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%I #32 Sep 07 2016 14:17:48
%S 0,2,0,0,3,0,0,0,4,0,0,0,0,5,0,0,0,0,0,6,0,0,0,0,0,0,7,0,0,0,0,0,0,0,
%T 8,0,0,0,0,0,0,0,0,9,0,0,0,0,0,0,0,0,0,10,0,0,0,0,0,0,0,0,0,0,11,0,0,
%U 0,0,0,0,0,0,0,0,0,12,0
%N Infinitesimal generator matrix for a diagonally-shifted Pascal matrix, binomial(n+m,k+m), for m=1, related to Laguerre(n,x,m).
%C Analogous to the infinitesimal Pascal matrix (m=0), A132440.
%C In general the matrix T begins (here m=1)
%C 0;
%C m+1,0;
%C 0, m+2, 0;
%C 0, 0, m+3, 0;
%C 0, 0, 0, m+4, 0;
%C Let LM(t) = exp(t*T) = limit [1 + t*T/n]^n as n tends to infinity.
%C Laguerre matrix(m) = [bin(n+m,k+m)] = LM(1) = exp(T) = [ revert of A074909 for m=1 ]. Truncating the series gives the n X n submatrices. In fact, the submatrices of T are nilpotent with [Tsub_n]^(n+1) = 0 for n=0,1,2,....
%C Inverse Lag matrix(m) = LM(-1) = exp(-T)
%C Umbrally LM[b(.)] = exp(b(.)*T) = [ bin(n+m,k+m) * b(n-k) ]
%C A(j) = T^j / j! equals the matrix [bin(n+m,k+m) * delta(n-k-j)] 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 j-th lower (or main for j=0) diagonal which equals that of the Laguerre(m) matrix. Hence the A(j)'s form a linearly independent basis for all matrices of the form [bin(n+m,k+m) d(n-k)].
%C For sequences with b(0) = 1, umbrally,
%C LM[b(.)] = exp(b(.)*T) = [ bin(n+m,k+m)] * b(n-k) ] .
%C [LM[b(.)]]^(-1) = exp(c(.)*T) = [ bin(n+m,k+m)] * c(n-k) ] where c = LPT(b) with LPT the list partition transform of A133314. Or,
%C [LM[b(.)]]^(-1) = exp[LPT(b(.))*T] = LPT[LM(b(.))] = LM[LPT(b(.))] = LM[c(.)] .
%C 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).
%C 1) b(0) = 0, b(n) = (n+m) * a(n-1),
%C 2) B(x) = x^(-m) (xDx) x^m A(x)
%C 3) B(x) = x * Lag(1,-:xD:,m) A(x) = x * [(m+1) + xD] A(x)
%C 4) EB(x) = D^(m) * (x) * D^(-m) EA(x) where D is the derivative w.r.t. x, (:xD:)^j = x^j*D^j, Lag(n,x,m) is the associated Laguerre polynomial and D^(-m) x^n / n! = x^(m+n) / (m+n)! are Riemann-Liouville integrals.
%C So the exponentiated operator can be characterized (with loose notation) as
%C 5) exp(t*T) A(x) = x^(-m) exp(t*xDx) x^m A(x) = [sum(n=0,1,...) (t*x)^n * Lag(n,-:xD:m)] A(x) = [exp{[t*u/(1-t*u)]*:xD:} / (1-t*u)^(m+1) ] A(x) (eval. at u=x) = A[x/(1-t*x)]/(1-t*x)^(m+1), a generalized Euler transformation for an o.g.f.,
%C 6) exp(t*T) EA(x) = D^(m) exp(t*x) D^(-m) EA(x) = [D/(D-1)]^m exp[(t+a(.))*x] = exp(t*x) [(t+D)/D]^m EA(x)
%C 7) exp(t*T) * a = LM(t) * a = [sum(k=0,...,n) bin(n+m,k+m)* t^(n-k) * a(k)], a vector array.
%C With t=1 and a(k) = (-x)^k / k!, then LM(1) * a = [Laguerre(n,x,m)], a vector array with index n and the o.g.f. A(x) becomes transformed into the e.g.f. for the associated Laguerre polynomials of order m.
%C The exponential formulas can be umbrally extended as in A132440. And, the formulas can be extended to non-integer m.
%H T. Copeland, <a href="http://tcjpn.wordpress.com/2012/11/29/infinigens-the-pascal-pyramid-and-the-witt-and-virasoro-algebras/">Infinitesimal Generators, the Pascal Pyramid, and the Witt and Virasoro Algebras</a>
%H G. Hetyei, <a href="http://arxiv.org/abs/0909.4352">Meixner polynomials of the second kind and quantum algebras representing su(1,1)</a>, arXiv preprint arXiv:0909.4352 [math.QA], 2009.
%H M. Janjic, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL12/Janjic/janjic22.html">Some classes of numbers and derivatives</a>, JIS 12 (2009) 09.8.3
%F 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
%F R P_n(x) = P_(n+1)(x), the matrix T represents the action of
%F R[(m+1)+ RL] 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). - _Tom Copeland_, Oct 25 2012
%t Table[PadLeft[{n, 0}, n], {n, 0, 12}] // Flatten (* _Jean-François Alcover_, Apr 30 2014 *)
%K easy,nonn,tabl
%O 0,2
%A _Tom Copeland_, Nov 15 2007, Nov 16 2007, Nov 27 2007
%E Missing 0 added to array by _Tom Copeland_, Aug 02 2013
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