A143495 -id:A143495 - cp's OEIS frontend

A028025 Expansion of 1/((1-3x)(1-4x)(1-5x)*(1-6x)).

1, 18, 205, 1890, 15421, 116298, 830845, 5709330, 38119741, 249026778, 1599719485, 10142356770, 63639854461, 396031348458, 2448208592125, 15053605980210, 92160458747581, 562225198873338, 3419937140824765

Offset: 0

N. J. A. Sloane

This gives the fourth column of the Sheffer triangle A143495 (3-restricted Stirling2 numbers). See the e.g.f. given below, and comments on the general case under A193685. - Wolfdieter Lang, Oct 08 2011

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (18,-119,342,-360).

Cf. A000453, A025211, A003468, A028165, A028200, A016109, A016075, A016094.

Cf. A143495, A193685.

CoefficientList[Series[1/((1-3x)(1-4x)(1-5x)(1-6x)),{x,0,30}],x] (* or *) LinearRecurrence[{18,-119,342,-360},{1,18,205,1890},30] (* Harvey P. Dale, Jan 29 2024 *)

Vec(1/((1-3*x)*(1-4*x)*(1-5*x)*(1-6*x))+O(x^99)) \\ Charles R Greathouse IV, Sep 26 2012

If we define f(m,j,x) = Sum_{k=j..m} binomial(m,k)*Stirling2(k,j)*x^(m-k) then a(n-3) = f(n,3,3), (n >= 3). - Milan Janjic, Apr 26 2009
a(n) = -5^(n+3)/2 + 2*4^(n+2)+ 6^(n+2) - 3^(n+2)/2. - R. J. Mathar, Mar 22 2011
O.g.f.: 1/((1-3*x)*(1-4*x)*(1-5*x)*(1-6*x)).
E.g.f.: (d^3/dx^3)(exp(3*x)*((exp(x)-1)^3)/3!). - Wolfdieter Lang, Oct 08 2011

A143492 Unsigned 3-Stirling numbers of the first kind.

Original entry on oeis.org

1, 3, 1, 12, 7, 1, 60, 47, 12, 1, 360, 342, 119, 18, 1, 2520, 2754, 1175, 245, 25, 1, 20160, 24552, 12154, 3135, 445, 33, 1, 181440, 241128, 133938, 40369, 7140, 742, 42, 1, 1814400, 2592720, 1580508, 537628, 111769, 14560, 1162, 52, 1, 19958400

Offset: 3

Peter Bala, Aug 20 2008

See A049458 for a signed version of this array. The unsigned 3-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 and 3 belong to distinct cycles. This is the case r = 3 of the unsigned r-Stirling numbers of the first kind. For other cases see abs(A008275) (r = 1), A143491 (r = 2) and A143493 (r = 4). See A143495 for the corresponding 3-Stirling numbers of the second kind. The theory of r-Stirling numbers of both kinds is developed in [Broder]. For details of the related 3-Lah numbers see A143498.
With offset n=0 and k=0, this is the Sheffer triangle (1/(1-x)^3, -log(1-x)) (in the umbral notation of S. Roman's book this would be called Sheffer for (exp(-3*t), 1-exp(-t))). See the e.g.f given below. Compare also with the e.g.f. for the signed version A049458. - Wolfdieter Lang, Oct 10 2011
With offset n=0 and k=0 : triangle T(n,k), read by rows, given by (3,1,4,2,5,3,6,4,7,5,8,6,...) DELTA (1,0,1,0,1,0,1,0,1,0,1,...) where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 31 2011

			Triangle begins
n\k|.....3.....4.....5.....6.....7.....8
========================================
3..|.....1
4..|.....3.....1
5..|....12.....7.....1
6..|....60....47....12.....1
7..|...360...342...119....18.....1
8..|..2520..2754..1175...245....25.....1
...
T(5,4) = 7. The permutations of {1,2,3,4,5} with 4 cycles such that 1, 2 and 3 belong to different cycles are: (14)(2)(3)(5), (15)(2)(3)(4), (24)(1)(3)(5), (25)(1)(3)(4), (34)(1)(2)(5), (35)(1)(2)(4) and (45)(1)(2)(3).

