A143496 -id:A143496 - cp's OEIS frontend

A143493 Unsigned 4-Stirling numbers of the first kind.

1, 4, 1, 20, 9, 1, 120, 74, 15, 1, 840, 638, 179, 22, 1, 6720, 5944, 2070, 355, 30, 1, 60480, 60216, 24574, 5265, 625, 39, 1, 604800, 662640, 305956, 77224, 11515, 1015, 49, 1, 6652800, 7893840, 4028156, 1155420, 203889, 22680, 1554, 60, 1, 79833600

Offset: 4

Peter Bala, Aug 20 2008

See A049459 for a signed version of the array. The unsigned 4-Stirling numbers of the first kind count the permutations of the set {1,2,...,n} into k disjoint cycles, with the restriction that the elements 1, 2, 3 and 4 belong to distinct cycles. This is the case r = 4 of the unsigned r-Stirling numbers of the first kind. For other cases see abs(A008275) (r = 1), A143491 (r = 2) and A143492 (r = 3). See A143496 for the corresponding triangle of 4-Stirling numbers of the second kind.
The theory of r-Stirling numbers of both kinds is developed in [Broder]. For details of the related 4-Lah numbers see A143499.
With offset n=0 and k=0, this is the Sheffer triangle (1/(1-x)^4,-log(1-x)) (in the umbral notation of S. Roman's book this would be called Sheffer for (exp(-4*t),1-exp(-t))). See the e.g.f given below. Compare also with the e.g.f. for the signed version A049459. - Wolfdieter Lang, Oct 10 2011
With offset n=0 and k=0: triangle T(n,k), read by rows, given by (4,1,5,2,6,3,7,4,8,5,9,6,...) DELTA (1,0,1,0,1,0,1,0,1,0,1,0,...) where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 31 2011

			Triangle begins
  n\k|     4     5     6     7     8     9
  ========================================
  4  |     1
  5  |     4     1
  6  |    20     9     1
  7  |   120    74    15     1
  8  |   840   638   179    22     1
  9  |  6720  5944  2070   355    30     1
  ...
T(6,5) = 9. The 9 permutations of {1,2,3,4,5,6} with 5 cycles such that 1, 2, 3 and 4 belong to different cycles are: (1,5)(2)(3)(4)(6), (1,6)(2)(3)(4)(5), (2,5)(1)(3)(4)(6), (2,6)(1)(3)(4)(5), (3,5)(1)(2)(4)(6), (3,6)(1)(2)(4)(5), (4,5)(1)(2)(3)(6), (4,6)(1)(2)(3)(5) and (5,6)(1)(2)(3)(4).

Broder Andrei Z., The r-Stirling numbers, Discrete Math. 49, 241-259 (1984)
A. Dzhumadildaev and D. Yeliussizov, Path decompositions of digraphs and their applications to Weyl algebra, arXiv preprint arXiv:1408.6764v1 [math.CO], 2014. [Version 1 contained many references to the OEIS, which were removed in Version 2. - _N. J. A. Sloane_, Mar 28 2015]
Askar Dzhumadil’daev and Damir Yeliussizov, Walks, partitions, and normal ordering, Electronic Journal of Combinatorics, 22(4) (2015), #P4.10.
Erich Neuwirth, Recursively defined combinatorial functions: Extending Galton's board, Discrete Math. 239 No. 1-3, 33-51 (2001).
Michael J. Schlosser and Meesue Yoo, Elliptic Rook and File Numbers, Electronic Journal of Combinatorics, 24(1) (2017), #P1.31.

