A062199 -id:A062199 - 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

A062140 Coefficient triangle of generalized Laguerre polynomials n!*L(n,4,x) (rising powers of x).

Original entry on oeis.org

1, 5, -1, 30, -12, 1, 210, -126, 21, -1, 1680, -1344, 336, -32, 1, 15120, -15120, 5040, -720, 45, -1, 151200, -181440, 75600, -14400, 1350, -60, 1, 1663200, -2328480, 1164240, -277200, 34650, -2310, 77, -1, 19958400, -31933440

Offset: 0

Wolfdieter Lang, Jun 19 2001

The row polynomials s(n,x) := n!*L(n,4,x)= sum(a(n,m)*x^m,m=0..n) have g.f. exp(-z*x/(1-z))/(1-z)^5. 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 p(n,x)=sum(|A008297(n,m)|*(-x)^m, m=1..n) and p(0,x)=1 (for Sheffer polynomials see A048854 for S. Roman reference).

			Triangle begins:
  {1};
  {5,-1};
  {30,-12,1};
  {210,-126,21,-1};
  ...
2!*L(2,4,x)=30-12*x+x^2.

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

For m=0..5 the (unsigned) columns give A001720(n+4), A062199, A062260-A062263. The row sums (signed) give A062265, the row sums (unsigned) give A062266.

Cf. A021009, A062137-A062139, A066667.

Flatten[Table[((-1)^m)*n!*Binomial[n+4,n-m]/m!,{n,0,11},{m,0,n}]] (* Indranil Ghosh, Feb 23 2017 *)

row(n) = Vecrev(n!*pollaguerre(n, 4)); \\ Michel Marcus, Feb 06 2021

import math
f=math.factorial
def C(n,r):
    return f(n)//f(r)//f(n-r)
i=0
for n in range(26):
    for m in range(n+1):
        print(i, (-1)**m*f(n)*C(n+4,n-m)//f(m))
        i+=1 # Indranil Ghosh, Feb 23 2017

T(n, m) = ((-1)^m)*n!*binomial(n+4, n-m)/m!.

E.g.f. for m-th column sequence: ((-x/(1-x))^m)/(m!*(1-x)^5), m >= 0.

A264781 Number T(n,k) of permutations of [n] with exactly k (possibly overlapping) occurrences of the consecutive pattern 45321; triangle T(n,k), n >= 0, 0 <= k <= max(0, floor((n-1)/4)), read by rows.

Original entry on oeis.org

1, 1, 2, 6, 24, 119, 1, 708, 12, 4914, 126, 38976, 1344, 347765, 15110, 5, 3447712, 180736, 352, 37598286, 2308548, 9966, 447294144, 31481472, 225984, 5764747515, 457520055, 4753185, 45, 80011430240, 7068885600, 97954080, 21280, 1189835682714, 115808906178

Offset: 0

Alois P. Heinz, Nov 24 2015

Consecutive patterns 12354, 21345, 54312 give the same triangle.

The attached Maple program gives a recurrence for the o.g.f. of each row in terms of u. Using that recurrence we may get any row or column from this irregular triangular array T(n,k). The recurrence follows from manipulation of the bivariate o.g.f./e.g.f. 1/W(u,z) = Sum_{n, k >= 0} T(n, k)*u^k*z^n/n!, whose reciprocal W(u,z) is the solution of the o.d.e. in Theorem 3.2 in Elizalde and Noy (2003) (with m = a = 3). - Petros Hadjicostas, Nov 05 2019

			T(5,1) = 1: 45321.
T(6,1) = 12: 156432, 256431, 356421, 453216, 456321, 463215, 546321, 563214, 564213, 564312, 564321, 645321.
T(9,2) = 5: 786549321, 796548321, 896547321, 897546321, 897645321.
Triangle T(n,k) begins:
00 :           1;
01 :           1;
02 :           2;
03 :           6;
04 :          24;
05 :         119,          1;
06 :         708,         12;
07 :        4914,        126;
08 :       38976,       1344;
09 :      347765,      15110,        5;
10 :     3447712,     180736,      352;
11 :    37598286,    2308548,     9966;
12 :   447294144,   31481472,   225984;
13 :  5764747515,  457520055,  4753185,    45;
14 : 80011430240, 7068885600, 97954080, 21280;