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, 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. A001710 - A001714 (column 3 - column 7), A001715 (row sums), A008275, A049458 (signed version), A143491, A143493, A143495, A143498.

with combinat: T := (n, k) -> (n-3)! * add(binomial(n-j-1,2)*abs(stirling1(j,k-3))/j!,j = k-3..n-3): for n from 3 to 12 do seq(T(n, k), k = 3..n) end do;

T(n,k) = (n-3)! * Sum_{j = k-3 .. n-3} C(n-j-1,2)*|Stirling1(j,k-3)|/j!.
Recurrence relation: T(n,k) = T(n-1,k-1) + (n-1)*T(n-1,k) for n > 3, with boundary conditions: T(n,2) = T(2,n) = 0, for all n; T(3,3) = 1; T(3,k) = 0 for k > 3.
Special cases:
T(n,3) = (n-1)!/2! for n >= 3.
T(n,4) = (n-1)!/2!*(1/3 + ... + 1/(n-1)) for n >= 3.
T(n,k) = Sum_{3 <= i_1 < ... < i_(n-k) < n} (i_1*i_2* ...*i_(n-k)). For example, T(6,4) = Sum_{3 <= i < j < 6} (i*j) = 3*4 + 3*5 + 4*5 = 47.
Row g.f.: Sum_{k = 3..n} T(n,k)*x^k = x^3*(x+3)*(x+4)* ... *(x+n-1).
E.g.f. for column (k+3): Sum_{n = k..inf} T(n+3,k+3)*x^n/n! = 1/k!*1/(1-x)^3 * (log(1/(1-x)))^k.
E.g.f.: (1/(1-t))^(x+3) = Sum_{n = 0..inf} Sum_{k = 0..n} T(n+3,k+3)*x^k*t^n/n! = 1 + (3+x)*t/1! + (12+7*x+x^2)*t^2/2! + ....
This array is the matrix product St1 * P^2, 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!/3! ( A001715 ). The alternating row sums are (n-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,i) = |f(n,i,3)|, for n=1,2,...;i=0...n. - Milan Janjic, Dec 21 2008

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.

A144697 Triangle of 3-Eulerian numbers.

Original entry on oeis.org

1, 1, 3, 1, 10, 9, 1, 25, 67, 27, 1, 56, 326, 376, 81, 1, 119, 1314, 3134, 1909, 243, 1, 246, 4775, 20420, 25215, 9094, 729, 1, 501, 16293, 115105, 248595, 180639, 41479, 2187, 1, 1012, 53388, 590764, 2048710, 2575404, 1193548, 183412, 6561

Offset: 3

Peter Bala, Sep 19 2008

This is the case r = 3 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 3-Eulerian numbers count the permutations in Permute(n,n-3) with k excedances (see the example section below for a numerical example).
For other cases see A008292 (r = 0 and r = 1), A144696 (r = 2), A144698 (r = 4) and A144699 (r = 5).
An alternative interpretation of the current array due to [Strosser] involves the 3-excedance statistic of a permutation (see also [Foata & Schutzenberger, Chapitre 4, Section 3]). We define a permutation p in Permute(n,n-3) to have a 3-excedance at position i (1 <= i <= n-3) if p(i) >= i + 3.
Given a permutation p in Permute(n,n-3), define ~p to be the permutation in Permute(n,n-3) that takes i to n+1 - p(n-i-2). The map ~ is a bijection of Permute(n,n-3) with the property that if p has (resp. does not have) an excedance in position i then ~p does not have (resp. has) a 3-excedance at position n-i-2. Hence ~ gives a bijection between the set of permutations with k excedances and the set of permutations with (n-k) 3-excedances. Thus reading the rows of this array in reverse order gives a triangle whose entries count the permutations in Permute(n,n-3) with k 3-excedances.
Example: Represent a permutation p:[n-3] -> [n] in Permute(n,n-3) by its image vector (p(1),...,p(n-3)). In Permute(10,7) the permutation p = (1,2,4,10,3,6,5) does not have an excedance in the first two positions (i = 1 and 2) or in the final three positions (i = 5, 6 and 7). The permutation ~p = (6,5,8,1,7,9,10) has 3-excedances only in the first three positions and the final two positions.