Cf. A001715 - A001719 (column 4 - column 8), A001720 (row sums), A008275, A049459 (signed version), A143491, A143492, A143496, A143499.

with combinat: T := (n, k) -> (n-4)! * add(binomial(n-j-1,3)*abs(stirling1(j,k-4))/j!,j = k-4..n-4): for n from 4 to 13 do seq(T(n, k), k = 4..n) end do;

T(n,k) = (n-4)! * Sum_{j = k-4 .. n-4} C(n-j-1,3)*|stirling1(j,k-4)|/j!.
Recurrence relation: T(n,k) = T(n-1,k-1) + (n-1)*T(n-1,k) for n > 4, with boundary conditions: T(n,3) = T(3,n) = 0 for all n; T(4,4) = 1; T(4,k) = 0 for k > 4.
Special cases:
T(n,4) = (n-1)!/3!.
T(n,5) = (n-1)!/3!*(1/4 + ... + 1/(n-1)).
T(n,k) = sum {4 <= i_1 < ...< i_(n-k) < n} (i_1*i_2* ...*i_(n-k)). For example, T(7,5) = Sum_{4 <= i < j < 7} (i*j) = 4*5 + 4*6 + 5*6 = 74.
Row g.f.: Sum_{k = 4..n} T(n,k)*x^k = x^4*(x+4)*(x+5)* ... *(x+n-1).
E.g.f. for column (k+4): Sum_{n = k..inf} T(n+4,k+4)*x^n/n! = 1/k!*1/(1-x)^4 * (log(1/(1-x)))^k.
E.g.f.: (1/(1-t))^(x+4) = Sum_{n = 0..inf} Sum_{k = 0..n} T(n+4,k+4)*x^k*t^n/n! = 1 + (4+x)*t/1! + (20+9*x+x^2)*t^2/2! + .... This array is the matrix product St1 * P^3, where St1 denotes the lower triangular array of unsigned Stirling numbers of the first kind, abs(A008275) and P denotes Pascal's triangle, A007318. The row sums are n!/4! ( A001720 ). The alternating row sums are (n-2)!/2!.
If we define f(n,i,a)=sum(binomial(n,k)*stirling1(n-k,i)*product(-a-j,j=0..k-1),k=0..n-i), then T(n+4,i) = |f(n,i,3)|, for n=1,2,...;i=0...n. - Milan Janjic, Dec 21 2008

A144698 Triangle of 4-Eulerian numbers.

Original entry on oeis.org

1, 1, 4, 1, 13, 16, 1, 32, 113, 64, 1, 71, 531, 821, 256, 1, 150, 2090, 6470, 5385, 1024, 1, 309, 7470, 40510, 65745, 33069, 4096, 1, 628, 25191, 221800, 612295, 592884, 194017, 16384, 1, 1267, 81853, 1113919, 4835875, 7843369, 4915423, 1101157, 65536

Offset: 4

Peter Bala, Sep 19 2008

This is the case r = 4 of the r-Eulerian numbers, denoted by A(r;n,k), defined as follows. Let [n] denote the ordered set {1,2,...,n} and let r be a nonnegative integer. Let Permute(n,n-r) denote the set of injective maps p:[n-r] -> [n], which we think of as permutations of n numbers taken n-r at a time. Clearly, |Permute(n,n-r)| = n!/r!. We say that the permutation p has an excedance at position i, 1 <= i <= n-r, if p(i) > i. Then the r-Eulerian number A(r;n,k) is defined as the number of permutations in Permute(n,n-r) with k excedances. Thus the 4-Eulerian numbers are the number of permutations in Permute(n,n-4) with k excedances. For other cases see A008292 (r = 0 and r = 1), A144696 (r = 2), A144697 (r = 3) and A144699 (r = 5).
An alternative interpretation of the current array due to [Strosser] involves the 4-excedance statistic of a permutation (see also [Foata & Schutzenberger, Chapter 4, Section 3]). We define a permutation p in Permute(n,n-4) to have a 4-excedance at position i (1 <= i <= n-4) if p(i) >= i + 4.
Given a permutation p in Permute(n,n-4), define ~p to be the permutation in Permute(n,n-4) that takes i to n+1 - p(n-i-3). The map ~ is a bijection of Permute(n,n-4) with the property that if p has (resp. does not have) an excedance in position i then ~p does not have (resp. has) a 4-excedance at position n-i-3. Hence ~ gives a bijection between the set of permutations with k excedances and the set of permutations with (n-k) 4-excedances. Thus reading the rows of this array in reverse order gives a triangle whose entries are the number of permutations in Permute(n,n-4) with k 4-excedances.
Example: Represent a permutation p:[n-4] -> [n] in Permute(n,n-4) by its image vector (p(1),...,p(n-4)). In Permute(10,6) the permutation p = (1,2,4,10,3,6) does not have an excedance in the first two positions (i = 1 and 2) or in the final two positions (i = 5 and 6). The permutation ~p = (5,8,1,7,9,10) has 4-excedances only in the first two positions and the final two positions.

