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Triangle of connection constants between the falling factorials (x)_(n) and (2*x)_(n).
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%I #34 Mar 28 2020 14:02:58

%S 1,0,2,0,2,4,0,0,12,8,0,0,12,48,16,0,0,0,120,160,32,0,0,0,120,720,480,

%T 64,0,0,0,0,1680,3360,1344,128,0,0,0,0,1680,13440,13440,3584,256,0,0,

%U 0,0,0,30240,80640,48384,9216,512

%N Triangle of connection constants between the falling factorials (x)_(n) and (2*x)_(n).

%C The falling factorial polynomials (x)_n := x*(x-1)*...*(x-n+1), n = 0,1,2,..., form a basis for the space of polynomials. Hence the polynomial (2*x)_n may be expressed as a linear combination of x_0, x_1,...,x_n; the coefficients in the expansion form the n-th row of the table. Some examples are given below.

%C This triangle is connected to two families of orthogonal polynomials, the Hermite polynomials H(n,x) A060821, and the Bessel polynomials y(n,x)A001498. The first few Hermite polynomials are

%C ... H(0,x) = 1

%C ... H(1,x) = 2*x

%C ... H(2,x) = -2+4*x^2

%C ... H(3,x) = -12*x+8*x^3

%C ... H(4,x) = 12-48*x^2+16*x^4.

%C The unsigned coefficients of H(n,x) give the nonzero entries of the n-th row of the triangle.

%C The Bessel polynomials y(n,x) begin

%C ... y(0,x) = 1

%C ... y(1,x) = 1+x

%C ... y(2,x) = 1+3*x+3*x^2

%C ... y(3,x) = 1+6*x+15*x^2+15*x^3.

%C The entries in the n-th column of this triangle are the coefficients of the scaled Bessel polynomials 2^n*y(n,x).

%C Also the Bell transform of g(n) = 2 if n<2 else 0. For the definition of the Bell transform see A264428. - _Peter Luschny_, Jan 19 2016

%D L. Comtet, Advanced Combinatorics, Reidel, 1974, page 158, exercise 7.

%F Defining relation: 2*x*(2*x-1)*...*(2*x-n+1) = sum {k=0..n} T(n, k)*x*(x-1)*...*(x-k+1)

%F Explicit formula: T(n,k) = (n!/k!)*binomial(k,n-k)*2^(2*k-n). [As defined by Comtet (see reference).]

%F Recurrence relation: T(n,k) = (2*k-n+1)*T(n-1,k)+2*T(n-1,k-1).

%F E.g.f.: exp(x*(t^2+2*t)) = 1 + (2*x)*t + (2*x+4*x^2)*t^2/2! + (12*x^2+8*x^3)*t^3/3! + ...

%F O.g.f. for m-th diagonal (starting at main diagonal m = 0): (2*m)!/m!*x^m/(1-2*x)^(2*m+1).

%F The triangle is the matrix product [2^k*s(n,k)]n,k>=0 * ([s(n,k)]n,k>=0)^(-1),

%F where s(n,k) denotes the signed Stirling number of the first kind.

%F Row sums are [1,2,6,20,76,...] = A000898.

%F Column sums are [1,4,28,296,...] = [2^n*A001515(n)] n>=0.

%e Triangle begins

%e n\k|...0.....1.....2.....3.....4.....5.....6

%e ============================================

%e 0..|...1

%e 1..|...0.....2

%e 2..|...0.....2.....4

%e 3..|...0.....0....12.....8

%e 4..|...0.....0....12....48....16

%e 5..|...0.....0.....0...120...160....32

%e 6..|...0.....0.....0...120...720...480....64

%e ..

%e Row 3:

%e (2*x)_3 = (2*x)*(2*x-1)*(2*x-2) = 8*x*(x-1)*(x-2) + 12*x*(x-1).

%e Row 4:

%e (2*x)_4 = (2*x)*(2*x-1)*(2*x-2)*(2*x-3) = 16*x*(x-1)*(x-2)*(x-3) +

%e 48*x*(x-1)*(x-2)+ 12*x*(x-1).

%e Examples of recurrence relation

%e T(4,4) = 5*T(3,4) + 2*T(3,3) = 5*0 + 2*8 = 16;

%e T(5,4) = 4*T(4,4) + 2*T(4,3) = 4*16 + 2*48 = 160;

%e T(6,4) = 3*T(5,4) + 2*T(5,3) = 3*160 + 2*120 = 720;

%e T(7,4) = 2*T(6,4) + 2*T(6,3) = 2*720 + 2*120 = 1680.

%p T := (n,k) -> (n!/k!)*binomial(k,n-k)*2^(2*k-n):

%p seq(seq(T(n, k), k=0..n), n=0..9);

%o (Sage) # uses[bell_matrix from A264428]

%o bell_matrix(lambda n: 2 if n<2 else 0, 12) # _Peter Luschny_, Jan 19 2016

%Y Cf. A000898 (row sums), A001498, A001515, A059343, A060821.

%K nonn,easy,tabl

%O 0,3

%A _Peter Bala_, Feb 20 2011