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%I #104 Sep 08 2022 08:45:31
%S 1,2,1,6,4,1,24,18,6,1,120,96,36,8,1,720,600,240,60,10,1,5040,4320,
%T 1800,480,90,12,1,40320,35280,15120,4200,840,126,14,1,362880,322560,
%U 141120,40320,8400,1344,168,16,1,3628800,3265920,1451520,423360,90720,15120
%N Lower triangular matrix T(n,j) for double application of an iterated mixed order Laguerre transform inverse to A132014. Coefficients of Laguerre polynomials (-1)^n * n! * L(n,-2-n,x).
%C 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).
%C 1) b(n) = n! Lag[n,(.)!*Lag[.,a1(.),-1],0], umbrally,
%C where a1(n) = n! Lag[n,(.)!*Lag[.,a(.),-1],0]
%C 2) b(n) = (-1)^n * n! * Lag(n,a(.),-2-n)
%C 3) b(n) = Sum_{j=0..n} (-1)^j * binomial(n,j) * binomial(-2,j) * j! * a(n-j)
%C 4) b(n) = Sum_{j=0..n} binomial(n,j) * (j+1)! * a(n-j)
%C 5) B(x) = (1-xDx))^(-2) A(x), formally
%C 6) B(x) = Sum_{j>=0} (-1)^j * binomial(-2,j) * (xDx)^j A(x)
%C = Sum_{j>=0} (j+1) * (xDx)^j A(x)
%C 7) B(x) = Sum_{j>=0} (j+1) * x^j * D^j * x^j A(x)
%C 8) B(x) = Sum_{j>=0} (j+1)! * x^j * Lag(j,-:xD:,0) A(x)
%C 9) EB(x) = Sum_{j>=0} x^j * Lag[j,(.)! * Lag[.,a1(.),-1],0]
%C 10) EB(x) = Sum_{j>=0} Lag[j,a1(.),-1] * (-x)^j / (1-x)^(j+1)
%C 11) EB(x) = Sum_{j>=0} x^n * Sum_{j=0..n} (j+1)!/j! * a(n-j) / (n-j)!
%C 12) EB(x) = Sum_{j>=0} (-x)^j * Lag[j,a(.),-2-j]
%C 13) EB(x) = exp(a(.)*x) / (1-x)^2 = (1-x)^(-2) * EA(x)
%C 14) T = A094587^2 = A132013^(-2) = A132014^(-1)
%C 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 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 formulas 4 and 8.
%C 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.)
%C These formulas are easily generalized for m applications of the basic operator n! Lag[n,(.)!*Lag[.,a(.),-1],0] by replacing 2 by m in formulas 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.
%C 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].
%C 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.
%C The e.g.f. for the row polynomials (see A132382) implies they form an Appell sequence (see Wikipedia). - _Tom Copeland_, Dec 03 2013
%C As noted in item 12 above and reiterated in the Bala formula below, the e.g.f. is e^(x*t)/(1-x)^2, and the Poisson-Charlier polynomials P_n(t,y) have the e.g.f. (1+x)^y e^(-xt) (Feinsilver, p. 5), so the row polynomials R_n(t) of this entry are (-1)^n P_n(t,-2). The associated Appell sequence IR_n(t) that is the umbral compositional inverse of this entry's polynomials has the e.g.f. (1-x)^2 e^(xt), i.e., the e.g.f. of A132014 (noted above), and, therefore, the row polynomials (-1)^n PC(t,2). As umbral compositional inverses, R_n(IR.(t)) = t^n = IR_n(R.(t)), where, by definition, P.(t)^n = P_n(t), is the umbral evaluation. - _Tom Copeland_, Jan 15 2016
%C T(n,k) is the number of ways to place (n-k) rooks in a 2 x (n-1) Ferrers board (or diagram) under the Goldman-Haglund i-row creation rook mode for i=2. Triangular recurrence relation is given by T(n,k) = T(n-1,k-1) + (n+1-k)*T(n-1,k). - _Ken Joffaniel M. Gonzales_, Jan 21 2016
%H Nathaniel Johnston, <a href="/A132159/b132159.txt">Rows 0..100, flattened</a>
%H Paul Barry, <a href="https://arxiv.org/abs/1804.06801">A note on number triangles that are almost their own production matrix</a>, arXiv:1804.06801 [math.CO], 2018.
%H P. Feinsilver, <a href="http://chanoir.math.siu.edu/MATH/Merida/PDF/Merida.pdf">Lie algebras, representations, and analytic semigroups through dual vector fields</a>
%H Jay Goldman and James Haglund, <a href="http://dx.doi.org/10.1006/jcta.2000.3113">Generalized rook polynomials</a>, J. Combin. Theory A 91 (2000), 509-530.
%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>
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Sheffer_sequence">Sheffer sequence</a>
%F T(n,k) = binomial(n,k)*c(n-k).
%F From _Peter Bala_, Jul 10 2008: (Start)
%F T(n,k) = binomial(n,k)*(n-k+1)!.
%F T(n,k) = (n-k+1)*T(n-1,k) + T(n-1,k-1).
%F E.g.f.: exp(x*y)/(1-y)^2 = 1 + (2+x)*y + (6+4*x+x^2)*y^2/2! + ... .
%F This array is the particular case P(2,1) of the generalized Pascal triangle P(a,b), a lower unit triangular matrix, shown below:
%F n\k|0....................1...............2.........3.....4
%F ----------------------------------------------------------
%F 0..|1.....................................................
