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%I #16 Sep 07 2016 14:19:14
%S 0,0,0,0,2,0,0,0,6,0,0,0,0,12,0,0,0,0,0,20,0,0,0,0,0,0,30,0,0,0,0,0,0,
%T 0,42,0,0,0,0,0,0,0,0,56,0,0,0,0,0,0,0,0,0,72,0,0,0,0,0,0,0,0,0,0,90,
%U 0,0,0,0,0,0,0,0,0,0,0,110,0
%N The infinitesimal Lah matrix: generator of unsigned A111596.
%C The matrix T begins
%C 0;
%C 0, 0;
%C 0, 2, 0;
%C 0, 0, 6, 0;
%C 0, 0, 0, 12, 0;
%C Along the nonvanishing diagonal the n-th term is (n+1)*(n).
%C Let LM(t) = exp(t*T) = limit [1 + t*T/n]^n as n tends to infinity.
%C Lah matrix = [ bin(n,k)*(n-1)!/(k-1)! ] = LM(1) = exp(T) = unsigned A111596. Truncating the series gives the n X n principal submatrices. In fact, the principal submatrices of T are nilpotent with [Tsub_n]^n = 0 for n=0,1,2,....
%C Inverse Lah matrix = LM(-1) = exp(-T)
%C Umbrally LM[b(.)] = exp(b(.)*T) = [ bin(n,k)*(n-1)!/(k-1)! * b(n-k) ]
%C A(j) = T^j / j! equals the matrix [ bin(n,k)*(n-1)!/(k-1)! * 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 Lah matrix. Hence the A(j)'s form a linearly independent basis for all matrices of the form [ bin(n,k)*(n-1)!/(k-1)! * d(n-k) ].
%C For sequences with b(0) = 1, umbrally,
%C LM[b(.)] = exp(b(.)*T) = [ bin(n,k)*(n-1)!/(k-1)! * b(n-k) ] .
%C [LM[b(.)]]^(-1) = exp(c(.)*T) = [ bin(n,k)*(n-1)!/(k-1)! * 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*(n-1) * a(n-1),
%C 2) B(x) = [ x^2 * D^2 * x ] A(x)
%C 3) B(x) = [ x^2 * 2 * Lag(2,-:xD:,0) x^(-1) ] A(x)
%C 4) EB(x) = [ D^(-1) * x * D^2 * x ] EA(x)
%C where D is the derivative w.r.t. x, (:xD:)^j = x^j * D^j and Lag(n,x,m) is the associated Laguerre polynomial of order m.
%C The exponentiated operator can be characterized (with loose notation) as
%C 5) exp(t*T) * a = LM(t) * a = [sum(k=0,...,n) bin(n-1,k-1) * (n! / k!) t^(n-k) * a(k) ] = [ t^n * n! * Lag(n,-a(.)/t,-1) ], a vector array. Note binomial(n-1,k-1) is 1 for n=k=0 and vanishes for n>0 and k=0 .
%C With t=1 and a(k) = (-x)^k, then LM(1) * a = [ n! * Laguerre(n,x,-1) ], a vector array with index n .
%C 6) exp(t*T) EA(x) = EB(x) = EA[ x / (1-x*t) ]
%C From the inverse operator (change t to -t), inverting amounts to substituting x/(1+x*t) for x in EB(x) in formula 6.
%C Compare analogous results in A132710.
%C T is also a shifted version of the infinitesimal Pascal matrix squared, i.e., T = (A132440^2) * A129185 . The non-vanishing diagonal of T is A002378.
%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 R^2*L^2*R
%F in the p_n(x) basis. For p_n(x) = x^n, L = D = d/dx and R = x.
%F For p_n(x) = x^n/n!, L = DxD and R = D^(-1). - _Tom Copeland_, Oct 25 2012
%t Table[PadLeft[{n*(n-1), 0}, n+1], {n, 0, 11}] // Flatten (* _Jean-François Alcover_, Apr 30 2014 *)
%K easy,nonn,tabl
%O 0,5
%A _Tom Copeland_, Nov 17 2007, Nov 27 2007, Nov 29 2007
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