OFFSET
0,5
COMMENTS
T is the transpose of A132440.
Let M(t) = exp(t*T) = limit [1 + t*T/n]^n as n tends to infinity.
Then M(1) = the transpose of the lower triangular Pascal matrix A007318, with inverse M(-1).
Given a polynomial sequence p_n(x) with p_0(x)=1 and the lowering and raising operators L and R defined by L P_n(x) = n * P_(n-1)(x) and
R P_n(x) = P_(n+1)(x), the matrix T represents the action of L in the p_n(x) basis. For p_n(x) = x^n, L = D = d/dx and R = x. For p_n(x) = x^n/n!, L = DxD and R = D^(-1).
See A132440 as an analog and more general discussion.
Sum_{n>=0} c_n T^n / n! = e^(c.T) gives the Maurer-Cartan form matrix for the one-dimensional Leibniz group defined by multiplication of a Taylor series by the formal Taylor series e^(c.x) (cf. Olver). - Tom Copeland, Nov 05 2015
From Tom Copeland, Jul 02 2018: (Start)
The transpose Psc^Trn of the lower triangular Pascal matrix Psc = A007318 gives the numerical coefficients of the Maurer-Cartan form matrix M of the Leibniz group Leibniz(n)(1,1) presented on p. 9 of the Olver paper. M = exp[c. * T] with (c.)^n = c_n and T the Lie infinitesimal generator of this entry. The columns e^T are the rows of the Pascal matrix A007318.
M can be obtained by multiplying each n-th column vector of Psc by c_n and then transposing the result; i.e., with the diagonal matrix H = Diag(c_0, c_1, c_2, ...), M = (Psc * H)^Trn = H * Psc^Trn.
M is a matrix representation of the differential operator S = e^{c.*D} with D = d/dx, which acting on x^m gives the Appell polynomial p_m(x) = (c. + x)^m, with (c.)^k = c_k an arbitrary indeterminate except for c_0 = 1. For example, S x^2 = (c. + x)^2 = c_0*x^2 + 2*c_1*x + c_2, and M * (0,0,1,0,0,...)^Trn = (c_2,2*c_1,c_0,0,0,...)^Trn = V, so V^Trn = (0,0,1,0,...) * M^Trn = (0,0,1,0,...) * Psc * H = (c_2,2*c_1,c_0,0,...).
The differential lowering and raising operators for the Appell sequence are given by L = D and R = x + dlog(S)/dD, with L p_n(x = n * p_(n-1)(x) and R p_n(x) = p_(n+1)(x).
(End)
LINKS
FORMULA
The matrix operation b = T*a can be characterized in several ways in terms of the coefficients a(n) and b(n), their o.g.f.s A(x) and B(x), or e.g.f.s EA(x) and EB(x):
1) b(n) = (n+1) * a(n+1),
2) B(x) = D A(x), or
3) EB(x) = DxD EA(x),
where D is the derivative w.r.t. x.
So the exponentiated operator can be characterized as
4) exp(t*T) A(x) = exp(t*D) A(x) = A(x+t),
5) exp(t*T) EA(x) = exp(t*DxD) EA(x) = exp[x*a/(1+t*a)]/(1+t*a),
= Sum_{n>=0} (1+t*a)^(-n-1) (x*a)^n/n!, where umbrally
a^n *(1+t*a)^(-n-1) = Sum_{j>0} binomial(n+j,j)a(n+j)t^j,
6) exp(t*T) EA(x) = Sum_{n>=0} a(n) t^n Lag(n,-x/t),
where Lag(n,x) are the Laguerre polynomials (A021009), or
7) [exp(t*T) * a]_n = [M(t) * a]_n
= Sum_{j>=0} binomial(n+j,j)a(n+j)t^j.
EXAMPLE
Matrix T begins
0,1;
0,0,2;
0,0,0,3;
0,0,0,0,4;
0,0,0,0,0,5;
0,0,0,0,0,0,6;
...
MATHEMATICA
Table[PadLeft[{n+1}, n+2], {n, 0, 11}] // Flatten (* Jean-François Alcover, Apr 30 2014 *)
CROSSREFS
KEYWORD
nonn,easy,tabf
AUTHOR
Tom Copeland, Oct 24 2012
STATUS
approved