Alois P. Heinz, Rows n = 0..170, flattened
A. Baxter, B. Nakamura, and D. Zeilberger, Automatic generation of theorems and proofs on enumerating consecutive Wilf-classes, 2011.
Sergi Elizalde and Marc Noy, Consecutive patterns in permutations, Adv. Appl. Math. 30 (2003), 110-125; see Theorem 3.2 (p. 116) with m = a = 3.
Petros Hadjicostas, Maple program for a recurrence.

Columns k=0-1 give: A202213, A264896.
Row sums give A000142.
T(4n+1,n) gives A007696.
Cf. A002265, A007696, A062199.

b:= proc(u, o, t) option remember; `if`(u+o=0, 1, add(
       b(u+j-1, o-j, `if`(u+j-30, -1, `if`(t=-1, -2, 0)))), j=1..u)))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n, 0$2)):
seq(T(n), n=0..17);

b[u_, o_, t_] := b[u, o, t] = If[u+o == 0, 1, Sum[
     b[u+j-1, o-j, If[u+j-3 < j, 0, j]], {j, 1, o}] + Expand[
     If[t == -2, x, 1]*Sum[b[u-j, o+j-1, If[j < t || t == -2, 0,
     If[t > 0, -1, If[t == -1, -2, 0]]]], {j, 1, u}]]];
T[n_] := CoefficientList[b[n, 0, 0], x];
T /@ Range[0, 17] // Flatten (* Jean-François Alcover, Feb 19 2021, after Alois P. Heinz *)

Sum_{k > 0} k * T(n,k) = A062199(n-5) for n > 4.

A062260 Third (unsigned) column sequence of triangle A062140 (generalized a=4 Laguerre).

Original entry on oeis.org

1, 21, 336, 5040, 75600, 1164240, 18627840, 311351040, 5448643200, 99891792000, 1917922406400, 38532804710400, 809188898918400, 17739910476288000, 405483668029440000, 9650511299100672000

Offset: 0

Wolfdieter Lang, Jun 19 2001

Harry J. Smith, Table of n, a(n) for n = 0..100
Index entries for sequences related to Laguerre polynomials

Cf. A001720, A062199.

[Factorial(n+2)*Binomial(n+6, 6)/2: n in [0..30]]; // G. C. Greubel, Feb 06 2018

a:=n->sum((n-j)*n!/6!, j=5..n): seq(a(n), n=6..21); # Zerinvary Lajos, Apr 29 2007

Table[(n + 2)! Binomial[n + 6, 6]/2, {n, 0, 20}] (* Wesley Ivan Hurt, Jan 23 2017 *)

{ f=1; for (n=0, 100, f*=n + 2; write("b062260.txt", n, " ", f*binomial(n + 6, 6)/2) ) } \\ Harry J. Smith, Aug 03 2009

[binomial(n,6)*factorial (n-4)/2 for n in range(6, 22)] # Zerinvary Lajos, Jul 07 2009

E.g.f.: (1+12*x+15*x^2)/(1-x)^9.
a(n) = A062140(n+2, 2) = (n+2)!*binomial(n+6, 6)/2!.
If we define f(n,i,x) = Sum_{k=1..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,-7), (n>=2). - Milan Janjic, Mar 01 2009
a(n) = binomial(n,6)*(n-4)!/2, n >= 6. - Zerinvary Lajos, Jul 07 2009

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).

A034865 a(n) = n!*(n-4)/2.