			Triangle begins
  ===+=============================================
  n\k|  0      1      2      3      4      5      6
  ===+=============================================
   4 |  1
   5 |  1      4
   6 |  1     13     16
   7 |  1     32    113     64
   8 |  1     71    531    821    256
   9 |  1    150   2090   6470   5385   1024
  10 |  1    309   7470  40510  65745  33069   4096
  ...
T(6,1) = 13: We represent a permutation p:[n-4] -> [n] in Permute(n,n-4) by its image vector (p(1),...,p(n-4)). The 13 permutations in Permute(6,2) having 1 excedance are (1,3), (1,4), (1,5), (1,6), (3,2), (4,2), (5,2), (6,2), (2,1), (3,1), (4,1), (5,1) and (6,1).

R. Strosser, Séminaire de théorie combinatoire, I.R.M.A., Université de Strasbourg, 1969-1970.

G. C. Greubel, Rows n = 4..54 of the triangle, flattened
J. F. Barbero G., J. Salas, and E. J. S. Villaseñor, Bivariate Generating Functions for a Class of Linear Recurrences. II. Applications, arXiv preprint arXiv:1307.5624 [math.CO], 2013-2015.
Ming-Jian Ding and Bao-Xuan Zhu, Some results related to Hurwitz stability of combinatorial polynomials, Advances in Applied Mathematics, Volume 152, (2024), 102591. See p. 9.
Sergi Elizalde, Descents on quasi-Stirling permutations, arXiv:2002.00985 [math.CO], 2020.
D. Foata and M. Schutzenberger, Théorie Géometrique des Polynômes Eulériens, arXiv:math/0508232 [math.CO], 2005; Lecture Notes in Math., no. 138, Springer Verlag, 1970.
L. Liu and Y. Wang, A unified approach to polynomial sequences with only real zeros, arXiv:math/0509207 [math.CO], 2005-2006.
Shi-Mei Ma, Some combinatorial sequences associated with context-free grammars, arXiv:1208.3104v2 [math.CO], 2012. - From _N. J. A. Sloane_, Aug 21 2012

Cf. A008292, A143493, A143496, A143499, A144696, A144697, A144699.

Cf. A001720 (row sums), A000302 (right diagonal).

m:=4; [(&+[(-1)^(k-j)*Binomial(n+1,k-j)*Binomial(j+m,m-1)*(j+1)^(n-m+1): j in [0..k]])/m: k in [0..n-m], n in [m..m+10]]; // G. C. Greubel, Jun 04 2022

with(combinat):
T:= (n,k) -> 1/4!*add((-1)^(k-j)*binomial(n+1,k-j)*(j+1)^(n-3)*(j+2)*(j+3)*(j+4),j = 0..k):
for n from 4 to 12 do
seq(T(n,k),k = 0..n-4)
end do;

T[n_, k_] /; 0 < k <= n-4 := T[n, k] = (k+1) T[n-1, k] + (n-k) T[n-1, k-1];
T[, 0] = 1; T[, _] = 0;
Table[T[n, k], {n, 4, 12}, {k, 0, n-4}] // Flatten (* Jean-François Alcover, Nov 11 2019 *)

m=4 # A144698
def T(n,k): return (1/m)*sum( (-1)^(k-j)*binomial(n+1,k-j)*binomial(j+m,m-1)*(j+1)^(n-m+1) for j in (0..k) )
flatten([[T(n,k) for k in (0..n-m)] for n in (m..m+10)]) # G. C. Greubel, Jun 04 2022

