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.

A094262 Triangle read by rows: T(n,k) is the number of rooted trees with k nodes which are disjoint sets of labels with union {1..n}. If a node has an empty set of labels then it must have at least two children.

Original entry on oeis.org

1, 1, 2, 1, 1, 6, 12, 10, 3, 1, 14, 61, 124, 131, 70, 15, 1, 30, 240, 890, 1830, 2226, 1600, 630, 105, 1, 62, 841, 5060, 16990, 35216, 47062, 40796, 22225, 6930, 945, 1, 126, 2772, 25410, 127953, 401436, 836976, 1196532, 1182195, 795718, 349020, 90090, 10395
Offset: 1

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Author

André F. Labossière, Jun 01 2004

Keywords

Comments

The original name for this sequence was "Triangle read by rows giving the coefficients of formulas generating each variety of S2(n,k) (Stirling numbers of 2nd kind). The p-th row (p>=1) contains T(i,p) for i=1 to 2*p-1, where T(i,p) satisfies Sum_{i=1..2*p-1} T(i,p) * C(n-p,i-1)".
The terms of the n-th diagonal sequence of the triangle of Stirling numbers of the second kind A008277, i.e., (Stirling2(N + n - 1,N)), N>=1, are given by a polynomial in N of degree 2*n - 2. This polynomial may be expressed as a linear combination of the falling factorial polynomials binomial(N - n,0), binomial(N - n,1), ... , binomial(N - n,2*n - 2). This table gives the coefficients in these expansions.
The formulas obtained are those for Stirling2(N+1,N) (A000217), Stirling2(N+2,N) (A001296), Stirling2(N+3,N) (A001297), Stirling2(N+4,N) (A001298), Stirling2(N+5,N) (A112494), Stirling2(N+6,N) (A144969) and so on.

Examples

			Row 5 contains 1,30,240,890,1830,2226,1600,630,105, so the formula generating Stirling2(n+4,n) numbers (A001298) will be the following: 1 + 30*(n-5) + 240*C(n-5,2) + 890*C(n-5,3) + 1830*C(n-5,4) + 2226*C(n-5,5) + 1600*C(n-5,6) + 630*C(n-5,7) + 105*C(n-5,8). For example, taking n = 9 gives Stirling2(13,9) = 359502.
Triangle starts:
  1;
  1,  2,   1;
  1,  6,  12,  10,    3;
  1, 14,  61, 124,  131,   70,   15;
  1, 30, 240, 890, 1830, 2226, 1600, 630, 105;
  ...
From _Peter Bala_, Jun 14 2016: (Start)
Connection with row polynomials of A134991:
  R(2,z) = (1 + z)^2*z
  R(3,z) = (1 + z)^2*(z + 3*z^2)
  R(4,z) = (1 + z)^4*(z + 10*z^2 + 15*z^3)
  R(5,z) = (1 + z)^5*(z + 25*z^2 + 105*z^3 + 105*z^4). (End)
From _Andrew Howroyd_, Mar 28 2025: (Start)
The T(3,3) = 12 trees up to relabeling have one of the following 3 forms:
     {}         {1}        {1}
    /  \       /   \        |
  {1} {2,3}   {2}  {3}     {2}
                            |
                           {3}
(End)
		

Crossrefs

Programs

  • Maple
    row_poly := n -> (1+z)^(n+1)*add(z^k*add((-1)^(m+k)*binomial(n+k,n+m)*Stirling2(n+m,m), m=0..k), k=0..n): T_row := n -> seq(coeff(row_poly(n),z,j),j=1..2*n+1):
    seq(T_row(n),n=0..6); # Peter Luschny, Jun 15 2016
  • Mathematica
    Clear[T, q, u]; T[0] = q[1];T[n_] := Sum[m*(u^2*q[m] + 2*u*q[m+1] + q[m+2])*D[T[n-1], q[m]], {m, 1, 2*n+1}]; row[n_] := List @@ Expand[T[n-1]] /. {u -> 1, q[] -> 1}; Table[row[n], {n, 1, 7}] // Flatten (* _Jean-François Alcover, Jun 12 2015 *)
  • PARI
    T(n)={my(g=serreverse(log(((1+1/y)*x+1)/exp(x + O(x*x^n))))); [Vecrev(p/y) | p<-Vec(serlaplace(g))]}
    { my(A=T(5)); for(i=1, #A, print(A[i])) } \\ Andrew Howroyd, Mar 28 2025

Formula

Apparently, a raising operator for bivariate polynomials P(n,u,z) having these coefficients is R = (u+z)^2 * z * d/dz with P(0,u,z) = z. E.g., R P(1,u,z) = R^2 P(0,u,z) = R^2 z = u^4 z + 6 u^3 z^2 + 12 u^2 z^3 + 10 u z^4 + 3 z^5 = P(2,u,z). See the Kazarian link. - Tom Copeland, Jun 12 2015
Reverse polynomials seem to be generated by 1 + exp[t*(x+1+z)^2*(1+z)d/dz]z evaluated at z = 0. - Tom Copeland, Jun 13 2015
From Peter Bala, Jun 14 2016: (Start)
T(n,k) = k*T(n,k) + 2*(k - 1)*T(n,k-1) + (k - 2)*T(n,k-2).
n-th diagonal of A008277: Stirling2(N + n - 1,N) = Sum_{k = 1..2*n - 1} T(n,k)*binomial(N - n,k - 1) for N = 1,2,3,....
Row polynomials R(n,z) = Sum_{k >= 1} k^(n+k-1)*( z/(1 + z)*exp(-z/(1 + z)) )^k/k!, n = 1,2,..., follows from the formula given in A008277 for the o.g.f.'s of the diagonals of the Stirling numbers of the second kind.
Consequently, R(n+1,z) = (1 + z)^2*z*d/dz(R(n,z)) for n >= 1 as conjectured above by Copeland.
R(n,z) = (1 + z)^n*P(n,z) where P(n,z) are the row polynomials of A134991.
R(n,z) = (1 + z)^(2*n+1)*B(n,z/(1 + z)), where B(n,z) are the row polynomials of the triangle of second-order Eulerian numbers A008517 (see Barbero et al., Section 6, equation 27). (End)
Based on the comment of Bala the row polynomials have the explicit form R(n, z) = (1+z)^(n+1)*Sum_{k=0..n}(z^k*Sum_{m=0..k}((-1)^(m+k)*binomial(n+k, n+m)* Stirling2(n+m,m))). - Peter Luschny, Jun 15 2016
E.g.f. G(x,y) satisfies G(x,y) = y*(exp(x)*exp(G(x,y)) - G(x,y) - 1). - Andrew Howroyd, Mar 28 2025

Extensions

Edited and Name changed by Peter Bala, Jun 16 2016
Name changed by Andrew Howroyd, Mar 28 2025