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%I #31 Feb 10 2022 22:10:51
%S 1,1,3,4,13,36,24,75,316,432,192,541,3060,6360,5760,1920,4683,33244,
%T 92880,127680,86400,23040,47293,403956,1418424,2620800,2688000,
%U 1451520,322560,545835,5449756,23051952,53548992,73785600,60318720,27095040,5160960,7087261,80985780,400813080,1122145920,1943867520,2133734400
%N Triangle of coefficients of a sequence of polynomials related to the enumeration of linear labeled rooted trees.
%C Define the sequence of polynomials {P(n,x)}n>=0 recursively by setting P(0,x) = 1, P(1,x) = 1 and P(n+1,x) = d/dx((1+x)*(1+2*x)*P(n,x)) for n >= 1. The first few values are P(2,x) = 3 + 4*x, P(3,x) = 13 + 36*x + 24*x^2 and P(4,x) = 75 + 316*x + 432*x^2 + 192*x^3.
%C This triangle shows the coefficients of the P(n,x) in ascending powers of x. The values of P(n,x) at an integer or half-integer value of x enumerate linear labeled rooted trees: in particular we have P(n,0) = A000670(n), P(n,1/2) = A050351(n), P(n,1) = A050352(n) and P(n,3/2) = A050353(n).
%C More generally, for m >= 2, P(n,m/2-1), n = 0,1,2,... counts m level linear labeled rooted trees (see the e.g.f. below and the comment of _Benoit Cloitre_ in A050351).
%H L. Liu and Y. Wang, <a href="https://arxiv.org/abs/math/0509207">A unified approach to polynomial sequences with only real zeros</a>, arXiv:math/0509207 [math.CO], 2005-2006.
%F T(n,k) = 2^k*Sum_{i = k+1..n} Stirling2(n,i)*i!*binomial(i-1,k).
%F Recurrence: T(n+1,k) = (k+1)*(2*T(n,k-1)+3*T(n,k)+T(n,k+1)).
%F E.g.f.: G(x,t) := 1 + (1-exp(t))/((2*x+1)*exp(t)-2*x-2) = Sum_{n>=0} P(n,x)*t^n/n! = 1 + t + (3 + 4*x)*t^2/2! + (13 + 36*x + 24*x^2)*t^3/3! + ....
%F Column k generating function: 2^k*((exp(x)-1)/(2-exp(x)))^(k+1) (apart from initial term 1 when k = 0).
%F The generating function G(x,t) satisfies the partial differential equation d/dx((1+x)*(1+2*x)*G(x,t)) - d/dt(G(x,t)) = 2*(2x+1). Hence the row polynomials P(n,x) satisfy the defining recurrence P(n+1,x) = d/dx((1+x)*(2+x)*P(n,x)), with P(0,x) = P(1,x) = 1.
%F Reflection property: P(n,x) = (-1)^n*P(n,-x-3/2).
%F The polynomial P(n,x) has all real zeros, lying in the interval [-1,-1/2] (apply [Liu et al, Theorem 1.1, Corollary 1.2] with f(x) = P(n,x-1/2) and g(x) = P'(n,x-1/2) and use the reflection property).
%F Row sums are A050352; Column 0: A000670; Column 1: 4*A083411; Main diagonal: A002866.
%e Triangle begins
%e n\k|......0.......1........2........3........4........5.......6
%e = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
%e .0.|......1
%e .1.|......1
%e .2.|......3.......4
%e .3.|.....13......36.......24
%e .4.|.....75.....316......432......192
%e .5.|....541....3060.....6360.....5760.....1920
%e .6.|...4683...33244....92880...127680....86400....23040
%e .7.|..47293..403956..1418424..2620800..2688000..1451520..322560
%e ..
%t T[0, 0] = T[1, 0] = 1; T[n_, k_] /; 0 <= k <= n-1 := T[n, k] = (k+1)*(2* T[n-1, k-1] + 3*T[n-1, k] + T[n-1, k+1]); T[_, _] = 0;
%t {1}~Join~Table[T[n, k], {n, 1, 9}, {k, 0, n-1}] // Flatten (* _Jean-François Alcover_, Nov 13 2019 *)
%Y Cf. A000670, A002866 (main diagonal), A050351, A050352, A050353, A083411 (1/4*column 1).
%K nonn,easy,tabf
%O 0,3
%A _Peter Bala_, Sep 01 2011
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