			Triangle begins
=================================================
n\k|..0......1......2......3......4......5......6
=================================================
3..|..1
4..|..1......3
5..|..1.....10......9
6..|..1.....25.....67.....27
7..|..1.....56....326....376.....81
8..|..1....119...1314...3134...1909....243
9..|..1....246...4775..20420..25215...9094....729
...
T(5,1) = 10: We represent a permutation p:[n-3] -> [n] in Permute(n,n-3) by its image vector (p(1),...,p(n-3)). The 10 permutations in Permute(5,2) having 1 excedance are (1,3), (1,4), (1,5), (3,2), (4,2), (5,2), (2,1), (3,1), (4,1) and (5,1).

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

G. C. Greubel, Rows n = 3..53 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.
Mark Conger, A refinement of the Eulerian polynomials and the joint distribution of pi(1) and Des(pi) in S_n, arXiv:math/0508112 [math.CO], 2005.
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 Géométrique 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, A060056, A143492, A143495, A144696, A144698, A144699.

Cf. A001715 (row sums), A000244 (right diagonal).

m:=3; [(&+[(-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..13]]; // G. C. Greubel, Jun 04 2022

with(combinat):
T:= (n,k) -> 1/3!*add((-1)^(k-j)*binomial(n+1,k-j)*(j+1)^(n-2)*(j+2)*(j+3),j = 0..k):
for n from 3 to 11 do
seq(T(n,k),k = 0..n-3)
end do;

T[n_, k_] /; 0 < k <= n-3 := 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, 3, 11}, {k, 0, n-3}] // Flatten (* Jean-François Alcover, Nov 11 2019 *)

m=3 # A144697
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..13)]) # G. C. Greubel, Jun 04 2022

