A062148 -id:A062148 - cp's OEIS frontend

A132159 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).

1, 2, 1, 6, 4, 1, 24, 18, 6, 1, 120, 96, 36, 8, 1, 720, 600, 240, 60, 10, 1, 5040, 4320, 1800, 480, 90, 12, 1, 40320, 35280, 15120, 4200, 840, 126, 14, 1, 362880, 322560, 141120, 40320, 8400, 1344, 168, 16, 1, 3628800, 3265920, 1451520, 423360, 90720, 15120

Offset: 0

Tom Copeland, Nov 01 2007

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 * binomial(n,j) * binomial(-2,j) * j! * a(n-j)
4) b(n) = Sum_{j=0..n} binomial(n,j) * (j+1)! * a(n-j)
5) B(x) = (1-xDx))^(-2) A(x), formally
6) B(x) = Sum_{j>=0} (-1)^j * binomial(-2,j) * (xDx)^j A(x)
= Sum_{j>=0} (j+1) * (xDx)^j A(x)
7) B(x) = Sum_{j>=0} (j+1) * x^j * D^j * x^j A(x)
8) B(x) = Sum_{j>=0} (j+1)! * x^j * Lag(j,-:xD:,0) A(x)
9) EB(x) = Sum_{j>=0} x^j * Lag[j,(.)! * Lag[.,a1(.),-1],0]
10) EB(x) = Sum_{j>=0} Lag[j,a1(.),-1] * (-x)^j / (1-x)^(j+1)
11) EB(x) = Sum_{j>=0} x^n * Sum_{j=0..n} (j+1)!/j! * a(n-j) / (n-j)!
12) EB(x) = Sum_{j>=0} (-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 formulas 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 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.
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.
The e.g.f. for the row polynomials (see A132382) implies they form an Appell sequence (see Wikipedia). - Tom Copeland, Dec 03 2013
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
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

			First few rows of the triangle are
    1;
    2,  1;
    6,  4,  1;
   24, 18,  6, 1;
  120, 96, 36, 8, 1;

Nathaniel Johnston, Rows 0..100, flattened
Paul Barry, A note on number triangles that are almost their own production matrix, arXiv:1804.06801 [math.CO], 2018.
P. Feinsilver, Lie algebras, representations, and analytic semigroups through dual vector fields
Jay Goldman and James Haglund, Generalized rook polynomials, J. Combin. Theory A 91 (2000), 509-530.
M. Janjic, Some classes of numbers and derivatives, JIS 12 (2009) 09.8.3.
Wikipedia, Appell sequence
Wikipedia, Sheffer sequence

Cf. A008277, A094587, A132013, A132382.

Columns: A000142 (k=0), A001563 (k=1), A001286 (k=2), A005990 (k=3), A061206 (k=4), A062199 (k=5), A062148 (k=6).

a132159 n k = a132159_tabl !! n !! k
a132159_row n = a132159_tabl !! n
a132159_tabl = map reverse a121757_tabl
-- Reinhard Zumkeller, Mar 06 2014

/* As triangle */ [[Binomial(n,k)*Factorial(n-k+1): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Feb 10 2016

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

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 *)

flatten([[binomial(n,k)*factorial(n-k+1) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, May 19 2021

T(n,k) = binomial(n,k)*c(n-k).
From Peter Bala, Jul 10 2008: (Start)
T(n,k) = binomial(n,k)*(n-k+1)!.
T(n,k) = (n-k+1)*T(n-1,k) + T(n-1,k-1).
E.g.f.: exp(x*y)/(1-y)^2 = 1 + (2+x)*y + (6+4*x+x^2)*y^2/2! + ... .
This array is the particular case P(2,1) of the generalized Pascal triangle P(a,b), a lower unit triangular matrix, shown below:
  n\k|0....................1...............2.........3.....4
  ----------------------------------------------------------
  0..|1.....................................................
  1..|a....................1................................
  2..|a(a+b)...............2a..............1................
  3..|a(a+b)(a+2b).........3a(a+b).........3a........1......
  4..|a(a+b)(a+2b)(a+3b)...4a(a+b)(a+2b)...6a(a+b)...4a....1
  ...
See A094587 for some general properties of these arrays.
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)
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
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
From Tom Copeland, Apr 21 2014: (Start)
T = (I-A132440)^(-2) = {2*I - exp[(A238385-I)]}^(-2) = unsigned exp[2*(I-A238385)] = exp[A005649(.)*(A238385-I)], umbrally, where I = identity matrix.
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).
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).
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.
The generalized Pascal triangle Bala mentions is a special case of the fundamental generalized factorial matrices in A133314. (End)
From Peter Bala, Jul 26 2021: (Start)
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 + ....
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.
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.
R(n,x) = A087981 (x = -2), A000255 (x = -1), A000142 (x = 0), A001339 (x = 1), A081923 (x = 2) and A081924 (x = 3). (End)