%F 1..|a....................1................................
%F 2..|a(a+b)...............2a..............1................
%F 3..|a(a+b)(a+2b).........3a(a+b).........3a........1......
%F 4..|a(a+b)(a+2b)(a+3b)...4a(a+b)(a+2b)...6a(a+b)...4a....1
%F ...
%F See A094587 for some general properties of these arrays.
%F Other cases recorded in the database include: P(1,0) = Pascal's triangle A007318, P(1,1) = A094587, P(2,0) = A038207, P(3,0) = A027465, P(1,3) = A136215 and P(2,3) = A136216. (End)
%F Let f(x) = (1/x^2)*exp(-x). The n-th row polynomial is R(n,x) = (-x)^n/f(x)*(d/dx)^n(f(x)), and satisfies the recurrence equation R(n+1,x) = (x+n+2)*R(n,x)-x*R'(n,x). Cf. A094587. - _Peter Bala_, Oct 28 2011
%F Exponential Riordan array [1/(1 - y)^2, y]. The row polynomials R(n,x) thus form a Sheffer sequence of polynomials with associated delta operator equal to d/dx. Thus d/dx(R(n,x)) = n*R(n-1,x). The Sheffer identity is R(n,x + y) = Sum_{k=0..n} binomial(n,k)*y^(n-k)*R(k,x). Define a polynomial sequence P(n,x) of binomial type by setting P(n,x) = Product_{k = 0..n-1} (2*x + k) with the convention that P(0,x) = 1. Then the present triangle is the triangle of connection constants when expressing the basis polynomials P(n,x + 1) in terms of the basis P(n,x). For example, row 3 is (24, 18, 6, 1) so P(3,x + 1) = (2*x + 2)*(2*x + 3)*(2*x + 4) = 24 + 18*(2*x) + 6*(2*x)*(2*x + 1) + (2*x)*(2*x + 1)*(2*x + 2). Matrix square of triangle A094587. - _Peter Bala_, Aug 29 2013
%F From _Tom Copeland_, Apr 21 2014: (Start)
%F T = (I-A132440)^(-2) = {2*I - exp[(A238385-I)]}^(-2) = unsigned exp[2*(I-A238385)] = exp[A005649(.)*(A238385-I)], umbrally, where I = identity matrix.
%F The e.g.f. is exp(x*y)*(1-y)^(-2), so the row polynomials form an Appell sequence with lowering operator D=d/dx and raising operator x+2/(1-D).
%F With L(n,m,x) = Laguerre polynomials of order m, the row polynomials are (-1)^n * n! * L(n,-2-n,x) = (-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 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.
%F The generalized Pascal triangle Bala mentions is a special case of the fundamental generalized factorial matrices in A133314. (End)
%F From _Peter Bala_, Jul 26 2021: (Start)
%F O.g.f: 1/y * Sum_{k >= 0} k!*( y/(1 - x*y) )^k = 1 + (2 + x)*y + (6 + 4*x + x^2)*y^2 + ....
%F First-order recurrence for the row polynomials: (n - x)*R(n,x) = n*(n - x + 1)*R(n-1,x) - x^(n+1) with R(0,x) = 1.
%F R(n,x) = (x + n + 1)*R(n-1,x) - (n - 1)*x*R(n-2,x) with R(0,x) = 1 and R(1,x) = 2 + x.
%F R(n,x) = A087981 (x = -2), A000255 (x = -1), A000142 (x = 0), A001339 (x = 1), A081923 (x = 2) and A081924 (x = 3). (End)
%e First few rows of the triangle are
%e 1;
%e 2, 1;
%e 6, 4, 1;
%e 24, 18, 6, 1;
%e 120, 96, 36, 8, 1;
%p T := proc(n,k) return binomial(n,k)*factorial(n-k+1): end: seq(seq(T(n,k),k=0..n),n=0..10); # _Nathaniel Johnston_, Sep 28 2011
%t nn=10;f[list_]:=Select[list,#>0&];Map[f,Range[0,nn]!CoefficientList[Series[Exp[y x]/(1-x)^2,{x,0,nn}],{x,y}]]//Grid (* _Geoffrey Critzer_, Feb 15 2013 *)
%o (Haskell)
%o a132159 n k = a132159_tabl !! n !! k
%o a132159_row n = a132159_tabl !! n
%o a132159_tabl = map reverse a121757_tabl
%o -- _Reinhard Zumkeller_, Mar 06 2014
%o (Magma) /* As triangle */ [[Binomial(n,k)*Factorial(n-k+1): k in [0..n]]: n in [0.. 15]]; // _Vincenzo Librandi_, Feb 10 2016
%o (Sage) flatten([[binomial(n,k)*factorial(n-k+1) for k in (0..n)] for n in (0..15)]) # _G. C. Greubel_, May 19 2021
%Y Cf. A008277, A094587, A132013, A132382.
%Y Columns: A000142 (k=0), A001563 (k=1), A001286 (k=2), A005990 (k=3), A061206 (k=4), A062199 (k=5), A062148 (k=6).
%K easy,nonn,tabl
%O 0,2
%A _Tom Copeland_, Nov 01 2007
%E Formula 3) in comments corrected by _Tom Copeland_, Apr 20 2014
%E Title modified by _Tom Copeland_, Apr 23 2014
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