|
COMMENT
|
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) = a(0), b(1) = a(n) - 2 a(0), b(n) = a(n) - 2n a(n-1) + n(n-1) a(n-2) for n > 0.
2) b(n) = n! Lag{n,(.)!*Lag[.,a1(.),0],-1}, umbrally, where a1(n) = n! Lag{n,(.)!*Lag[.,a(.),0],-1} .
3) b(n) = n! sum(j=0,...,min(2,n)) (-1)^j * Binom(n,j)*a(n-j)/(n-j)!
4) b(n) = (-1)^n n! Lag(n,a(.),2-n)
5) B(x) = (1-xDx))^2 A(x)
6) B(x) = sum(j=0,1,2) {(-1)^j * Binom(2,j)*j!*x^j*Lag(j,-:xD:,0)} A(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.
7) EB(x) = (1-x)^2 EA(x)
8) T = S^2 = A132013^2 = A094587^(-2) = A132159^(-1).
c = (1,-2,2,0,0,...) is the sequence associated to T under the list partition transform and associated operations described in A133314. c are also the coefficients in formula 6. Thus T(n,k) = binomial(n,k)*c(n-k).
The reciprocal sequence to c is d = (1!,2!,3!,4!,...), so the inverse of T is TI(n,k) = binomial(n,k)*d(n-k) = A132159.
These formulae are easily generalized for m applications of the basic operator n! Lag[n,(.)!*Lag[.,a(.),0],-1] by replacing 2 by m in formulae 3, 4, 5, 6 and 7.
The generalized c are given by the generalized coefficients of 6, i.e.,
c(n) = (-1)^n * Binom(m,n)*n! = (-1)^n * m!/(m-n)! .
The generalized d are given by the array at and below the term SI(m-1,m-1) in SI(n,k) = Binom(n,k) * (n-k)!, the inverse of S ; i.e.,
d(n) = SI(m-1+n,m-1) = Binom(m-1+n,m-1) * n! = (m-1+n)!/(m-1)! .
As an aside, this shows that the signed falling factorials and the rising factorials form reciprocal pairs under the list partition transform of A133314.
Row sums of T = [formula 3 with all a(n) = 1] = [binomial transform of c] = [coefficients of B(x) with A(x) = 1/(1-x)] = (1,-1,-1,1,5,11,19,...),
with e.g.f. = [EB(x) with EA(x) = exp(x)] = (1-x)^2 * exp(x) = exp(x)*exp(c(.)*x) = exp[(1+c(.))*x].
Alternating row sums of T = [formula 3 with all a(n) = (-1)^n] = [finite differences of c] = [coefficients of B(x) with A(x) = 1/(1+x)] = (1,-3,7,-13,21,-31,...) = (-1)^n A002061(n+1),
with e.g.f. = [EB(x) with EA(x) = exp(-x)] = (1-x)^2 * exp(-x) = exp(- x)*exp(c(.)*x) = exp[-(1-c(.))*x].
|
