

A132792


The infinitesimal Lah matrix: generator of unsigned A111596.


3



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, 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, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 110, 0
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,5


COMMENTS

The matrix T begins
0;
0, 0;
0, 2, 0;
0, 0, 6, 0;
0, 0, 0, 12, 0;
Along the nonvanishing diagonal the nth term is (n+1)*(n).
Let LM(t) = exp(t*T) = limit [1 + t*T/n]^n as n tends to infinity.
Lah matrix = [ bin(n,k)*(n1)!/(k1)! ] = 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,....
Inverse Lah matrix = LM(1) = exp(T)
Umbrally LM[b(.)] = exp(b(.)*T) = [ bin(n,k)*(n1)!/(k1)! * b(nk) ]
A(j) = T^j / j! equals the matrix [ bin(n,k)*(n1)!/(k1)! * 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 Lah matrix. Hence the A(j)'s form a linearly independent basis for all matrices of the form [ bin(n,k)*(n1)!/(k1)! * d(nk) ].
For sequences with b(0) = 1, umbrally,
LM[b(.)] = exp(b(.)*T) = [ bin(n,k)*(n1)!/(k1)! * b(nk) ] .
[LM[b(.)]]^(1) = exp(c(.)*T) = [ bin(n,k)*(n1)!/(k1)! * c(nk) ] where c = LPT(b) with LPT the list partition transform of A133314. Or,
[LM[b(.)]]^(1) = exp[LPT(b(.))*T] = LPT[LM(b(.))] = LM[LPT(b(.))] = LM[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).
1) b(0) = 0, b(n) = n*(n1) * a(n1),
2) B(x) = [ x^2 * D^2 * x ] A(x)
3) B(x) = [ x^2 * 2 * Lag(2,:xD:,0) x^(1) ] A(x)
4) EB(x) = [ D^(1) * x * D^2 * x ] EA(x)
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.
The exponentiated operator can be characterized (with loose notation) as
5) exp(t*T) * a = LM(t) * a = [sum(k=0,...,n) bin(n1,k1) * (n! / k!) t^(nk) * a(k) ] = [ t^n * n! * Lag(n,a(.)/t,1) ], a vector array. Note binomial(n1,k1) is 1 for n=k=0 and vanishes for n>0 and k=0 .
With t=1 and a(k) = (x)^k, then LM(1) * a = [ n! * Laguerre(n,x,1) ], a vector array with index n .
6) exp(t*T) EA(x) = EB(x) = EA[ x / (1x*t) ]
From the inverse operator (change t to t), inverting amounts to substituting x/(1+x*t) for x in EB(x) in formula 6.
Compare analogous results in A132710.
T is also a shifted version of the infinitesimal Pascal matrix squared, i.e., T = (A132440^2) * A129185 . The nonvanishing diagonal of T is A002378.


LINKS

Table of n, a(n) for n=0..77.
M. Janjic, Some classes of numbers and derivatives, JIS 12 (2009) 09.8.3


FORMULA

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_(n1)(x) and
R P_n(x) = P_(n+1)(x), the matrix T represents the action of R^2*L^2*R
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


MATHEMATICA

Table[PadLeft[{n*(n1), 0}, n+1], {n, 0, 11}] // Flatten (* JeanFrançois Alcover, Apr 30 2014 *)


CROSSREFS

Sequence in context: A113044 A082399 A051883 * A136572 A262679 A326390
Adjacent sequences: A132789 A132790 A132791 * A132793 A132794 A132795


KEYWORD

easy,nonn,tabl


AUTHOR

Tom Copeland, Nov 17 2007, Nov 27 2007, Nov 29 2007


STATUS

approved