Formula 3) in comments corrected by Tom Copeland, Apr 20 2014

Title modified by Tom Copeland, Apr 23 2014

A062138 Coefficient triangle of generalized Laguerre polynomials n!*L(n,5,x)(rising powers of x).

Original entry on oeis.org

1, 6, -1, 42, -14, 1, 336, -168, 24, -1, 3024, -2016, 432, -36, 1, 30240, -25200, 7200, -900, 50, -1, 332640, -332640, 118800, -19800, 1650, -66, 1, 3991680, -4656960, 1995840, -415800, 46200, -2772, 84, -1, 51891840, -69189120

Offset: 0

Wolfdieter Lang, Jun 19 2001

The row polynomials s(n,x) := n!*L(n,5,x)= sum(a(n,m)*x^m,m=0..n) have e.g.f. exp(-z*x/(1-z))/(1-z)^6. They are Sheffer polynomials satisfying the binomial convolution identity s(n,x+y) = sum(binomial(n,k)*s(k,x)*p(n-k,y),k=0..n), with polynomials sum(|A008297(n,m)|*(-x)^m, m=1..n), n >= 1 and p(0,x)=1 (for Sheffer polynomials see A048854 for S. Roman reference).
These polynomials appear in the radial part of the l=2 (d-wave) eigen functions for the discrete energy levels of the H-atom. See Messiah reference.
For m=0..5 the (unsigned) column sequences (without leading zeros) are: A001725(n+5), A062148-A062152. Row sums (signed) give A062191; row sums (unsigned) give A062192.
The unsigned version of this triangle is the triangle of unsigned 3-Lah numbers A143498. - Peter Bala, Aug 25 2008

			Triangle begins:
  {1};
  {6, -1};
  {42, -14, 1};
  {336, -168, 24, -1};
  ...
2!*L(2, 5, x) = 42-14*x+x^2.

A. Messiah, Quantum mechanics, vol. 1, p. 419, eq.(XI.18a), North Holland, 1969.

Indranil Ghosh, Rows 0..125, flattened
Index entries for sequences related to Laguerre polynomials

Cf. A021009, A062137, A062139, A062140, A066667.
For m=0..5 the (unsigned) column sequences (without leading zeros) are: A001725(n+5), A062148, A062149, A062150, A062151, A062152.
Row sums (signed) give A062191, row sums (unsigned) give A062192.
Cf. A143498.

Flatten[Table[((-1)^m)*n!*Binomial[n+5,n-m]/m!,{n,0,8},{m,0,n}]] (* Indranil Ghosh, Feb 24 2017 *)

tabl(nn) = {for (n=0, nn, for (m=0, n, print1(((-1)^m)*n!*binomial(n+5, n-m)/m!, ", "); ); print(); ); } \\ Indranil Ghosh, Feb 24 2017

row(n) = Vecrev(n!*pollaguerre(n, 5)); \\ Michel Marcus, Feb 06 2021

import math
f=math.factorial
def C(n, r):return f(n)//f(r)//f(n-r)
i=-1
for n in range(26):
    for m in range(n+1):
        i += 1
        print(str(i)+" "+str(((-1)**m)*f(n)*C(n+5, n-m)//f(m))) # Indranil Ghosh, Feb 24 2017

T(n, m) = ((-1)^m)*n!*binomial(n+5, n-m)/m!.

E.g.f. for m-th column: ((-x/(1-x))^m)/(m!*(1-x)^6), m >= 0.