T(n,k) = (1/4!)*Sum_{j = 0..k} (-1)^(k-j)*binomial(n+1,k-j)*(j+1)^(n-3)*(j+2)*(j+3)*(j+4).
T(n,n-k) = (1/4!)*Sum_{j = 4..k} (-1)^(k-j)*binomial(n+1,k-j)*j^(n-3)*(j-1)*(j-2)*(j-3).
Recurrence relation:
T(n,k) = (k + 1)*T(n-1,k) + (n-k)*T(n-1,k-1) with boundary conditions T(n,0) = 1 for n >= 4, T(4,k) = 0 for k >= 1. Special cases: T(n,n-4) = 4^(n-4); T(n,n-5) = 5^(n-3) - 4^(n-3) - (n-3)*4^(n-4).
E.g.f. (with suitable offsets): 1/4*[(1 - x)/(1 - x*exp(t - t*x))]^4 = 1/4 + x*t + (x + 4*x^2)*t^2/2! + (x + 13*x^2 + 16*x^3)*t^3/3! + ... .
The row generating polynomials R_n(x) satisfy the recurrence R_(n+1)(x) = (n*x + 1)*R_n(x) + x*(1 - x)*d/dx(R_n(x)) with R_4(x) = 1. It follows that the polynomials R_n(x) for n >= 5 have only real zeros (apply Corollary 1.2. of [Liu and Wang]).
The (n+3)-th row generating polynomial = (1/4!)*Sum_{k = 1..n} (k+3)!*Stirling2(n,k)*x^(k-1)*(1-x)^(n-k).
For n >= 4,
1/4*(x*d/dx)^(n-3) (1/(1-x)^4) = x/(1-x)^(n+1) * Sum_{k = 0..n-4} T(n,k)*x^k,
1/4*(x*d/dx)^(n-3) (x^4/(1-x)^4) = 1/(1-x)^(n+1) * Sum_{k = 4..n} T(n,n-k)*x^k,
1/(1-x)^(n+1) * Sum {k = 0..n-4} T(n,k)*x^k = (1/4!) * Sum_{m = 0..inf} (m+1)^(n-3)*(m+2)*(m+3)*(m+4)*x^m,
1/(1-x)^(n+1) * Sum {k = 4..n} T(n,n-k)*x^k = (1/4!) * Sum_{m = 4..inf} m^(n-3)*(m-1)*(m-2)*(m-3)*x^m,
Worpitzky-type identities:
Sum_{k = 0..n-4} T(n,k)*binomial(x+k,n) = (1/4!)*x^(n-3)*(x-1)*(x-2)*(x-3).
Sum_{k = 4..n} T(n,n-k)* binomial(x+k,n) = (1/4!)*(x+1)^(n-3)*(x+2)*(x+3)*(x+4).
Relation with Stirling numbers (Frobenius-type identities):
T(n+3,k-1) = (1/4!) * Sum_{j = 0..k} (-1)^(k-j)* (j+3)!* binomial(n-j,k-j)*Stirling2(n,j) for n,k >= 1;
T(n+3,k-1) = 1/4! * Sum_{j = 0..n-k} (-1)^(n-k-j)*(j+3)!* binomial(n-j,k)*S(4;n+4,j+4) for n,k >= 1 and
T(n+4,k) = 1/4! * Sum_{j = 0..n-k} (-1)^(n-k-j)*(j+4)!* binomial(n-j,k)*S(4;n+4,j+4) for n,k >= 0, where S(4;n,k) denotes the 4-Stirling numbers of the second kind A143496(n,k).
For n >=4, the shifted row polynomial t*R(n,t) = (1/4)*D^(n-3)(f(x,t)) evaluated at x = 0, where D is the operator (1-t)*(1+x)*d/dx and f(x,t) = (1+x*t/(t-1))^(-4). - Peter Bala, Apr 22 2012

A016103 Expansion of 1/((1-4x)(1-5x)(1-6x)).

Original entry on oeis.org

1, 15, 151, 1275, 9751, 70035, 481951, 3216795, 20991751, 134667555, 852639151, 5343198315, 33212784151, 205111785075, 1260114546751, 7708980203835, 46999640806951, 285743822630595, 1733261544204751

Offset: 0

Robert G. Wilson v

2*a(n-2) = 6^n - 2*5^n + 4^n is the number of 3 X n {0,1}-matrices such that: (a) first and second row have a common 1, (b) first and third row have a common 1, (c) second and third row have no common 1. - Andi Fugard and Vladeta Jovovic, Jul 26 2008

This is the third column of the Sheffer triangle A143496 (4-restricted Stirling2 numbers). See A193685 for general comments. - Wolfdieter Lang, Oct 08 2011

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Andi Fugard, Counting first-order models (with n individuals) of syllogisms.
Index entries for linear recurrences with constant coefficients, signature (15,-74,120).

