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%I #72 Jul 27 2021 21:21:35
%S 1,-1,1,0,-2,1,0,0,-3,1,0,0,0,-4,1,0,0,0,0,-5,1,0,0,0,0,0,-6,1,0,0,0,
%T 0,0,0,-7,1,0,0,0,0,0,0,0,-8,1,0,0,0,0,0,0,0,0,-9,1,0,0,0,0,0,0,0,0,0,
%U -10,1,0,0,0,0,0,0,0,0,0,0,-11,1,0,0,0,0,0,0,0,0,0,0,0,-12,1
%N T(n,j) for an iterated mixed order Laguerre transform. Coefficients of the normalized generalized Laguerre polynomials (-1)^n*n!*L(n,1-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(n) = a(n) - n*a(n-1) for n > 0
%C 2) b(n) = n! Lag{n,(.)!*Lag[.,a(.),0],-1}, umbrally
%C 3) b(n) = n! Sum_{j=0..min(1,n)} (-1)^j * binomial(n,j)*a(n-j)/(n-j)!
%C 4) b(n) = (-1)^n n! Lag(n,a(.),1-n)
%C 5) B(x) = (1-xDx) A(x) = [1-x*Lag(1,-xD:,0)] A(x)
%C 6) EB(x) = (1-x) EA(x),
%C where D is the derivative w.r.t. x and Lag(n,x,m) is the associated Laguerre polynomial of order m. These formulas are easily generalized for repeated applications of the operator.
%C c = (1,-1,0,0,0,...) is the sequence associated to T under the list partition transform and the associated operations described in A133314. The reciprocal sequence is d = (0!,1!,2!,3!,4!,...).
%C Consequently, the inverse of T is TI(n,k) = binomial(n,k)*d(n-k) = A094587, which has the property that the terms at and below TI(m,m) are the associated sequence under the list partition transform for the inverse for T^(m+1) for m=0,1,2,3,... .
%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)] = A024000 = (1,0,-1,-2,-3,...), with e.g.f. = [EB(x) with EA(x) = exp(x)] = (1-x) * 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,-2,3,-4,...), with e.g.f. = [EB(x) with EA(x) = exp(-x)] = (1-x) * exp(-x) = exp(-x)*exp(c(.)*x) = exp[-(1-c(.))*x].
%C An e.g.f. for the o.g.f.s for repeated applications of T on A(x) is given by
%C exp[t*(1-xDx)] A(x) = e^t * Sum_{n=0,1,...} (-t*x)^n * Lag(n,-:xD:,0) A(x)
%C = e^t * exp{[-t*u/(1+t*u)]*:xD:} / (1+t*u) A(x) (eval. at u=x)
%C = e^t * A[x/(1+t*x)]/(1+t*x) .
%C See A132014 for more notes on repeated applications.
%H G. C. Greubel, <a href="/A132013/b132013.txt">Rows n=0..100 of triangle, flattened</a>
%H T. Copeland, <a href="http://tcjpn.wordpress.com/2016/11/06/compositional-inverse-operators-and-sheffer-sequences/">Compositional inverse operators and Sheffer sequences</a>, 2016.
%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), with the sequence c defined in the comments.
%F E.g.f.: exp(x*y)(1-x), which implies the row polynomials form an Appell sequence. More relations can be found in A132382. - _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,1-n,x) = -x^(n-1)*L(1,n-1,x) =
%F (-1)^n*(1/(1-n)!)*K(-n,1-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 - 1/(1-D).
%F T = I - A132440 = 2*I - exp[A238385-I] = signed exp[A238385-I], where I = identity matrix.
%F Operationally, (-1)^n*n!*L(n,1-n,-:xD:) = -x^(n-1)*:Dx:^n*x^(1-n) = (-1)^n*x^(-1)*:xD:^n*x = (-1)^n*n!*binomial(xD+1,n) = (-1)^n*n!*binomial(1,n)*K(-n,1-n+1,-:xD:) where :AB:^n = A^n*B^n for any two operators. Cf. A235706. (End)
%F The unsigned row polynomials have e.g.f. (1+t)e^(xt) = exp(t*p.(x)), umbrally, and p_n(x) = (1+D) x^n. With q_n(x) the row polynomials of A094587, p_n(x) = u_n(q.(v.(x))), umbrally, where u_n(x) = (-1)^n v_n(-x) = (-1)^n Lah_n(x), the Lah polynomials with e.g.f. exp[x*t/(t-1)]. This has the matrix form unsigned [T] = [p] = [u]*[q]*[v]. Conversely, q_n(x) = v_n (p.(u.(x))). - _Tom Copeland_, Nov 10 2016
%F n-th row polynomial: n!*Sum_{k = 0..n} (-1)^k*binomial(n,k)*Lag(k,1,x). - _Peter Bala_, Jul 25 2021
%e First few rows of the triangle are
%e 1;
%e -1, 1;
%e 0, -2, 1;
%e 0, 0, -3, 1;
%e 0, 0, 0, -4, 1;
%e 0, 0, 0, 0, -5, 1;
%e 0, 0, 0, 0, 0, -6, 1;
%e 0, 0, 0, 0, 0, 0, -7, 1;
%p c := n -> `if`(n=0,1,`if`(n=1,-1,0)):
%p T := (n,k) -> binomial(n,k)*c(n-k); # _Peter Luschny_, Nov 14 2016
%t Table[PadLeft[{-n, 1}, n+1], {n, 0, 13}] // Flatten (* _Jean-François Alcover_, Apr 29 2014 *)
%o (PARI) row(n) = Vecrev((-1)^n*n!*pollaguerre(n, 1-n)); \\ _Michel Marcus_, Jul 26 2021
%Y Cf. A235706, A132382, A132159, A094587, A132014, A154955, A235706.
%K easy,sign,tabl
%O 0,5
%A _Tom Copeland_, Oct 30 2007
%E Title modified by _Tom Copeland_, Apr 21 2014
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