T(n,k) = (1/3!)*Sum_{j = 0..k} (-1)^(k-j)*binomial(n+1,k-j)*(j+1)^(n-2)*(j+2)*(j+3);
T(n,n-k) = (1/3!)*Sum_{j = 3..k} (-1)^(k-j)*binomial(n+1,k-j)*j^(n-2)*(j-1)*(j-2).
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 >= 3, T(3,k) = 0 for k >= 1. Special cases: T(n,n-3) = 3^(n-3); T(n,n-4) = A086443 (n-2).
E.g.f. (with suitable offsets): (1/3)*((1 - x)/(1 - x*exp(t - t*x)))^3 = 1/3 + x*t + (x + 3*x^2)*t^2/2! + (x + 10*x^2 + 9*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_3(x) = 1. It follows that the polynomials R_n(x) for n >= 4 have only real zeros (apply Corollary 1.2. of [Liu and Wang]).
The (n+2)-th row generating polynomial = (1/3!)*Sum_{k = 1..n} (k+2)!*Stirling2(n,k)*x^(k-1)*(1-x)^(n-k).
For n >= 3,
(1/3)*(x*d/dx)^(n-2) (1/(1-x)^3) = (x/(1-x)^(n+1)) * Sum_{k = 0..n-3} T(n,k)*x^k,
(1/3)*(x*d/dx)^(n-2) (x^3/(1-x)^3) = (1/(1-x)^(n+1)) * Sum_{k = 3..n} T(n,n-k)*x^k,
(1/(1-x)^(n+1)) * Sum_{k = 0..n-3} T(n,k)*x^k = (1/3!) * Sum_{m >= 0} (m+1)^(n-2)*(m+2)*(m+3)*x^m,
(1/(1-x)^(n+1)) * Sum_{k = 3..n} T(n,n-k)*x^k = (1/3!) * Sum_{m >= 3} m^(n-2)*(m-1)*(m-2)*x^m.
Worpitzky-type identities:
Sum_{k = 0..n-3} T(n,k)* binomial(x+k,n) = (1/3!)*x^(n-2)*(x-1)*(x-2);
Sum_{k = 3..n} T(n,n-k)* binomial(x+k,n) = (1/3!)*(x+1)^(n-2)*(x+2)*(x+3).
Relation with Stirling numbers (Frobenius-type identities):
T(n+2,k-1) = (1/3!) * Sum_{j = 0..k} (-1)^(k-j)* (j+2)!* binomial(n-j,k-j)*Stirling2(n,j) for n,k >= 1;
T(n+2,k-1) = (1/3!) * Sum_{j = 0..n-k} (-1)^(n-k-j)* (j+2)!* binomial(n-j,k)*S(3;n+3,j+3) for n,k >= 1 and
T(n+3,k) = (1/3!) * Sum_{j = 0..n-k} (-1)^(n-k-j)*(j+3)!* binomial(n-j,k)*S(3;n+3,j+3) for n,k >= 0, where S(3;n,k) denotes the 3-Stirling numbers A143495(n,k).
The row polynomials of this array are related to the 2-Eulerian polynomials (see A144696). For example, (1/3)*x*d/dx (x^3*(1 + 7*x + 4*x^2)/(1-x)^5) = x^3*(1 + 10*x + 9*x^2)/(1-x)^6 and (1/3)*x*d/dx (x^3*(1 + 18*x + 33*x^2 + 8*x^3)/(1-x)^6) = x^3*(1 + 25*x + 67*x^2 + 27*x^3)/(1-x)^7.
For n >=3, the shifted row polynomial t*R(n,t) = (1/3)*D^(n-2)(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))^(-3). - Peter Bala, Apr 22 2012

A232473 3-Fubini numbers.

Original entry on oeis.org

6, 42, 342, 3210, 34326, 413322, 5544342, 82077450, 1330064406, 23428165002, 445828910742, 9116951060490, 199412878763286, 4646087794988682, 114884369365147542, 3005053671533400330, 82905724863616146966, 2406054103612912660362, 73277364784409578094742, 2336825320400166931304970

Offset: 3

N. J. A. Sloane, Nov 27 2013

Vincenzo Librandi, Table of n, a(n) for n = 3..200
Andrei Z. Broder, The r-Stirling numbers, Discrete Math. 49, 241-259 (1984).
I. Mezo, Periodicity of the last digits of some combinatorial sequences, arXiv preprint arXiv:1308.1637 [math.CO], 2013.
Benjamin Schreyer, Rigged Horse Numbers and their Modular Periodicity, arXiv:2409.03799 [math.CO], 2024. See p. 12.

Cf. A008277, A143494, A143495, A000110, A005493, A005494, A000670, A226738, A232472, A232473, A232474.

r:=3; r_Fubini:=func;
[r_Fubini(n, r): n in [r..22]]; // Bruno Berselli, Mar 30 2016

# r-Stirling numbers of second kind (e.g. A008277, A143494, A143495):
T := (n,k,r) -> (1/(k-r)!)*add ((-1)^(k+i+r)*binomial(k-r,i)*(i+r)^(n-r),i = 0..k-r):
# r-Bell numbers (e.g. A000110, A005493, A005494):
B := (n,r) -> add(T(n,k,r),k=r..n);
SB := r -> [seq(B(n,r),n=r..30)];
SB(2);
# r-Fubini numbers (e.g. A000670, A232472, A232473, A232474):
F := (n,r) -> add((k)!*T(n,k,r),k=r..n);
SF := r -> [seq(F(n,r),n=r..30)];
SF(3);

