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COMMENT
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The matrix operation b = T*a can be characterized several ways
in terms of the coefficients a(n) and b(n), their o.g.f.'s A(x) and B(x), or their e.g.f.'s EA(x) and EB(x).
1) b(n) = n! Lag[n,(.)!*Lag[.,a1(.),-1],0] , umbrally,
where a1(n) = n! Lag[n,(.)!*Lag[.,a(.),-1],0]
2) b(n) = (-1)^n * n! * Lag(n,a(.),-2-n)
3) b(n) = sum(j=0,...,n) (-1)^j * Binom(n,j) * Binom(-2,-2-j) * a(n-j)
4) b(n) = sum(j=0,...,n) Binom(n,j) * (j+1)! * a(n-j)
5) B(x) = (1-xDx))^(-2) A(x) , formally
6) B(x) = sum(j=0,1,...) (-1)^j * bin(-2,j) * (xDx)^j A(x)
= sum(j=0,1,...) (j+1) * (xDx)^j A(x)
7) B(x) = sum(j=0,1,...) (j+1) * x^j * D^j * x^j A(x)
8) B(x) = sum(j=0,1,...) (j+1)! * x^j * Lag(j,-:xD:,0) A(x)
9) EB(x) = sum(j=0,1,...) x^j * Lag[j,(.)! * Lag[.,a1(.),-1],0]
10) EB(x) = sum(j=0,1,...) Lag[j,a1(.),-1] * (-x)^j / (1-x)^(j+1)
11) EB(x) = sum(n=0,1,...) x^n * sum(j=0,...,n) (j+1)!/j! * a(n-j) / (n-j)!
12) EB(x) = sum(j=0,1,...) (-x)^j * Lag[j,a(.),-2-j]
13) EB(x) = exp(a(.)*x) / (1-x)^2 = (1-x)^(-2) * EA(x)
14) T = A094587^2 = A132013^(-2) = A132014^(-1)
where Lag(n,x,m) are the Laguerre polynomials of order m, D the derivative w.r.t. x, and (:xD:)^j = x^j * D^j. Truncating the D operator series at the j = n term gives an o.g.f. for b(0) through b(n).
c = (1!,2!,3!,4!,...) is the sequence associated to T under the list partition transform and associated operations described in A133314. Thus T(n,k) = binomial(n,k)*c(n-k) . c are also the coefficients in formulae 4 and 8.
The reciprocal sequence to c is d = (1,-2,2,0,0,0,...), so the inverse of T is TI(n,k) = binomial(n,k)*d(n-k) = A132014. (A121757 is the reverse of T.)
These formulae are easily generalized for m applications of the basic operator n! Lag[n,(.)!*Lag[.,a(.),-1],0] by replacing 2 by m in formulae 2, 3, 5, 6, 12, 13, and 14, or (j+1)! by (m-1+j)!/(m-1)! in 4, 8 and 11. For further discussion of repeated applications of T, see A132014.
The row sums of T = [formula 4 with a(n) all 1] = [binomial transform of c] = [coefficients of B(x) with A(x) = 1/(1-x)] = A001339. Therefore the e.g.f. of A001339 = [formula 13 with a(n) all 1] = exp(x)*(1-x)^(-2) = exp(x)*exp[c(.)*x)] = exp[(1+c(.))*x].
Note the reciprocal is 1/{exp[(1+c(.))*x]} = exp(-x)*(1-x)^2 = e.g.f. of signed A002061 with leading 1 removed] , which makes A001339 and the signed, shifted A002061 reciprocal arrays under the list partition transform of A133314.
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