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%I #56 May 08 2018 15:11:55
%S 1,1,2,1,1,6,12,10,3,1,14,61,124,131,70,15,1,30,240,890,1830,2226,
%T 1600,630,105,1,62,841,5060,16990,35216,47062,40796,22225,6930,945,1,
%U 126,2772,25410,127953,401436,836976,1196532,1182195,795718,349020,90090,10395
%N Triangle related to the diagonals of the triangle of Stirling numbers of the second kind A008277.
%C 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)".
%C 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.
%C 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.
%H M. Abramowitz and I. A. Stegun, eds., <a href="http://www.convertit.com/Go/ConvertIt/Reference/AMS55.ASP">Handbook of Mathematical Functions</a>, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
%H Milton Abramowitz and Irene A. Stegun, eds., <a href="http://www.convertit.com/Go/ConvertIt/Reference/AMS55.ASP">Handbook of Mathematical Functions</a> (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.
%H J. Fernando Barbero G., Jesús Salas, Eduardo J. S. Villaseñor, <a href="http://www.combinatorics.org/ojs/index.php/eljc/article/view/v22i3p37">Generalized Stirling permutations and forests: Higher-order Eulerian and Ward numbers</a>, Electronic Journal of Combinatorics 22(3) (2015), #P3.37.
%H M. Kazarian, <a href="http://arxiv.org/abs/0809.3263">KP hierarchy for Hodge integrals</a>, p. 2, arxiv:0809.3263 [math.AG], 18 Sep 2008. [From _Tom Copeland_, Jun 12 2015]
%H Eric Weisstein's World of Mathematics, <a href="http://www.mathworld.wolfram.com/StirlingNumberoftheSecondKind.html">Stirling numbers of the 2nd kind</a>.
%F 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
%F 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
%F From _Peter Bala_, Jun 14 2016: (Start)
%F T(n,k) = k*T(n,k) + 2*(k - 1)*T(n,k-1) + (k - 2)*T(n,k-2).
%F 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,....
%F 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.
%F Consequently, R(n+1,z) = (1 + z)^2*z*d/dz(R(n,z)) for n >= 1 as conjectured above by Copeland.
%F R(n,z) = (1 + z)^n*P(n,z) where P(n,z) are the row polynomials of A134991.
%F 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)
%F 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 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.
%e Triangle starts:
%e 1;
%e 1, 2, 1;
%e 1, 6, 12, 10, 3;
%e 1, 14, 61, 124, 131, 70, 15;
%e 1, 30, 240, 890, 1830, 2226, 1600, 630, 105;
%e ...
%e From _Peter Bala_, Jun 14 2016: (Start)
%e Connection with row polynomials of A134991:
%e R(2,z) = (1 + z)^2*z
%e R(3,z) = (1 + z)^2*(z + 3*z^2)
%e R(4,z) = (1 + z)^4*(z + 10*z^2 + 15*z^3)
%e R(5,z) = (1 + z)^5*(z + 25*z^2 + 105*z^3 + 105*z^4). (End)
%p 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):
%p seq(T_row(n),n=0..6); # _Peter Luschny_, Jun 15 2016
%t 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 *)
%Y Cf. A008277, A000217, A001296, A001297, A001298, A094216, A008275, A008517, A134991, A112494, A144969.
%K easy,nonn,tabf
%O 1,3
%A _André F. Labossière_, Jun 01 2004
%E Edited and Name changed by _Peter Bala_, Jun 16 2016
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