Original entry on oeis.org

0, 60, 720, 7560, 80640, 907200, 10886400, 139708800, 1916006400, 28021593600, 435891456000, 7192209024000, 125536739328000, 2311968282624000, 44816615940096000, 912338253066240000, 19463216065413120000, 434273008459530240000, 10116006549998469120000

Offset: 4

N. J. A. Sloane

nonn

Vincenzo Librandi, Table of n, a(n) for n = 4..200
J. Riordan, Enumeration of trees by height and diameter, IBM J. Res. Dev. 4 (1960), 473-478.
Index entries for sequences related to factorial numbers

[Factorial(n)*(n-4)/2: n in [4..22]]; // Vincenzo Librandi, May 25 2017

Table[n!*(n-4)/2,{n,4,30}] (* Harvey P. Dale, Apr 23 2012 *)

x='x+O('x^30); concat([0], Vec(serlaplace(x^5/(2*(1-x)^2)))) \\ G. C. Greubel, Feb 16 2018

a(n) = 60 * A062199(n-5).

E.g.f.: x^5/(2*(1 - x)^2). - Ilya Gutkovskiy, May 23 2017

A062261 Fourth (unsigned) column sequence of triangle A062140 (generalized a=4 Laguerre).

Original entry on oeis.org

1, 32, 720, 14400, 277200, 5322240, 103783680, 2075673600, 42810768000, 913296384000, 20183850086400, 462393656524800, 10981849342464000, 270322445352960000, 6893222356500480000, 181981070211612672000

Offset: 0

Wolfdieter Lang, Jun 19 2001

Harry J. Smith, Table of n, a(n) for n=0..100
Index entries for sequences related to Laguerre polynomials

Cf. A001720, A062199, A062260.

[Factorial(n+3)*Binomial(n+7,7)/6: n in [0..30]]; // G. C. Greubel, May 13 2018

Table[(n+3)!*Binomial[n+7, 7]/3!, {n, 0, 30}] (* G. C. Greubel, May 13 2018 *)

{ f=2; for (n=0, 100, f*=n + 3; write("b062261.txt", n, " ", f*binomial(n + 7, 7)/6) ) } \\ Harry J. Smith, Aug 03 2009

E.g.f.: (1+21*x+63*x^2+35*x^3)/(1-x)^11.
a(n) = A062140(n+3, 3).
a(n) = (n+3)!*binomial(n+7, 7)/3!.
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-3)=(-1)^(n-1)*f(n,3,-8), (n>=3). - Milan Janjic, Mar 01 2009

A062262 Fifth (unsigned) column sequence of triangle A062140 (generalized a=4 Laguerre).

Original entry on oeis.org

1, 45, 1350, 34650, 831600, 19459440, 454053600, 10702692000, 256864608000, 6307453152000, 158947819430400, 4118193503424000, 109818493424640000, 3015784780968960000, 85303626661693440000

Offset: 0

Wolfdieter Lang, Jun 19 2001

Harry J. Smith, Table of n, a(n) for n = 0..100
Index entries for sequences related to Laguerre polynomials

Cf. A001720, A062199, A062260, A062261.

[Factorial(n+4)*Binomial(n+8,8)/24: n in [0..30]]; // G. C. Greubel, May 13 2018

Table[(n+4)!*Binomial[n+8, 8]/4!, {n, 0, 30}] (* G. C. Greubel, May 13 2018 *)
With[{nn=20},CoefficientList[Series[(1+32x+168x^2+224x^3+70x^4)/(1-x)^13,{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, Mar 18 2025 *)

{ f=6; for (n=0, 100, f*=n + 4; write("b062262.txt", n, " ", f*binomial(n + 8, 8)/24) ) } \\ Harry J. Smith, Aug 03 2009

E.g.f.: (1+32*x+168*x^2+224*x^3+70*x^4)/(1-x)^13.
a(n) = A062140(n+4, 4).
a(n) = (n+4)!*binomial(n+8, 8)/4!.
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-4) = (-1)^n*f(n,4,-9), (n>=4). - Milan Janjic, Mar 01 2009

A062263 Sixth (unsigned) column of triangle A062140 (generalized a=4 Laguerre).