Fubini[n_, r_] := Sum[k!*Sum[(-1)^(i+k+r)*(i+r)^(n-r)/(i!*(k-i-r)!), {i, 0, k-r}], {k, r, n}]; Table[Fubini[n, 3], {n, 3, 22}] (* Jean-François Alcover, Mar 30 2016 *)

From Peter Bala, Dec 16 2020: (Start)
a(n+3) = Sum_{k = 0..n} (k+3)!/k!*( Sum{i = 0..k} (-1)^(k-i)*binomial(k,i)*(i+3)^n ).
a(n+3) = Sum_{k = 0..n} 3^(n-k)*binomial(n,k)*( Sum_{i = 0..k} Stirling2(k,i)*(i+3)! ).
E.g.f. with offset 0: 6*exp(3*z)/(2 - exp(z))^4 = 6 + 42*z + 342*z^2/2! + 3210*z^3/3! + .... (End)
a(n) ~ n! / (2 * log(2)^(n+1)). - Vaclav Kotesovec, Dec 17 2020

A232474 4-Fubini numbers.

Original entry on oeis.org

24, 216, 2184, 24696, 310344, 4304376, 65444424, 1083832056, 19437971784, 375544415736, 7779464328264, 172062025581816, 4047849158698824, 100946105980181496, 2660400563437957704, 73890563849015945976, 2157336929022064219464, 66059202473570840113656, 2116993226046938197020744

Offset: 4

N. J. A. Sloane, Nov 27 2013

Vincenzo Librandi, Table of n, a(n) for n = 4..200
Andrei Z. Broder, The r-Stirling numbers, Discrete Math. 49, 241-259 (1984).
I. Mezo, Periodicity of the last digits of some combinatorial sequences, arXiv preprint arXiv:1308.1637 [math.CO], 2013.
Benjamin Schreyer, Rigged Horse Numbers and their Modular Periodicity, arXiv:2409.03799 [math.CO], 2024. See p. 12.

Cf. A008277, A143494, A143495, A000110, A005493, A005494, A000670, A232472, A232473, A232474.

r:=4; r_Fubini:=func;
[r_Fubini(n, r): n in [r..22]]; // Bruno Berselli, Mar 30 2016

# r-Stirling numbers of second kind (e.g. A008277, A143494, A143495):
T := (n,k,r) -> (1/(k-r)!)*add ((-1)^(k+i+r)*binomial(k-r,i)*(i+r)^(n-r),i = 0..k-r):
# r-Bell numbers (e.g. A000110, A005493, A005494):
B := (n,r) -> add(T(n,k,r),k=r..n);
SB := r -> [seq(B(n,r),n=r..30)];
SB(2);
# r-Fubini numbers (e.g. A000670, A232472, A232473, A232474):
F := (n,r) -> add((k)!*T(n,k,r),k=r..n);
SF := r -> [seq(F(n,r),n=r..30)];
SF(4);

Fubini[n_, r_] := Sum[k!*Sum[(-1)^(i+k+r)*(i+r)^(n-r)/(i!*(k-i-r)!), {i, 0, k-r}], {k, r, n}]; Table[Fubini[n, 4], {n, 4, 22}] (* Jean-François Alcover, Mar 30 2016 *)

From Peter Bala, Dec 16 2020: (Start)
a(n+4) = Sum_{k = 0..n} (k+4)!/k!*( Sum{i = 0..k} (-1)^(k-i)*binomial(k,i)*(i+4)^n ).
a(n+4) = Sum_{k = 0..n} 4^(n-k)*binomial(n,k)*( Sum_{i = 0..k} Stirling2(k,i)*(i+4)! ).
E.g.f. with offset 0: 24*exp(4*z)/(2 - exp(z))^5 = 24 + 216*z + 2184*z^2/2! + 24696*z^3/3! + .... (End)
a(n) ~ n! / (2 * log(2)^(n+1)). - Vaclav Kotesovec, Dec 17 2020

A016753 Expansion of 1/((1-3x)(1-4x)(1-5*x)).