A143498 Triangle of unsigned 3-Lah numbers.

Original entry on oeis.org

1, 6, 1, 42, 14, 1, 336, 168, 24, 1, 3024, 2016, 432, 36, 1, 30240, 25200, 7200, 900, 50, 1, 332640, 332640, 118800, 19800, 1650, 66, 1, 3991680, 4656960, 1995840, 415800, 46200, 2772, 84, 1, 51891840, 69189120, 34594560, 8648640, 1201200, 96096, 4368

Offset: 3

Peter Bala, Aug 25 2008

For a signed version of this triangle see A062138. This is the case r = 3 of the unsigned r-Lah numbers L(r;n,k). The unsigned 3-Lah numbers count the partitions of the set {1,2,...,n} into k ordered lists with the restriction that the elements 1, 2 and 3 belong to different lists. For other cases see A105278 (r = 1), A143497 (r = 2 and comments on the general case) and A143499 (r = 4).

The unsigned 3-Lah numbers are related to the 3-Stirling numbers: the lower triangular array of unsigned 3-Lah numbers may be expressed as the matrix product St1(3) * St2(3), where St1(3) = A143492 and St2(3) = A143495 are the arrays of 3-Stirling numbers of the first and second kind respectively. An alternative factorization for the array is as St1 * P^4 * St2, where P denotes Pascal's triangle, A007318, St1 is the triangle of unsigned Stirling numbers of the first kind, abs(A008275) and St2 denotes the triangle of Stirling numbers of the second kind, A008277.

			Triangle begins
n\k|......3......4......5......6......7......8
==============================================
3..|......1
4..|......6......1
5..|.....42.....14......1
6..|....336....168.....24......1
7..|...3024...2016....432.....36......1
8..|..30240..25200...7200....900.....50......1
...
T(4,3) = 6. The partitions of {1,2,3,4} into 3 ordered lists, such that the elements 1, 2 and 3 lie in different lists, are: {1}{2}{3,4} and {1}{2}{4,3}, {1}{3}{2,4} and {1}{3}{4,2}, {2}{3}{1,4} and {2}{3}{4,1}.

Erich Neuwirth, Recursively defined combinatorial functions: Extending Galton's board, Discrete Math. 239 No. 1-3, 33-51 (2001).
G. Nyul, G. Rácz, The r-Lah numbers, Discrete Mathematics, 338 (2015), 1660-1666.
Michael J. Schlosser and Meesue Yoo, Elliptic Rook and File Numbers, Electronic Journal of Combinatorics, 24(1) (2017), #P1.31.
M. Shattuck, Generalized r-Lah numbers, arXiv:1412.8721 [math.CO], 2014.

Cf. A001725 (column 3), A007318, A008275, A008277, A062138, A062148 - A062152 (column 4 to column 8), A062191 (alt. row sums), A062192 (row sums), A105278 (unsigned Lah numbers), A143492, A143495, A143497, A143499.

/* As triangle */ [[Factorial(n-3)/Factorial(k-3)*Binomial(n+2, k+2): k in [3..n]]: n in [3.. 15]]; // Vincenzo Librandi, Nov 27 2018

with combinat: T := (n, k) -> (n-3)!/(k-3)!*binomial(n+2,k+2): for n from 3 to 12 do seq(T(n, k), k = 3..n) end do;

T[n_, k_] := (n-3)!/(k-3)!*Binomial[n+2, k+2]; Table[T[n, k], {n, 3, 10}, {k, 3, n}] // Flatten (* Amiram Eldar, Nov 26 2018 *)

