
COMMENTS

Matrix T begins
0;
1,0;
0,2,0;
0,0,3,0;
0,0,0,4,0;
Let M(t) = exp(t*T) = limit [1 + t*T/n]^n as n tends to infinity.
Pascal matrix = [ binomial(n,k) ] = M(1) = exp(T), truncating the series gives the n X n submatrices.
Inverse Pascal matrix = M(1) = exp(T) = matrix for inverse binomial transform.
A(j) = T^j / j! equals the matrix [bin(n,k) * delta(nkj)] where delta(n) = 1 if n=0 and vanishes otherwise (Kronecker delta); i.e., A(j) is a matrix with all the terms 0 except for the jth lower (or main for j=0) diagonal which equals that of the Pascal triangle. Hence the A(j)'s form a linearly independent basis for all matrices of the form [binomial(n,k) d(nk)] which include as a subset the invertible associated matrices of the list partition transform (LPT) of A133314.
For sequences with b(0) = 1, umbrally,
M[b(.)] = exp(b(.)*T) = [ binomial(n,k) * b(nk) ] = matrices associated to b by LPT.
[M[b(.)]]^(1) = exp(c(.)*T) = [ binomial(n,k) * c(nk) ] = matrices associated to c, where c = LPT(b) . Or,
[M[b(.)]]^(1) = exp[LPT(b(.))*T] = LPT[M(b(.))] = M[LPT(b(.))]= M[c(.)].
This is related to xDx, the iterated Laguerre transform and the general Euler transformation of a sequence through the comments in A132013 and A132014 and the relation [sum(k=0,...,n) binomial(n,k) * b(nk) * d(k)] = M(b)*d, (nth term). See also A132382.
If b(n,x) is a binomial type Sheffer sequence, then M[b(.,x)]*s(y) = s(x+y) when s(y) = (s(0,y),s(1,y),s(2,y),...) is an array for a Sheffer sequence with the same delta operator as b(n,x) and [M[b(.,x)]]^(1) is given by the formulae above with b(n) replaced by b(n,x) as b(0,x)=1 for a binomial type Sheffer sequence.
T = I  A132013 and conversely A132013 = I  T, which is the matrix representation for the iterated mixed order Laguerre transform characterized in A132013 (and A132014).
(IT)^m generates the group [A132013]^m for m= 0,1,2,.. discussed in A132014.
The inverse is 1/(IT) = I+T+T^2+T^3+... = [A132013]^(1) = A094587 with the associated sequence (0!,1!,2!,3!,...) under the LPT.
And 1/(IT)^2 = I+2*T+3*T^2+4*T^3+... = [A132013]^(2) = A132159 with the associated sequence (1!,2!,3!,4!,...) under the LPT.
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(0) = 0, b(n) = n * a(n1),
2) B(x) = xDx A(x)
3) B(x) = x * Lag(1,:xD:) A(x)
4) EB(x) = x * EA(x) where D is the derivative w.r.t. x, (:xD:)^j = x^j*D^j and Lag(n,x) is the Laguerre polynomial.
So the exponentiated operator can be characterized as
5) exp(t*T) A(x) = exp(t*xDx) A(x) = [sum(n=0,1,...) (t*x)^n * Lag(n,:xD:)] A(x) = [exp{[t*u/(1t*u)]*:xD:} / (1t*u) ] A(x) (eval. at u=x) = A[x/(1t*x)]/(1t*x), a generalized Euler transformation for an o.g.f.,
6) exp(t*T) EA(x) = exp(t*x)*EA(x) = exp[(t+a(.))*x], gen. Euler trf. for an e.g.f.
7) exp(t*T) * a = M(t) * a = [sum(k=0,...,n) binomial(n,k) * t^(nk) * a(k)].
The umbral extension of formulae 5, 6 and 7 gives formally
8) exp[c(.)*T] A(x) = exp(c(.)*xDx) A(x) = [sum(n=0,1,...) (c(.)*x)^n * Lag(n,:xD:)] A(x) = [exp{[c(.)*u/(1c(.)*u)]*:xD:} / (1c(.)*u) ] A(x) (eval. at u=x) = A[x/(1c(.)*x)]/(1c(.)*x), where the umbral evaluation should be applied only after a power series in c is obtained,
9) exp[c(.)*T] EA(x) = exp(c(.)*x)*EA(x) = exp[(c(.)+a(.))*x]
10) exp[c(.)*T] * a = M[c(.)] * a = [sum(k=0,...,n) binomial(n,k) * c(nk) * a(k)] .
The n X n principal submatrix of T is nilpotent, in particular, [Tsub_n]^(n+1) = 0, n=0,1,2,3,....
Note (xDx)^n = x^n D^n x^n = x^n n! (:Dx:)^n/n! = x^n n! Lag(n,:xD:) .
The operator xDx is an important, classical operator explored by among others Dattoli, AlSalam, Carlitz and Stokes and even earlier investigators.
For a recent treatment of xDx, DxD and more general operators see the paper "Laguerretype derivatives: Dobinski relations and combinatorial identities".  Karol A. Penson, Sep 15 2009
See Copeland's link for generalized Laguerre functions and connection to fractional differintegrals in exercises through (:Dx:)^a/a!=(D^a x^a)/a!.  Tom Copeland, Nov 17 2011
From Tom Copeland, Apr 25 2014: (Start)
Conjugation or "similarity" transformations of [dP]=A132440 have an operator interpretation (cf. A074909 and A238363):
In general, select two operators A and B such that A^n = F1(n,B) and B^n = F2(n,A); then A^n =F1(n,F2(.,A)) and B^n = F2(n,F1(.,B)), evaluated umbrally, i.e., F1(n,F2(.,x))=F2(n,F1(.,x))=x^n, implying the polynomials F1 and F2 are an umbral compositional inverse pair.
One such pair are the Bell polynomials Bell(n,x) and falling factorials (x)_n with Bell(n,:xD:)=(xD)^n and (xD)_n=:xD:^n (cf. A074909). Another are the Laguerre polynomials LN(n,x)= n!*Lag(n,x) (A021009), which are umbrally selfinverse, with LN(n,:xD)=:Dx:^n and LN(n,:Dx:)= (:xD:)^n with :Dx:^n=D^n*x^n.
Evaluating, for n>=0, the operator derivative d(B^n)/dA = d(F2(n,A))/dA in the basis B^n, i.e., with A^n finally replaced by F1(n,B), or A^n=F1(.,B)^n=F1(n,B), is equivalent to the matrix conjugation
A) [F2]*[dP]*[F1]
B) = [F2]*[dP]*[F2]^(1)
C) = [F1]^(1)*[dP]*[F1],
where [F1] is the lower triangular matrix with the nth row the coefficients of F1(n,x) and analogously for [F2].
So, given the row vector Rv=(c0 c1 c2 c3 ...) and the column vector Cv(x)=(1 x x^2 x^3 ...)^Transpose, form the power series V(x)=Rv*Cv(x).
D) dV(B)/dA = Rv * [F2]*[dP]*[F1] * Cv(B).
E) With A=D and B=D, F1(n,x)=F2(n,x)=x^n and [F1]=[F2]=I. Then d(B^n)/dA = d(D^n)/dD = n * D^(n1); therefore, consistently [F2]*[dP]*[F1] = [dP] and dV(D)/dD = Rv * [dP] * Cv(D). (End)