Original entry on oeis.org

1, 12, 97, 660, 4081, 23772, 133057, 724260, 3863761, 20308332, 105558817, 544039860, 2785713841, 14192221692, 72020501377, 364354427460, 1838822866321, 9262446387852, 46585947584737

Offset: 0

N. J. A. Sloane

As (0,0,1,12,97,...) this is the fourth binomial transform of cosh(x)-1. It is the binomial transform of A016269, when this has two leading zeros. Its e.g.f. is then exp(4x)cosh(x) - exp(4x). - Paul Barry, May 13 2003
This gives the third column of the Sheffer triangle A143495 (3-restricted Stirling2 numbers). See the e.g.f. below, and A193685 for comments on the general case. - Wolfdieter Lang, Oct 08 2011
From Kevin Long, Mar 25 2017: (Start)
In the power set poset 2^(n+2), a(n) gives the number of size 3 subposets {A,B,C} such that A subset of C, B subset of C, and A||B. By symmetry, it also counts the size 3 subposets {A,B,C} such that C subset of A, C subset of B, and A||B.
By the power set poset, I mean the subsets of [n+2] ordered by inclusion. A||B means A and B are incomparable.
The result can be proved by showing that the formula holds. 5^n counts triples (A,B,C) of subsets of [n] where A subset of  C and B subset of C, since for each x in [n], it is either in C only, in A and C, in B and C, in all three, or in none. However, this also counts the cases where A subset of B and where B subset of A, and we want A||B.
Each case can be counted by 4^n, since if A subset of B⊆C, then each element x of [n] is either in all three, in B and C, in only C, or in none. Hence we subtract 2*4^n from 5^n. These two cases intersect, however, when A = B subset of C, which can be counted by 3^n, since each element x of [n] can be either in all three sets, in only C, or in none.
For the purposes of inclusion-exclusion, we add these sets back in to get 5^n-2*4^n+3^n to count all triples (A,B,C) where A subset of C, B subset of C, and A||B. We want sets, not triples, so this double-counts the sets since interchanging A and B give the same set, so we divide this by 2. Hence the formula for a(n) counts these subposets for 2^(n+2).  (End)

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (12,-47,60).

Cf. A000244, A005061, A016753, A132495, A193685, A245019.

[(5^(n+2) - 2*4^(n+2) + 3^(n+2))/2: n in [0..30]]; // G. C. Greubel, Sep 15 2018

CoefficientList[ Series[ 1/((1 - 3x)(1 - 4x)(1 - 5x)), {x, 0, 25} ], x ]
LinearRecurrence[{12,-47,60}, {1, 12, 97}, 30] (* G. C. Greubel, Sep 15 2018 *)

Vec(1/((1-3*x)*(1-4*x)*(1-5*x))+O(x^99)) \\ Charles R Greathouse IV, Sep 26 2012

a(n) = 5^(n+2)/2 - 4^(n+2) + 3^(n+2)/2. - Paul Barry, May 13 2003
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,2,3), (n >= 2). - Milan Janjic, Apr 26 2009
a(n) = 9*a(n-1) - 20*a(n-2) + 3^n, n >= 2. - Vincenzo Librandi, Mar 20 2011
O.g.f.: 1/((1-3*x)*(1-4*x)*(1-5*x)).
E.g.f.: (d^2/dx^2) (exp(3*x)*((exp(x)-1)^2)/2!). - Wolfdieter Lang, Oct 08 2011
a(n) = A245019(n+2)/2. - Kevin Long, Mar 24 2017

A347204 a(n) = a(f(n)/2) + a(floor((n+f(n))/2)) for n > 0 with a(0) = 1 where f(n) = A129760(n).