Original entry on oeis.org

1, 60, 2310, 73920, 2162160, 60540480, 1664863200, 45664819200, 1261490630400, 35321737651200, 1006669523059200, 29284931579904000, 871226714502144000, 26538906072526848000, 828392996692445184000

Offset: 0

Wolfdieter Lang, Jun 19 2001

Harry J. Smith, Table of n, a(n) for n = 0..100
Index entries for sequences related to Laguerre polynomials

Cf. A001720, A062140, A062199, A062260, A062261, A062262.

[Factorial(n+5)*Binomial(n+9, 9)/Factorial(5): n in [0..20]]; // G. C. Greubel, May 12 2018

Table[(n+5)!*Binomial[n+9,9]/5!, {n, 0, 20}] (* G. c. Greubel, May 12 2018 *)

{ f=24; for (n=0, 100, f*=n + 5; write("b062263.txt", n, " ", f*binomial(n + 9, 9)/120) ) } \\ Harry J. Smith, Aug 03 2009

E.g.f.: N(4;5, x)/(1-x)^15, with N(4;5, x) := Sum_{k=0..5} A062264(5, k)* x^k = 1 + 45*x + 360*x^2 + 840*x^3 + 630*x^4 + 226*x^5.
a(n) = A062140(n+5, 5).
a(n) = (n+5)!*binomial(n+9, 9)/5!.
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-9) = (-1)^(n-1)*f(n,9,-6), (n>=9). - Milan Janjic, Mar 01 2009

A341743 T(n,k) is the number of labeled Eulerian graphs with n vertices and k edges (according to Harary and Palmer) or the number of labeled connected Eulerian graphs with n vertices and k edges (according to Mallows and Sloane); irregular triangle T, read by rows (n >= 0 and 0 <= k <= n*(n-1)/2).

Original entry on oeis.org

0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 12, 15, 10, 0, 0, 1, 0, 0, 0, 0, 0, 0, 60, 180, 195, 120, 90, 60, 15, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 360, 1890, 3675, 4830, 5061, 4410, 3255, 1935, 805, 252, 105, 35, 0, 0, 1

Offset: 0

Petros Hadjicostas, Feb 18 2021

The value T(0,0) = 0 has no physical meaning. It is there because it makes the formula for the bivariate e.g.f.-o.g.f. (shown below) work.
Since T(n,k) counts even connected graphs with n vertices and k edges, for n >= 2, each vertex must have at least two edges, so k >= n. Hence, T(n,k) = 0 for 0 <= k < n.
We have T(n,n) = (n-1)!/2 for n >= 3 because T(n,n) counts the different labelings of cyclic graphs with n vertices and n edges, and we have (n-1)! cyclic permutations of the numbers 1, 2, ..., n. We divide by 2 because we get the same labeling if we flip the cyclic graph over (like a bracelet).
We have T(n, n*(n-1)/2) = 1, if n is odd, because the complete graph on n vertices is even (each vertex has degree n-1) and has only one non-isomorphic labeling.
We have T(n, n*(n-1)/2 - s) = 0 for s = 0, 1, 2, ..., (n/2)-1, if n is even, because in an even graph with n vertices we cannot have more than n*(n-2)/2 = n*(n-1)/2 - n/2 edges.
Finally, we have T(n, n*(n-2)/2) = A001147(n/2), if n is even >= 4, because any labeling in an even graph with n vertices and n*(n-2)/2 edges corresponds to a perfect matching in a complete graph with n vertices (by considering the pairs of vertices that are not connected).
See the comments for A058878 about the different (and sometimes confusing) terminology regarding even and (connected or not) Euler graphs.