Cf. A051588, A000302, A005060, A003468.

m:=25; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(1/((1-4*x)*(1-5*x)*(1-6*x)))); // Vincenzo Librandi, Jun 24 2013

I:=[1, 15, 151]; [n le 3 select I[n] else 15*Self(n-1)-74*Self(n-2)+120*Self(n-3): n in [1..20]]; // Vincenzo Librandi, Jun 24 2013

CoefficientList[Series[1 / ((1 - 4 x) (1 - 5 x) (1 - 6 x)), {x, 0, 20}], x] (* Vincenzo Librandi, Jun 24 2013 *)

Vec(1/((1-4*x)*(1-5*x)*(1-6*x))+O(x^99)) \\ Charles R Greathouse IV, Sep 26 2012

a(n) = 2^(3 + 2*n) + 2^(1 + n) * 3^(2 + n) - 5^(2 + n). - Andi Fugard, Jul 22 2008
If we define f(m,j,x) = Sum_{k=j..m} binomial(m,k)*Stirling2(k,j)*x^(m-k) then a(n-2) = f(n,2,4), n >= 2. - Milan Janjic, Apr 26 2009
O.g.f.: 1/((1-4*x)*(1-5*x)*(1-6*x)).
E.g.f.: (d^2/dx^2)(exp(4*x)*((exp(x)-1)^2)/2!). See the Sheffer triangle comment above. - Wolfdieter Lang, Oct 08 2011
a(n) = 15*a(n-1) - 74*a(n-2) + 120*a(n-3). - Vincenzo Librandi, Jun 24 2013

A143499 Triangle of unsigned 4-Lah numbers.

Original entry on oeis.org

1, 8, 1, 72, 18, 1, 720, 270, 30, 1, 7920, 3960, 660, 44, 1, 95040, 59400, 13200, 1320, 60, 1, 1235520, 926640, 257400, 34320, 2340, 78, 1, 17297280, 15135120, 5045040, 840840, 76440, 3822, 98, 1, 259459200, 259459200, 100900800, 20180160, 2293200

Offset: 4

Peter Bala, Aug 25 2008

This is the case r = 4 of the unsigned r-Lah numbers L(r;n,k). The unsigned 4-Lah numbers count the partitions of the set {1,2,...,n} into k ordered lists with the restriction that the elements 1, 2, 3 and 4 belong to different lists. For other cases see A105278 (r = 1), A143497 (r = 2 and comments on the general case) and A143498 (r = 3).

The unsigned 4-Lah numbers are related to the 4-Stirling numbers: the lower triangular array of unsigned 4-Lah numbers may be expressed as the matrix product St1(4) * St2(4), where St1(4) = A143493 and St2(4) = A143496 are the arrays of 4-restricted Stirling numbers of the first and second kind respectively. An alternative factorization for the array is as St1 * P^6 * 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|......4......5......6......7......8......9
==============================================
4..|......1
5..|......8......1
6..|.....72.....18......1
7..|....720....270.....30......1
8..|...7920...3960....660.....44......1
9..|..95040..59400..13200...1320.....60......1
...
T(5,4) = 8. The partitions of {1,2,3,4,5} into 4 ordered lists, such that the elements 1, 2, 3 and 4 lie in different lists, are: {1}{2}{3}{4,5} and {1}{2}{3}{5,4}, {1}{2}{4}{3,5} and {1}{2}{4}{5,3}, {1}{3}{4}{2,5} and {1}{3}{4}{5,2}, {2}{3}{4}{1,5} and {2}{3}{4}{5,1}.