Original entry on oeis.org

1, 2, 3, 5, 4, 7, 10, 15, 5, 9, 13, 20, 17, 27, 37, 52, 6, 11, 16, 25, 21, 34, 47, 67, 26, 43, 60, 87, 77, 114, 151, 203, 7, 13, 19, 30, 25, 41, 57, 82, 31, 52, 73, 107, 94, 141, 188, 255, 37, 63, 89, 132, 115, 175, 235, 322, 141, 218, 295, 409, 372, 523, 674

Offset: 0

Mikhail Kurkov, Aug 23 2021 [verification needed]

Modulo 2 binomial transform of A243499(n).

Antti Karttunen, Table of n, a(n) for n = 0..8191
Kevin Ryde, PARI/GP Code

Cf. A000110, A000120, A002720, A005493, A005494, A007814, A008277, A035009, A045379, A053645, A129760, A143494, A143495, A196834, A265649, A295989.

Similar recurrences: A124758, A243499, A284005, A329369, A341392.

function a = A347204(max_n)
    a(1) = 1;
    a(2) = 2;
    for nloop = 3:max_n
        n = nloop-1;
        s = 0;
        for k = 0:floor(log2(n))-1
            s = s + a(1+A053645(n)-2^k*(mod(floor(n/(2^k)),2)));
        end
        a(nloop) = 2*a(A053645(n)+1) + s;
    end
end
function a_n = A053645(n)
    a_n = n - 2^floor(log2(n));
end % Thomas Scheuerle, Oct 25 2021

f[n_] := BitAnd[n, n - 1]; a[0] = 1; a[n_] := a[n] = a[f[n]/2] + a[Floor[(n + f[n])/2]]; Array[a, 100, 0] (* Amiram Eldar, Nov 19 2021 *)

f(n) = bitand(n, n-1); \\ A129760
a(n) = if (n<=1, n+1, if (n%2, a(n\2)+a(n-1), a(f(n/2)) + a(n/2+f(n/2)))); \\ Michel Marcus, Oct 25 2021

PARI
```
\\ Also see links.
```

A129760(n) = bitand(n, n-1);
memoA347204 = Map();
A347204(n) = if (n<=1, n+1, my(v); if(mapisdefined(memoA347204,n,&v), v, v = if(n%2, A347204(n\2)+A347204(n-1), A347204(A129760(n/2)) + A347204(n/2+A129760(n/2))); mapput(memoA347204,n,v); (v))); \\ (Memoized version of Michel Marcus's program given above) - Antti Karttunen, Nov 20 2021

