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Revision History for A132159

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Showing entries 1-10 | older changes
Lower triangular matrix T(n,j) for double application of an iterated mixed order Laguerre transform inverse to A132014. Coefficients of Laguerre polynomials (-1)^n * n! * L(n,-2-n,x).
(history; published version)
#104 by Charles R Greathouse IV at Thu Sep 08 08:45:31 EDT 2022
PROG

(MAGMAMagma) /* As triangle */ [[Binomial(n, k)*Factorial(n-k+1): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Feb 10 2016

Discussion
Thu Sep 08
08:45
OEIS Server: https://oeis.org/edit/global/2944
#103 by Susanna Cuyler at Sun Aug 01 03:50:14 EDT 2021
STATUS

proposed

approved

#102 by Jon E. Schoenfield at Wed Jul 28 22:48:20 EDT 2021
STATUS

editing

proposed

#101 by Jon E. Schoenfield at Wed Jul 28 22:48:16 EDT 2021
FORMULA

Exponential Riordan array [1/(1 - y)^2, y]. The row polynomials R(n,x) thus form a Sheffer sequence of polynomials with associated delta operator equal to d/dx. Thus d/dx(R(n,x)) = n*R(n-1,x). The Sheffer identity is R(n,x + y) = Sum_{k=0..n} binomial(n,k)*y^(n-k)*R(k,x). Define a polynomial sequence P(n,x) of binomial type by setting P(n,x) = Product_ {k = 0..n-1} (2*x + k) with the convention that P(0,x) = 1. Then the present triangle is the triangle of connection constants when expressing the basis polynomials P(n,x + 1) in terms of the basis P(n,x). For example, row 3 is (24, 18, 6, 1) so P(3,x + 1) = (2*x + 2)*(2*x + 3)*(2*x + 4) = 24 + 18*(2*x) + 6*(2*x)*(2*x + 1) + (2*x)*(2*x + 1)*(2*x + 2). Matrix square of triangle A094587. - Peter Bala, Aug 29 2013

STATUS

proposed

editing

#100 by Peter Bala at Wed Jul 28 08:46:30 EDT 2021
STATUS

editing

proposed

#99 by Peter Bala at Mon Jul 26 10:54:50 EDT 2021
FORMULA

First-order recurrence for the row polynomials: (n - x)*R(n,x) = n*(n - x + 1)*R(n-1,x) - x^(n+1) with R(0,x) = 1.

#98 by Peter Bala at Mon Jul 26 10:53:15 EDT 2021
FORMULA

O.g.f: 1/y * Sum_{k >= 0} k!*( y/(1 - x*y) )^k = 1 + (2 + x)*y + (6 + 4*x + x^2)*y^2 + ....

R(n,x) = (x + n + 1)*R(n-1,x) - (n - 1)*x*R(n-2,x) with R(0,x) = 1 and R(1,x) = 2 + x.

#97 by Peter Bala at Mon Jul 26 10:31:59 EDT 2021
FORMULA

R(n,x) = (x+n+1)*R(n-1,x) - (n-1)*x*R(n-2,x) wih with R(0,x) = 1 and R(1,x) = 2 + x.

#96 by Peter Bala at Mon Jul 26 10:21:24 EDT 2021
FORMULA

From Peter Bala, Jul 26 2021: (Start)

First-order recurrence for the row polynomials: (n - x)*R(n,x) = n*(n - x + 1)*R(n-1,x) - x^(n+1) with R(0,x) = 1. - _Peter Bala_, Jul 25 2021

R(n,x) = (x+n+1)*R(n-1,x) - (n-1)*x*R(n-2,x) wih R(0,x) = 1 and R(1,x) = 2 + x.

R(n,x) = A087981 (x = -2), A000255 (x = -1), A000142 (x = 0), A001339 (x = 1), A081923 (x = 2) and A081924 (x = 3). (End)

#95 by Peter Bala at Mon Jul 26 04:15:29 EDT 2021
FORMULA

Exponential Riordan array [1/(1 - y)^2, y]. The row polynomials R(n,x) thus form a Sheffer sequence of polynomials with associated delta operator equal to d/dx. Thus d/dx(R(n,x)) = n*R(n-1,x). The Sheffer identity is R(n,x + y) = Sum_{k=0..n} binomial(n,k)*y^(n-k)*R(k,x). Define a polynomial sequence P(n,x) of binomial type by setting P(n,x) = product Product_ {k = 0..n-1} (2*x + k) with the convention that P(0,x) = 1. Then the present triangle is the triangle of connection constants when expressing the basis polynomials P(n,x + 1) in terms of the basis P(n,x). For example, row 3 is (24, 18, 6, 1) so P(3,x + 1) = (2*x + 2)*(2*x + 3)*(2*x + 4) = 24 + 18*(2*x) + 6*(2*x)*(2*x + 1) + (2*x)*(2*x + 1)*(2*x + 2). Matrix square of triangle A094587. - Peter Bala, Aug 29 2013