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. A007318, A008275, A008277, A049388 (column 4), A105278 (unsigned Lah numbers), A143493, A143496, A143497, A143498.

with combinat: T := (n, k) -> (n-4)!/(k-4)!*binomial(n+3,k+3): for n from 4 to 13 do seq(T(n, k), k = 4..n) end do;

T[n_, k_] := (n-4)!/(k-4)!*Binomial[n+3, k+3]; Table[T[n, k], {n, 4, 10}, {k, 4, n}] // Flatten (* Amiram Eldar, Nov 26 2018 *)

T(n,k) = (n-4)!/(k-4)!*binomial(n+3,k+3), n,k >= 4.
Recurrence: T(n,k) = (n+k-1)*T(n-1,k) + T(n-1,k-1) for n,k >= 4, with the boundary conditions: T(n,k) = 0 if n < 4 or k < 4; T(4,4) = 1.
E.g.f. for column k: Sum_{n >= k} T(n,k)*t^n/(n-4)! = 1/(k-4)!*t^k/(1-t)^(k+4) for k >= 4.
E.g.f: Sum_{n = 4..inf} Sum_{k = 4..n} T(n,k)*x^k*t^n/(n-4)! = (x*t)^4/(1-t)^8*exp(x*t/(1-t)) = (x*t)^4*(1 + (8+x)*t +(72+18*x+x^2)*t^2/2! + ...).
Generalized Lah identity: (x+7)*(x+8)*...*(x+n+2) = Sum_{k = 4..n} T(n,k)*(x-1)*(x-2)*...*(x-k+4).
The polynomials 1/n!*Sum_{k = 4..n+4} T(n+4,k)*(-x)^(k-4) for n >= 0 are the generalized Laguerre polynomials Laguerre(n,7,x).
Array = A132493* A143496 = abs(A008275) * ( A007318 )^6 * A008277 (apply Theorem 10 of [Neuwirth]). Array equals exp(D), where D is the array with the quadratic sequence (8,18,30,44, ... ) on the main subdiagonal and zeros everywhere else.

A143399 Expansion of x^k/Product_{t=k..2k} (1-tx) for k=4.

Original entry on oeis.org

0, 0, 0, 0, 1, 30, 545, 7770, 95781, 1071630, 11192665, 111095490, 1060634861, 9822843030, 88799732385, 787259974410, 6869327386741, 59158464019230, 503954741177705, 4254156112792530, 35637875826743421, 296621138907400230, 2455329298857576625

Offset: 0

Alois P. Heinz, Aug 12 2008

nonn

a(n) is also the number of forests of 4 labeled rooted trees of height at most 1 with n labels, where any root may contain >= 1 labels.

This gives also the fifth column of the Sheffer triangle A143496 (4-restricted Stirling2 numbers). See the e.g.f. given below. See also A193685 for Sheffer comments and the hint for the proof in the o.g.f. formula there. - Wolfdieter Lang, Oct 08 2011

Alois P. Heinz, Table of n, a(n) for n = 0..350
Index entries for sequences related to rooted trees
Index entries for linear recurrences with constant coefficients, signature (30, -355, 2070, -5944, 6720).

4th column of A143395.

a:= proc(k::nonnegint) local M; M := Matrix(k+1, (i,j)-> if (i=j-1) then 1 elif j=1 then [seq(-1* coeff(product(1-t*x, t=k..2*k), x, u), u=1..k+1)][i] else 0 fi); p-> (M^p)[1, k+1] end(4): seq(a(n), n=0..30);

LinearRecurrence[{30,-355,2070,-5944,6720},{0,0,0,0,1},30] (* Harvey P. Dale, Mar 12 2013 *)

G.f.: x^4/((1-4x)(1-5x)(1-6x)(1-7x)(1-8x)).
a(n) = 30a(n-1) -355a(n-2) +2070a(n-3) -5944a(n-4) +6720a(n-5).
E.g.f.: exp(4*x)*((exp(x)-1)^4)/4!. - Wolfdieter Lang, Oct 08 2011

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A143493 Unsigned 4-Stirling numbers of the first kind.

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A144698 Triangle of 4-Eulerian numbers.

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A016103 Expansion of 1/((1-4x)(1-5x)(1-6x)).

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A143499 Triangle of unsigned 4-Lah numbers.

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A143399 Expansion of x^k/Product_{t=k..2k} (1-tx) for k=4.

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