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%I #67 Jul 27 2021 21:21:44
%S 1,-2,1,2,-4,1,0,6,-6,1,0,0,12,-8,1,0,0,0,20,-10,1,0,0,0,0,30,-12,1,0,
%T 0,0,0,0,42,-14,1,0,0,0,0,0,0,56,-16,1,0,0,0,0,0,0,0,72,-18,1,0,0,0,0,
%U 0,0,0,0,90,-20,1,0,0,0,0,0,0,0,0,0,110,-22,1,0,0,0,0,0,0,0,0,0,0,132,-24,1
%N T(n,j) for double application of an iterated mixed order Laguerre transform: Coefficients of Laguerre polynomial (-1)^n*n!*L(n,2-n,x).
%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) = 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.
%C 2) b(n) = n! Lag{n,(.)!*Lag[.,a1(.),0],-1}, umbrally, where a1(n) = n! Lag{n,(.)!*Lag[.,a(.),0],-1}.
%C 3) b(n) = n! Sum_{j=0..min(2,n)} (-1)^j * binomial(n,j)*a(n-j)/(n-j)!
%C 4) b(n) = (-1)^n n! Lag(n,a(.),2-n)
%C 5) B(x) = (1-xDx)^2 A(x)
%C 6) B(x) = Sum_{j=0..2} {(-1)^j * binomial(2,j)*j!*x^j*Lag(j,-:xD:,0)} A(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 7) EB(x) = (1-x)^2 EA(x)
%C 8) T = S^2 = A132013^2 = A094587^(-2) = A132159^(-1).
%C 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).
%C 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.
%C These formulas are easily generalized for m applications of the basic operator n! Lag[n,(.)!*Lag[.,a(.),0],-1] by replacing 2 with m in formulas 3, 4, 5, 6 and 7.
%C The generalized c are given by the generalized coefficients of 6, i.e.,
%C c(n) = (-1)^n * binomial(m,n)*n! = (-1)^n * m!/(m-n)!.
%C The generalized d are given by the array at and below the term SI(m-1,m-1) in SI(n,k) = binomial(n,k) * (n-k)!, the inverse of S; i.e.,
%C d(n) = SI(m-1+n,m-1) = binomial(m-1+n,m-1) * n! = (m-1+n)!/(m-1)!.
%C As an aside, this shows that the signed falling factorials and the rising factorials form reciprocal pairs under the list partition transform of A133314.
%C 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,...),
%C 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].
%C 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),
%C 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].
%C See A132159 for a relation to the Poisson-Charlier polynomials. - _Tom Copeland_, Jan 15 2016
%H G. C. Greubel, <a href="/A132014/b132014.txt">Rows n=0..100 of triangle, flattened</a>
%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.
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Appell_sequence">Appell sequence</a>
%F T(n,k) = binomial(n,k)*c(n-k).
%F E.g.f. for row polynomials: exp(x*y)(1-x)^2. Implies the row polynomials form an Appell sequence (see Wikipedia). - _Tom Copeland_, Dec 03 2013
%F From _Tom Copeland_, Apr 21 2014: (Start)
%F Change notation letting L(n,m,x) = Lag(n,x,m).
%F Row polynomials: (-1)^n*n!*L(n,2-n,x) = (-1)^n*(-x)^(n-2)*2!*L(2,n-2,x) =
%F (-1)^n*(2!/(2-n)!)*K(-n,2-n+1,x) where K is Kummer's confluent hypergeometric function (as a limit of n+s as s tends to zero).
%F For the row polynomials, the lowering operator = d/dx and the raising operator = x - 2/(1-D).
%F T = (I - A132440)^2 = [2*I - exp(A238385-I)]^2 = signed exp[2*(A238385-I)], where I = identity matrix.
%F Operationally, (-1)^n*n!*L(n,2-n,-:xD:) = (-1)^n*x^(n-2)*:Dx:^n*x^(2-n) = (-1)^n*x^(-2)*:xD:^n*x^2 = (-1)^n*n!*binomial(xD+2,n) = (-1)^n*n!*binomial(2,n)*K(-n,2-n+1,-:xD:) where :AB:^n = A^n*B^n for any two operators. Cf. A235706. (End)
%F n-th row polynomial: n!*Sum_{k = 0..n} (-1)^(n-k)*binomial(n,k)*Lag(k,2,x). - _Peter Bala_, Jul 25 2021
%e First few rows of the triangle are
%e 1;
%e -2, 1;
%e 2, -4, 1;
%e 0, 6, -6, 1;
%e 0, 0, 12, -8, 1;
%e 0, 0, 0, 20, -10, 1;
%t m = 12; s = Exp[x*y]*(1 - x)^2 + O[x]^(m + 2) + O[y]^(m + 2); T[n_, k_] := SeriesCoefficient[s, {x, 0, n}, {y, 0, k}]*n!; T[0, 0] = 1; Table[T[n, k], {n, 0, m}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Jul 09 2015 *)
%o (PARI) row(n) = Vecrev((-1)^n*n!*pollaguerre(n, 2-n)); \\ _Michel Marcus_, Feb 06 2021
%Y Cf. A132382, A110330, A132159, A094587, A132013, A235706.
%K easy,sign,tabl
%O 0,2
%A _Tom Copeland_, Oct 30 2007, Nov 05 2007, Nov 11 2007
%E Title modified by _Tom Copeland_, Apr 21 2014
%E Missing term -18 inserted in 10th row by _Jean-François Alcover_, Jul 09 2015
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