002), p. 223-239. See 234. [Tom Copeland, Jan 03 2016]
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
EXAMPLE
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.
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 *)
PROG
(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
CROSSREFS
Row sums give A001147(k+1) = (2*k+1)!!, k>=0.
Sequence in context: A110608 A318397 A190015 * A113374 A136470 A363747
KEYWORD
nonn,easy,tabl
AUTHOR
Wolfdieter Lang, Sep 12 2005
STATUS
approved

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Last modified March 19 03:33 EDT 2024. Contains 370952 sequences. (Running on oeis4.)