A235706 - OEIS

%I #50 Sep 08 2022 08:46:06

%S 1,3,1,6,6,1,6,18,9,1,0,24,36,12,1,0,0,60,60,15,1,0,0,0,120,90,18,1,0,

%T 0,0,0,210,126,21,1,0,0,0,0,0,336,168,24,1,0,0,0,0,0,0,504,216,27,1,0,

%U 0,0,0,0,0,0,720,270,30,1

%N (I + A132440)^3: Coefficients for normalized generalized Laguerre polynomials n!*Lag(n, 3-n, -x).

%C The associated Laguerre polynomials n!*Lag(n,3-n,-x) are related to the rook polynomials of a rectangular 3 X n chessboard by R(3,n,x) = n!*x^n*Lag(n,3-n,-1/x), which are also the matching polynomials, or generating function of the number of k-edge matchings, of the complete bipartite graph K(m,n), or biclique (cf. Wikipedia for details).

%C The formulas here and below can be naturally extended with 3 replaced by any positive integer m. For m = 1 and 2, see unsigned A132013 and A132014. The formulas there can be extrapolated to apply to this matrix.

%H Michel Marcus, <a href="/A235706/b235706.txt">Rows n = 0..50 of triangle, flattened</a>

%F T(n,k) = binomial(3,n-k)*n!/k! = binomial(n,k)*3!/(3-n+k)!.

%F E.g.f.: exp(y*x)(1+y)^3, so this is an Appell sequence of polynomials with lowering operator L= D= d/dx and raising operator R = x + 3/(1+D).

%F E.g.f. of inverse matrix is exp(x*y)/(1+y)^3.

%F Multiply the n-th diagonal of the Pascal matrix A007318 by d(0)=1, d(1)=3, d(2)=6, d(3)=6, and d(n)=0 for n>3 to obtain T.

%F Row polynomials: n!*Lag(n,3-n,-x) = x^(n-3)*3!*Lag(3,n-3,-x) =

%F (3!/(3-n)!)*K(-n,3-n+1,-x) where K is Kummer's confluent hypergeometric function (as a limit of n+c as c tends to zero).

%F T = (I + A132440)^3 = exp[3*(A238385-I)]. I = identity matrix.

%F Operationally, n!Lag(n,3-n,-:xD:) = x^(n-3)*:Dx:^n*x^(3-n) = x^(-3)*:xD:^n*x^3 = n!*binomial(xD+3,n) = n!*binomial(3,n)*K(-n,3-n+1,-:xD:) where :AB:^n = A^n*B^n for any two operators.

%F n-th row polynomial: n!*Sum_{k = 0..n} (-1)^(n-k)*binomial(n,k)*Lag(k,3,-x). - _Peter Bala_, Jul 25 2021

%e Triangle begins:

%e   1;

%e   3,  1;

%e   6,  6,  1;

%e   6, 18,  9,  1;

%e   0, 24, 36, 12,  1;

%e   0,  0, 60, 60, 15, 1;

%e   ...

%t Table[Binomial[3, n - k] n! / k!, {n, 0, 9}, {k, 0, n}]//Flatten (* _Vincenzo Librandi_, Jul 28 2017 *)

%o (PARI) T(n,k) = binomial(3,n-k)*n!/k!

%o tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n, k), ", ")); print); \\ _Michel Marcus_, Jul 28 2017

%o (PARI) row(n) = Vecrev(n!*pollaguerre(n, 3-n, -x)); \\ _Michel Marcus_, Feb 06 2021

%o (Magma) /* As triangle */ [[Binomial(3, n-k)*Factorial(n)/Factorial(k): k in [0..n]]: n in [0.. 15]]; // _Vincenzo Librandi_, Jul 28 2017

%Y Cf. A007318, A008306 for a generalization, A132013, A132014, A132440, A238363, A238385.

%Y ....................................

%Y With 0th row: 1

%Y n-th row: n!*Lag(n,3-n,-x)

%Y ....................................

%Y 1st: 1!*Lag(1,2,-x)  = A062139(1,k,-x)

%Y 2nd: 2!*Lag(2,1,-x)  = A105278(2,k,x)

%Y 3rd: 3!*Lag(3,0,-x)  = A021009(3,k,-x)

%Y 4th: 4!*Lag(4,-1,-x) = A111596(4,k,-x)

%Y 5th: 5!*Lag(5,-2,-x) = cf. x^2*A062139(3,k,x)

%Y 6th: 6!*Lag(6,-3,-x) = cf. x^3*A062137(3,k,-x)

%Y ....................................

%Y n-th row: x^(n-3)*3!*Lag(3,n-3,-x)

%Y ....................................

%Y 1st: x^(-2)*3!Lag(3,-2,-x) = cf. x^(-2)*[x^2*A062139(1,k,x)]

%Y 2nd: x^(-1)*3!Lag(3,-1,-x) = x^(-1)*A111596(3,k,-x)

%Y 3rd: x^0*3!Lag(3,0,-x)     = x^0*A021009(3,k,-x)

%Y 4th: x^1*3!Lag(3,1,-x)     = x^1*A105278(3,k,x)

%Y 5th: x^2*3!Lag(3,2,-x)     = x^2*A062139(3,k,-x)

%Y 6th: x^3*3!Lag(3,3,-x)     = x^3*A062137(3,k,-x)

%K nonn,tabl,easy

%O 0,2

%A _Tom Copeland_, Apr 20 2014