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.

A112486 Coefficient triangle for polynomials used for e.g.f.s for unsigned Stirling1 diagonals.

Original entry on oeis.org

1, 1, 1, 2, 5, 3, 6, 26, 35, 15, 24, 154, 340, 315, 105, 120, 1044, 3304, 4900, 3465, 945, 720, 8028, 33740, 70532, 78750, 45045, 10395, 5040, 69264, 367884, 1008980, 1571570, 1406790, 675675, 135135, 40320, 663696, 4302216, 14777620, 29957620
Offset: 0

Views

Author

Wolfdieter Lang, Sep 12 2005

Keywords

Comments

The k-th diagonal of |A008275| appears as the k-th column in |A008276| with k-1 leading zeros.
The recurrence, given below, is derived from (d/dx)g1(k,x) - g1(k,x)= x*(d/dx)g1(k-1,x) + g1(k-1,x), k >= 1, with input g(-1,x):=0 and initial condition g1(k,0)=1, k >= 0. This differential recurrence for the e.g.f. g1(k,x) follows from the one for unsigned Stirling1 numbers.
The column sequences start with A000142 (factorials), A001705, A112487- A112491, for m=0,...,5.
The main diagonal gives (2*k-1)!! = A001147(k), k >= 1.
This computation was inspired by the Bender article (see links), where the Stirling polynomials are discussed.
The e.g.f. for the k-th diagonal, k >= 1, of the unsigned Stirling1 triangle |A008275| with k-1 leading zeros is g1(k-1,x) = exp(x)*Sum_{m=0..k-1} a(k,m)*(x^(k-1+m))/(k-1+m)!.
a(k,n) = number of lists with entries from [n] such that (i) each element of [n] occurs at least once and at most twice, (ii) for each i that occurs twice, all entries between the two occurrences of i are > i, and (iii) exactly k elements of [n] occur twice. Example: a(1,2)=5 counts 112, 121, 122, 211, 221, and a(2,2)=3 counts 1122,1221,2211. - David Callan, Nov 21 2011

Examples

			Triangle begins:
    1;
    1,    1;
    2,    5,     3;
    6,   26,    35,    15;
   24,  154,   340,   315,   105;
  120, 1044,  3304,  4900,  3465,   945;
  720, 8028, 33740, 70532, 78750, 45045, 10395;
k=3 column of |A008276| is [0,0,2,11,35,85,175,...] (see A000914), its e.g.f. exp(x)*(2*x^2/2! + 5* x^3/3! + 3*x^4/4!).
		

Crossrefs

Cf. A112007 (triangle for o.g.f.s for unsigned Stirling1 diagonals). A112487 (row sums).

Programs

  • Maple
    A112486 := proc(n,k)
        if n < 0 or k<0 or  k> n then
            0 ;
        elif n = 0 then
            1 ;
        else
            (n+k)*procname(n-1,k)+(n+k-1)*procname(n-1,k-1) ;
        end if;
    end proc: # R. J. Mathar, Dec 19 2013
  • Mathematica
    A112486 [n_, k_] := A112486[n, k] = Which[n<0 || k<0 || k>n, 0, n == 0, 1, True, (n+k)*A112486[n-1, k]+(n+k-1)*A112486[n-1, k-1]]; Table[A112486[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 05 2014, after R. J. Mathar *)

Formula

a(k, m) = (k+m)*a(k-1, m) + (k+m-1)*a(k-1, m-1) for k >= m >= 0, a(0, 0)=1, a(k, -1):=0, a(k, m)=0 if k < m.
From Tom Copeland, Oct 05 2011: (Start)
With polynomials
P(0,t) = 0
P(1,t) = 1
P(2,t) = -(1 + t)
P(3,t) = 2 + 5 t + 3 t^2
P(4,t) = -( 6 + 26 t + 35 t^2 + 15 t^3)
P(5,t) = 24 + 154 t +340 t^2 + 315 t^3 + 105 t^4
Apparently, P(n,t) = (-1)^(n+1) PW[n,-(1+t)] where PW are the Ward polynomials A134991. If so, an e.g.f. for the polynomials is
A(x,t) = -(x+t+1)/t - LW{-((t+1)/t) exp[-(x+t+1)/t]}, where LW(x) is a suitable branch of the Lambert W Fct. (e.g., see A135338). The comp. inverse in x (about x = 0) is B(x) = x + (t+1) [exp(x) - x - 1]. See A112487 for special case t = 1. These results are a special case of A134685 with u(x) = B(x), i.e., u_1=1 and (u_n)=(1+t) for n>0.
Let h(x,t) = 1/(dB(x)/dx) = 1/[1+(1+t)*(exp(x)-1)], an e.g.f. in x for row polynomials in t of signed A028246 , then P(n,t), is given by
(h(x,t)*d/dx)^n x, evaluated at x=0, i.e., A(x,t)=exp(x*h(u,t)*d/du) u, evaluated at u=0. Also, dA(x,t)/dx = h(A(x,t),t).
The e.g.f. A(x,t) = -v * Sum_{j>=1} D(j-1,u) (-z)^j / j! where u=-(x+t+1)/t, v=1+u, z=(1+t*v)/(t*v^2) and D(j-1,u) are the polynomials of A042977. dA/dx = -1/[t*(v-A)].(End)
A133314 applied to the derivative of A(x,t) implies (a.+b.)^n = 0^n, for (b_n)=P(n+1,t) and (a_0)=1, (a_1)=t+1, and (a_n)=t*P(n,t) otherwise. E.g., umbrally, (a.+b.)^2 = a_2*b_0 + 2 a_1*b_1 + a_0*b_2 =0. - Tom Copeland, Oct 08 2011
The row polynomials R(n,x) may be calculated using R(n,x) = 1/x^(n+1)*D^n(x), where D is the operator (x^2+x^3)*d/dx. - Peter Bala, Jul 23 2012
For n>0, Sum_{k=0..n} a(n,k)*(-1/(1+W(t)))^(n+k+1) = (t d/dt)^(n+1) W(t), where W(t) is Lambert W function. For t=-x, this gives Sum_{k>=1} k^(k+n)*x^k/k! = - Sum_{k=0..n} a(n,k)*(-1/(1+W(-x)))^(n+k+1). - Max Alekseyev, Nov 21 2019
Conjecture: row polynomials are R(n,x) = Sum_{i=0..n} Sum_{j=0..i} Sum_{k=0..j} (n+i)!*Stirling2(n+j-k,j-k)*x^k*(x+1)^(j-k)*(-1)^(n+j+k)/((n+j-k)!*(i-j)!*k!). - Mikhail Kurkov, Apr 21 2025