			Irregular triangle T(n,k) (with rows n >= 0 and columns k = 0..n*(n-1)/2) begins
  0;
  1;
  0, 0;
  0, 0, 0, 1;
  0, 0, 0, 0, 3,  0,  0;
  0, 0, 0, 0, 0, 12, 15,  10,   0,   0,  1;
  0, 0, 0, 0, 0,  0, 60, 180, 195, 120, 90, 60, 15, 0, 0, 0;
  ...
T(5,5) = 12 because we have (5-1)!/2 = 12 non-isomorphic labelings of the following Eulerian graph with 5 vertices and 5 edges:
        *
       /  \
      /    \
     /      \
    *        *
     \      /
      \    /
       *--*
T(5,6) = 15 because we have 5*3 = 15 non-isomorphic labelings of the following Eulerian graph with 5 vertices and 6 edges:
         *______*
        /|\    /
       / | \  /
      *  |  \/
       \ |  *
        \|
         *
In the above graph, we have 5 choices for the vertex that is common to both triangles and using the other 4 numbers 1, 2, 3, 4 we have the following 3 possible labelings of the other 4 vertices: {{1,2},{3,4}}, {{1,3},{2,4}}, {{1,4},{2,3}}.
T(5,7) = 10 because we have C(5,2) = 10 non-isomorphic labelings of the following Eulerian graph with 5 vertices and 7 edges:
V = {a,b,c,d,e} and E = {{a,b}, {a,c}, {a,d}, {a,e}, {b,c}, {b,d}, {b,e}}.
T(5,10) = 1 because all labelings of the complete graph with 5 vertices (and C(5,2) = 10 edges) are isomorphic.
There are no other (unlabeled) Eulerian graphs with 5 vertices: A003049(5) = 4. (In the name of A003049, the phrase "connected Euler graphs" is according to Mallows and Sloane (1975). According to Harary and Palmer (1973), we only need to say "Euler graphs" because, for them, an Euler graph is connected and even.)

F. Harary and E. M. Palmer, Graphical Enumeration, Academic Press, NY, 1973; see Eqs. (1.4.7), (1.4.18), and (1.4.19) on pp. 11-16.

P. J. Cameron, Cohomological aspects of two-graphs, Math. Zeit., 157 (1977), 101-119.
C. L. Mallows and N. J. A. Sloane (1975), Two-graphs, switching classes and Euler graphs are equal in number, SIAM Journal on Applied Mathematics, 28(4) (1975), 876-880.
math.stackexchange.com, Is it possible disconnected graph has euler circuit? [sic], August 30, 2015.
Ronald C. Read, Euler graphs on labelled nodes, Canadian Journal of Mathematics, 14 (1962), 482-486; see the discussion in Section 4, following Eq. (8) on p. 486.

Cf. A001147, A001710, A003049, A033678, A058878, A062199.

# Slow program based on Eqs. (1.4.7), (1.4.18), and (1.4.19) in Harary and Palmer (1973).
w := proc(n, y) local m: expand(simplify(2^(-n)*(y + 1)^(1/2*n*(n - 1))*add(binomial(n, m)*((1 - y)/(y + 1))^(m*(n - m)), m = 0 .. n))): end proc:
u := proc(x, y, M) local n: add(w(n, y)*x^n/n!, n = 0 .. M): end proc:
T := proc(n, k) coeftayl(log(u(x, y, n + 2)), [x, y] = [0, 0], [n, k])*n!: end proc:
# Another, slightly faster, program based on one of the recurrences:
S := proc(n, k) local s, t: add(binomial(n, s)*add((-1)^t*binomial(s*(n - s), t)*binomial(binomial(s, 2) + binomial(n - s, 2), k - t), t = 0 .. k), s = 0 .. n)/2^n: end proc: # A058878
T := proc(n, k) local x, s, t: option remember: if n = 0 then x := 0: end if: if 1/2*n*(n - 1) < k then x := 0: end if: if 1 <= n and 0 <= k and k <= 1/2*n*(n - 1) then x := S(n, k) - add(add(binomial(n - 1, s)*T(s + 1, t)*S(n - 1 - s, k - t), t = 0 .. k), s = 0 .. n - 2): end if: x: end proc:
# Third program based on another recurrence (the S(n,k) is as above):
T1 := proc(n, k) local x, s, t: option remember: if k = 0 and (n = 0 or 2 <= n) then x := 0: end if: if n = 1 and k = 0 then x := 1: end if; if 1/2*n*(n - 1) < k then x := 0: end if: if 2 <= n and 1 <= k and k <= 1/2*n*(n - 1) then x := S(n, k) - add(add((t + 1)*binomial(n, s)*T1(s, t + 1)*S(n - s, k - 1 - t)/k, t = 0 .. k - 2), s = 0 .. n) - add(binomial(n, s)*T1(s, k), s = 0 .. n - 1): end if: x: end proc:

