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%I #58 Aug 28 2019 15:23:22
%S 1,1,2,1,10,4,1,30,72,8,1,74,516,464,16,1,166,2584,7016,2864,32,1,354,
%T 10740,64240,84480,17376,64,1,734,40008,450280,1321760,949056,104704,
%U 128,1,1498,139108,2681296,14713840,24198976,10223488,629248,256,1,3030,462264,14341992
%N Triangle read by rows, coefficients of the polynomials P(k, x) = (1/2) Sum_{p=0..k-1} Stirling2(k, p+1)*x^p*(1-4*x)^(k-1-p)*(2*p+2)!/(p+1)!.
%C Previous name: Table of coefficients of row polynomials of certain o.g.f.s.
%C The o.g.f.s G(k, x) for the k-family of sequences S(k, n):= Sum_{p=0..n} p^k*binomial(2*p, p)*binomial(2*(n-p), n-p), k=0,1,... (convolution of two sequences involving the central binomial coefficients) are 1/(1-4*x) for k=0 and 2*x*P(k, x)/(1-4*x)^(k+1) for k=1,2,..., with the row polynomials P(k, x) = Sum_{m=0..k-1} a(n,m)*x^m).
%C The author was led to compute the sums S(k, n) by a question asked by M. Greiter, Jun 27 2008.
%C In order to keep the index k>=1 of Sigma(k, n) also for the polynomials P(k, x), their degree is then k-1.
%H Vincenzo Librandi, <a href="/A142963/b142963.txt">Table of n, a(n) for n = 1..210</a>
%H Wolfdieter Lang, <a href="/A142963/a142963_1.txt">First 10 rows and more.</a>
%H L. Liu, Y. Wang, <a href="https://arxiv.org/abs/math/0509207">A unified approach to polynomial sequences with only real zeros</a>, arXiv:math/0509207v5 [math.CO], 2005-2006.
%F G(k, x) = Sum_{p=0..k} S2(k, p)*((2*p)!/p!)*x^p/(1-4*x)^(p+1), k >= 0 (here k >= 1), with the Stirling2 triangle S2(k, p):=A048993(k, p). (Proof from the product of the o.g.f.s of the two convoluted sequences and the normal ordering (x^d_x)^k = Sum_{p=0..k} S2(k, p)*x^p*d_x^p, with the derivative operator d_x.)
%F a(k,m) = [x^m]P(k, x) = [x^m] ((1-4*x)^(k+1))*G(k,x)/(2*x), k>=1, m=0,1,...,k-1.
%F For the triangle coefficients the following relation holds: T(n,m) = (m+1)*T(n-1,m) + (4*n-4*m-2)*T(n-1,m-1) with T(n,m=0) = 1 and T(n,m=n-1) = 2^(n-1), n >= 1 and 0 <= m <= n-1. - _Johannes W. Meijer_, Feb 20 2009
%F From _Peter Bala_, Jan 18 2018: (Start)
%F (x*d/dx)^n (1/(sqrt(1 - 4*x)) = 2*x*P(n,x)/sqrt(1 - 4*x)^(n+1/2) for n >= 1.
%F x*P(n,x)/(1 - 4*x)^(n+1/2) = (1/2)*Sum_{k >= 1} binomial(2*k,k)* k^n*x^k for n >= 1.
%F P(n+1,x) = ((4*n - 2)*x + 1)*P(n,x) - x*(4*x - 1)*d/dx(P(n,x)).
%F Hence the polynomial P(n,x) has all real zeros by Liu et al., Theorem 1.1, Corollary 1.2. (End)
%e Triangle starts:
%e [1]
%e [1, 2]
%e [1, 10, 4]
%e [1, 30, 72, 8]
%e [1, 74, 516, 464, 16]
%e [1, 166, 2584, 7016, 2864, 32]
%e [1, 354, 10740, 64240, 84480, 17376, 64]
%e [1, 734, 40008, 450280, 1321760, 949056, 104704, 128]
%e ...
%e P(3,x) = 1+10*x+4*x^2.
%e G(3,x) = 2*x*(1+10*x+4*x^2)/(1-4*x)^4.
%p A142963 := proc(n,m): if n=m+1 then 2^(n-1); elif m=0 then 1 ; elif m<0 or m>n-1 then 0; else (m+1)*procname(n-1, m)+(4*n-4*m-2)*procname(n-1, m-1); end if; end proc: seq(seq(A142963(n,m), m=0..n-1), n=1..9); # _Johannes W. Meijer_, Sep 28 2011
%p # Alternatively (assumes offset 0):
%p p := (n,x) -> (1/2)*add(Stirling2(n+1,k+1)*x^k*(1-4*x)^(n-k)*(2*k+2)!/(k+1)!, k=0..n): for n from 0 to 7 do [n], PolynomialTools:-CoefficientList(p(n,x), x) od;
%p # _Peter Luschny_, Jun 18 2017
%t t[_, 0] = 1; t[n_, m_] /; m == n-1 := 2^m; t[n_, m_] := (m+1)*t[n-1, m] + (4*n-4*m-2)*t[n-1, m-1]; Table[t[n, m], {n, 1, 10}, {m, 0, n-1}] // Flatten (* _Jean-François Alcover_, Jun 21 2013, after _Johannes W. Meijer_ *)
%Y Left hand column sequences 2*A142964, 4*A142965, 8*A142966, 16*A142968.
%Y Row sums A142967.
%Y From _Johannes W. Meijer_, Feb 20 2009: (Start)
%Y A156919 and this sequence can be mapped onto A156920.
%Y Cf. A156921, A156925, A156927, A156933.
%Y Right hand column sequences 2^n*A000340, 2^n*A156922, 2^n*A156923, 2^n*A156924. (End)
%Y Cf. A142961, A142962.
%K nonn,easy,tabl
%O 1,3
%A _Wolfdieter Lang_, Sep 15 2008
%E Minor edits by _Johannes W. Meijer_, Sep 28 2011
%E A more precise name by _Peter Luschny_, Jun 18 2017
%E Name reformulated with offset corrected, edited by _Wolfdieter Lang_, Aug 23 2019