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A112007 Coefficient triangle for polynomials used for o.g.f.s for unsigned Stirling1 diagonals. 29

%I #59 Mar 27 2020 20:11:28

%S 1,2,1,6,8,1,24,58,22,1,120,444,328,52,1,720,3708,4400,1452,114,1,

%T 5040,33984,58140,32120,5610,240,1,40320,341136,785304,644020,195800,

%U 19950,494,1,362880,3733920,11026296,12440064,5765500,1062500,67260,1004,1

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

%C This is the row reversed second-order Eulerian triangle A008517(k+1,k+1-m). For references see A008517.

%C 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.

%C 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.

%C The column sequences start with A000142 (factorials), A002538, A002539, A112008, A112485.

%C This o.g.f. computation was inspired by Bender et al. article where the Stirling polynomials have been rediscussed.

%C 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

%H Robert Israel, <a href="/A112007/b112007.txt">Table of n, a(n) for n = 0..10010</a> (rows 0 to 140, flattened)

%H Peter Bala, <a href="/A112007/a112007_Bala.txt">Diagonals of triangles with generating function exp(t*F(x))</a>

%H C. M. Bender, D. C. Brody and B. K. Meister, <a href="https://arxiv.org/abs/math-ph/0509008">Bernoulli-like polynomials associated with Stirling Numbers</a>, arXiv:math-ph/0509008 [math-ph], 2005.

%H E. Burlachenko, <a href="https://arxiv.org/abs/1907.12272">Composition polynomials of the RNA matrix and B-composition polynomials of the Riordan pseudo-involution</a>, arXiv:1907.12272 [math.NT], 2019.

%H Tom Copeland, <a href="http://tcjpn.wordpress.com/2015/12/21/generators-inversion-and-matrix-binomial-and-integral-transforms/">Generators, Inversion, and Matrix, Binomial, and Integral Transforms</a>

%H Wolfdieter Lang, <a href="/A112007/a112007.txt">First 10 rows.</a>

%H Wolfdieter Lang, <a href="https://arxiv.org/abs/1708.01421">On Generating functions of Diagonals Sequences of Sheffer and Riordan Number Triangles</a>, arXiv:1708.01421 [math.NT], August 2017.

%H R. Paris, <a href="http://dx.doi.org/10.1016/S0377-0427(02)00553-8">A uniform asymptotic expansion for the incomplete gamma function</a>, Journal of Computational and Applied Mathematics, 148 (2002), p. 223-239. See 234. [_Tom Copeland_, Jan 03 2016]

%H Andrew Elvey Price, Alan D. Sokal, <a href="https://arxiv.org/abs/2001.01468">Phylogenetic trees, augmented perfect matchings, and a Thron-type continued fraction (T-fraction) for the Ward polynomials</a>, arXiv:2001.01468 [math.CO], 2020.

%F 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.

%F 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

%F 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

%e Triangle begins:

%e 1;

%e 2, 1;

%e 6, 8, 1;

%e 24, 58, 22, 1;

%e 120, 444, 328, 52, 1;

%e ...

%e 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.

%p 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:

%p a(0,0):= 1:

%p seq(seq(a(k,m),m=0..k),k=0..10); # _Robert Israel_, Jul 20 2017

%t 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_ *)

%o (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));

%o for(k=0, 10, for(m=0, k, print1(a(k, m),", "))) \\ _Indranil Ghosh_, Jul 21 2017

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

%K nonn,easy,tabl

%O 0,2

%A _Wolfdieter Lang_, Sep 12 2005

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