S[n_, k_] := S[n, k] = Sum[Binomial[n, s]*Sum[(-1)^t* Binomial[s*(n-s), t]*Binomial[Binomial[s, 2] + Binomial[n-s, 2], k-t], {t, 0, k}], {s, 0, n}]/2^n;
T[n_, k_] := T[n, k] = If[n == 0 || k > n(n-1)/2, 0, S[n, k] - Sum[Binomial[n-1, s]*T[s+1, t]*S[n-1-s, k-t], {t, 0, k}, {s, 0, n-2}]];
Table[T[n, k], {n, 0, 8}, {k, 0, n(n-1)/2}] // Flatten (* Jean-François Alcover, Feb 14 2023, after 2nd Maple program *)

Sum_{k=0..n} T(n,k) = A033678(n) for n >= 1.
Bivariate e.g.f.-o.g.f.: Sum_{n,k>=0} T(n,k)*(x^n/n!)*y^k = log(Sum_{n,k>=0} A058878(n,k)*(x^n/n!)*y^k) = log(Sum_{n >= 0} (x^n/n!)*[o.g.f. of n-th row of A058878](y)).
Sum_{s=0..n} Sum_{t=0..k} binomial(n,s) * T(s+1,t) * A058878(n-s,k-t) = A058878(n+1,k) for n >= 0 and 0 <= k <= n*(n+1)/2.
Sum_{s=0..n} Sum_{t=0..k} ((t+1)/(k+1)) * binomial(n,s) * T(s,t+1) * A058878(n-s,k-t) = A058878(n,k+1) for n >= 2 and 0 <= k <= n*(n-1)/2 - 1
T(n,k) = A058878(n,k) - Sum_{s=0..n-2} Sum_{t=0..k} binomial(n-1,s) * T(s+1,t) * A058878(n-1-s,k-t) for n >= 1 and 0 <= k <= n*(n-1)/2, and T(n,k) = 0 otherwise.
T(n,k) = A058878(n,k) - Sum_{s=0..n} Sum_{t=0..k-2} ((t+1)/(k+1)) * binomial(n,s) * T(s,t+1) * A058878(n-s,k-1-t) - Sum_{s=0..n-1} binomial(n,s) * T(s,k) for n >= 2 and 1 <= k <= n*(n-1)/2 (with T(1,0) = 1 and T(n,k) = 0 otherwise).
T(n,k) = 0 for n >= 2 and 0 <= k <= n-1.
T(n,n) = A001710(n-1) = (n-1)!/2 for n >= 3.
Conjecture: T(n,n+1) = n!*(n-4)/8 = 15*A062199(n-5) for n >= 4 (with A062199(-1) = 0).
T(n, n*(n-1)/2) = 1 if n is odd.
T(n, k) = 0 if n is even and n*(n-1)/2 - n/2 + 1 <= k < n*(n-1)/2.
T(n, n*(n-2)/2) = A001147(n/2) if n is even >= 4.

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).

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A062140 Coefficient triangle of generalized Laguerre polynomials n!*L(n,4,x) (rising powers of x).

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A264781 Number T(n,k) of permutations of [n] with exactly k (possibly overlapping) occurrences of the consecutive pattern 45321; triangle T(n,k), n >= 0, 0 <= k <= max(0, floor((n-1)/4)), read by rows.

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A062260 Third (unsigned) column sequence of triangle A062140 (generalized a=4 Laguerre).

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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.

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A034865 a(n) = n!*(n-4)/2.

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A062261 Fourth (unsigned) column sequence of triangle A062140 (generalized a=4 Laguerre).

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