OFFSET
1,3
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.
LINKS
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Milton Abramowitz and Irene A. Stegun, eds., Handbook of Mathematical Functions (with Formulas, Graphs and Mathematical Tables), U.S. Dept. of Commerce, National Bureau of Standards, Applied Math. Series 55, 1964, 1046 pages (9th Printing: November 1970) - Combinatorial Analysis, Table 24.4, Stirling Numbers of the Second Kind (author: Francis L. Miksa), p. 835.
J. Fernando Barbero G., Jesús Salas, Eduardo J. S. Villaseñor, Generalized Stirling permutations and forests: Higher-order Eulerian and Ward numbers, Electronic Journal of Combinatorics 22(3) (2015), #P3.37.
M. Kazarian, KP hierarchy for Hodge integrals, p. 2, arxiv:0809.3263 [math.AG], 18 Sep 2008. [From Tom Copeland, Jun 12 2015]
Eric Weisstein's World of Mathematics, Stirling numbers of the 2nd kind.
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
EXAMPLE
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)
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 *)
CROSSREFS
KEYWORD
easy,nonn,tabf
AUTHOR
André F. Labossière, Jun 01 2004
EXTENSIONS
Edited and Name changed by Peter Bala, Jun 16 2016
STATUS
approved