T(n,k) = (n-3)!/(k-3)!*binomial(n+2,k+2) for n,k >= 3.
Recurrence: T(n,k) = (n+k-1)*T(n-1,k) + T(n-1,k-1) for n,k >= 3, with the boundary conditions: T(n,k) = 0 if n < 3 or k < 3; T(3,3) = 1.
E.g.f. for column k: Sum_{n >= k} T(n,k)*t^n/(n-3)! = 1/(k-3)!*t^k/(1-t)^(k+3) for k >= 3.
E.g.f: Sum_{n = 3..inf} Sum_{k = 3..n} T(n,k)*x^k*t^n/(n-3)! = (x*t)^3/(1-t)^6*exp(x*t/(1-t)) = (x*t)^3*(1 + (6+x)*t +(42+14*x+x^2)*t^2/2! + ... ).
Generalized Lah identity: (x+5)*(x+6)*...*(x+n+1) = Sum_{k = 3..n} T(n,k)*(x-1)*(x-2)*...*(x-k+3).
The polynomials 1/n!*Sum_{k = 3..n+3} T(n+3,k)*(-x)^(k-3) for n >= 0 are the generalized Laguerre polynomials Laguerre(n,5,x). See A062138.
Array = A143492 * A143495 = abs(A008275) * ( A007318 )^4 * A008277 (apply Theorem 10 of [Neuwirth]). Array equals exp(D), where D is the array with the quadratic sequence (6,14,24,36, ... ) on the main subdiagonal and zeros everywhere else.

A167556 A triangle related to the GF(z) formulas of the rows of the ED1 array A167546.

Original entry on oeis.org

1, 1, 2, 2, 6, 2, 6, 24, 4, 8, 24, 120, 0, 48, 24, 120, 720, -120, 384, 72, 144, 720, 5040, -1680, 3696, -432, 1296, 720, 5040, 40320, -20160, 40320, -15840, 17280, 2880, 5760, 40320, 362880, -241920, 483840, -311040, 288000, -46080, 69120, 40320

Offset: 1

Johannes W. Meijer, Nov 10 2009

The GF(z) formulas given below correspond to the first ten rows of the ED1 array A167546. The polynomials in their numerators lead to the triangle given above.

			Row 1: GF(z) = 1/(1-z).
Row 2: GF(z) = (1 + 2*z)/(1-z)^2.
Row 3: GF(z) = (2 + 6*z + 2*z^2)/(1-z)^3.
Row 4: GF(z) = (6 + 24*z + 4*z^2 + 8*z^3)/(1-z)^4.
Row 5: GF(z) = (24 + 120*z + 0*z^2 + 48*z^3 + 24*z^4)/(1-z)^5.
Row 6: GF(z) = (120 + 720*z - 120*z^2 + 384*z^3 + 72*z^4 + 144*z^5)/ (1-z)^6.
Row 7: GF(z) = (720 + 5040*z - 1680*z^2 + 3696*z^3 - 432*z^4 + 1296*z^5 + 720*z^6)/(1-z)^7.
Row 8: GF(z) = (5040 + 40320*z - 20160*z^2 + 40320*z^3 - 15840*z^4 + 17280*z^5 + 2880*z^6 + 5760*z^7)/(1-z)^8.
Row 9: GF(z) = (40320 +362880*z -241920*z^2 + 483840*z^3 - 311040*z^4 + 288000*z^5 - 46080*z^6 + 69120*z^7 + 40320*z^8)/(1-z)^9.
Row 10: GF(z) = (362880 +3628800*z -3024000*z^2 +6289920*z^3 -5495040*z^4 + 5276160*z^5 - 2131200*z^6 + 1382400*z^7 + 201600*z^8 + 403200*z^9)/(1-z)^10;

A167546 is the ED1 array.
A000142, A000142 (n=>2) and 120*A062148 (with three extra terms at the beginning of the sequence) equal the first three left hand triangle columns.
A098557(n) and A098557(n)*A064455(n) equal the first two right hand triangle columns.
A007680 equals the row sums.

A324224 Total number T(n,k) of 1's in falling diagonals with index k in all n X n permutation matrices divided by |k|!; triangle T(n,k), n>=1, 1-n<=k<=n-1, read by rows.

Original entry on oeis.org

1, 1, 2, 1, 1, 4, 6, 4, 1, 1, 6, 18, 24, 18, 6, 1, 1, 8, 36, 96, 120, 96, 36, 8, 1, 1, 10, 60, 240, 600, 720, 600, 240, 60, 10, 1, 1, 12, 90, 480, 1800, 4320, 5040, 4320, 1800, 480, 90, 12, 1, 1, 14, 126, 840, 4200, 15120, 35280, 40320, 35280, 15120, 4200, 840, 126, 14, 1

Offset: 1

Alois P. Heinz, Feb 18 2019

			Triangle T(n,k) begins:
  :                                 1                              ;
  :                           1,    2,    1                        ;
  :                     1,    4,    6,    4,    1                  ;
  :               1,    6,   18,   24,   18,    6,   1             ;
  :          1,   8,   36,   96,  120,   96,   36,   8,  1         ;
  :      1, 10,  60,  240,  600,  720,  600,  240,  60, 10,  1     ;
  :  1, 12, 90, 480, 1800, 4320, 5040, 4320, 1800, 480, 90, 12, 1  ;

Alois P. Heinz, Rows n = 1..100, flattened
Wikipedia, Permutation
Wikipedia, Permutation matrix

Columns k=0-6 give (offsets may differ): A000142, A001563, A001286, A005990, A061206, A062199, A062148.
Row sums give A306495(n-1).
Cf. A132159 (right part of triangle), A306234, A324225.

b:= proc(s, c) option remember; (n-> `if`(n=0, c,
      add(b(s minus {i}, c+x^(n-i)), i=s)))(nops(s))
    end:
T:= n-> (p-> seq(coeff(p, x, i)/abs(i)!, i=1-n..n-1))(b({$1..n}, 0)):
seq(T(n), n=1..8);
# second Maple program:
egf:= k-> (t-> x^t/t!*hypergeom([2, t], [t+1], x))(abs(k)+1):
T:= (n, k)-> n! * coeff(series(egf(k), x, n+1), x, n):
seq(seq(T(n, k), k=1-n..n-1), n=1..8);
# third Maple program:
T:= (n, k)-> (t-> `if`(t

T[n_, k_] := With[{t = Abs[k]}, If[tJean-François Alcover, Mar 25 2021, after 3rd Maple program *)

T(n,k) = T(n,-k).
T(n,k) = (n-t)*(n-1)!/t! if t < n with t = |k|, T(n,k) = 0 otherwise.
T(n,k) = 1/|k|! * A324225(n,k).
E.g.f. of column k: x^t/t! * hypergeom([2, t], [t+1], x) with t = |k|+1.
Sum_{k=1-n..n-1} T(n,k) = A306495(n-1).

A062149 Third column sequence of triangle A062138 (generalized a=5 Laguerre).

Original entry on oeis.org

1, 24, 432, 7200, 118800, 1995840, 34594560, 622702080, 11675664000, 228324096000, 4657811558400, 99084354969600, 2196369868492800, 50685458503680000, 1216451004088320000, 30330178368602112000

Offset: 0

Wolfdieter Lang, Jun 19 2001

			a(2) = (2+2)! * binomial(2+7,7) / 2! = (24 * 36) / 2 = 432. - _Indranil Ghosh_, Feb 24 2017

Indranil Ghosh, Table of n, a(n) for n = 0..400
Index entries for sequences related to Laguerre polynomials

Cf. A062148.

[Factorial(n+2)*Binomial(n+7,7)/2: n in [0..20]]; // G. C. Greubel, May 12 2018

a:=n-> (n+2)!*binomial(n+7, 7)/2!: seq(a(n), n=0..22); # Zerinvary Lajos, Apr 29 2007

Table[(n+2)!*Binomial[n+7,7]/2!,{n,0,15}] (* Indranil Ghosh, Feb 24 2017 *)

a(n)=(n+2)!*binomial(n+7, 7)/2! \\ Indranil Ghosh, Feb 24 2017

import math
f=math.factorial
def C(n, r):return f(n)/f(r)/f(n-r)
def A062149(n): return f(n+2)*C(n+7, 7)/f(2) # Indranil Ghosh, Feb 24 2017

E.g.f.: (1 + 14*x + 21*x^2)/(1 - x)^10.
a(n) = A062138(n+2, 2).
a(n) = (n+2)!*binomial(n+7, 7)/2!.
If we define f(n,i,x) = Sum_{k=i..n} Sum_{j=i..k} binomial(k,j) * Stirling1(n,k) * Stirling2(j,i) * x^(k-j), then a(n-2) = (-1)^n*f(n,2,-8), (n >= 2). - Milan Janjic, Mar 01 2009

cp's OEIS Frontend

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A062149 Third column sequence of triangle A062138 (generalized a=5 Laguerre).

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