a(n) = a(n - 2^f(n)) + (1 + f(n))*a((n - 2^f(n))/2) for n > 0 with a(0) = 1 where f(n) = A007814(n).
a(2n+1) = a(n) + a(2n) for n >= 0.
a(2n) = a(n - 2^f(n)) + a(2n - 2^f(n)) for n > 0 with a(0) = 1 where f(n) = A007814(n).
a(n) = 2*a(f(n)) + Sum_{k=0..floor(log_2(n))-1} a(f(n) - 2^k*T(n,k)) for n > 1 with a(0) = 1, a(1) = 2, and where f(n) = A053645(n), T(n,k) = floor(n/2^k) mod 2.
Sum_{k=0..2^n - 1} a(k) = A035009(n+1) for n >= 0.
a((4^n - 1)/3) = A002720(n) for n >= 0.
a(2^n - 1) = A000110(n+1),
a(2*(2^n - 1)) = A005493(n),
a(2^2*(2^n - 1)) = A005494(n),
a(2^3*(2^n - 1)) = A045379(n),
a(2^4*(2^n - 1)) = A196834(n),
a(2^m*(2^n-1)) = T(n,m+1) is the n-th (m+1)-Bell number for n >= 0, m >= 0 where T(n,m) = m*T(n-1,m) + Sum_{k=0..n-1} binomial(n-1,k)*T(k,m) with T(0,m) = 1.
a(n) = Sum_{j=0..2^A000120(n)-1} A243499(A295989(n,j)) for n >= 0. Also A243499(n) = Sum_{j=0..2^f(n)-1} (-1)^(f(n)-f(j)) a(A295989(n,j)) for n >= 0 where f(n) = A000120(n). In other words, a(n) = Sum_{j=0..n} (binomial(n,j) mod 2)*A243499(j) and A243499(n) = Sum_{j=0..n} (-1)^(f(n)-f(j))*(binomial(n,j) mod 2)*a(j) for n >= 0 where f(n) = A000120(n).
Generalization:
b(n, x) = (1/x)*b((n - 2^f(n))/2, x) + (-1)^n*b(floor((2n - 2^f(n))/2), x) for n > 0 with b(0, x) = 1 where f(n) = A007814(n).
Sum_{k=0..2^n - 1} b(k, x) = (1/x)^n for n >= 0.
b((4^n - 1)/3, x) = (1/x)^n*n!*L_{n}(x) for n >= 0 where L_{n}(x) is the n-th Laguerre polynomial.
b((8^n - 1)/7, x) = (1/x)^n*Sum_{k=0..n} (-x)^k*A265649(n, k) for n >= 0.
b(2^n - 1, x) = (1/x)^n*Sum_{k=0..n} (-x)^k*A008277(n+1, k+1),
b(2*(2^n - 1), x) = (1/x)^n*Sum_{k=0..n} (-x)^k*A143494(n+2, k+2),
b(2^2*(2^n - 1), x) = (1/x)^n*Sum_{k=0..n} (-x)^k*A143495(n+3, k+3),
b(2^m*(2^n - 1), x) = (1/x)^n*Sum_{k=0..n} (-x)^k*T(n+m+1, k+m+1, m+1) for n >= 0, m >= 0 where T(n,k,m) is m-Stirling numbers of the second kind.

A219374 Triangle of F(n,r) of r-geometric numbers, 1 <= r <= n.

Original entry on oeis.org

1, 3, 2, 13, 10, 6, 75, 62, 42, 24, 541, 466, 342, 216, 120, 4683, 4142, 3210, 2184, 1320, 720, 47293, 42610, 34326, 24696, 15960, 9360, 5040, 545835, 498542, 413322, 310344, 211560, 131760, 75600, 40320, 7087261, 6541426, 5544342, 4304376, 3063000, 2005200, 1214640, 685440, 362880, 102247563, 95160302

Offset: 1

R. J. Mathar, Nov 19 2012

			[1] 1;
[2] 3, 2;
[3] 13, 10, 6;
[4] 75, 62, 42, 24;
[5] 541, 466, 342, 216, 120;
[6] 4683, 4142, 3210, 2184, 1320, 720;

S. Alex Bradt, Jennifer Elder, Pamela E. Harris, Gordon Rojas Kirby, Eva Reutercrona, Yuxuan (Susan) Wang, and Juliet Whidden, Unit interval parking functions and the r-Fubini numbers, arXiv:2401.06937 [math.CO], 2024. See Table 1 at page 9.
A. Dil and V. Kurt, Polynomials related to harmonic numbers and evaluation of harmonic number series, II, Appl. An. Disc. Math. 5 (2011) 212-229, section 3.2.

Cf. A008277 {n over k}_1, A143494 {n over k}_2, A143495 {n over k}_3, A000670 (first column).

Stirr := proc(n,k,r)
    option remember;
    if n < r then
        0;
    elif n = r then
        if k = r then
            1 ;
        else
            0 ;
        end if;
    else
        procname(n-1,k-1,r) + k*procname(n-1,k,r) ;
    end if;
end proc:
A := proc(n,r)
    add( k!*Stirr(n,k,r),k=0..n) ;
end proc:
seq(seq( A(n,r),r=1..n),n=1..12) ;

Stirr[n_, k_, r_] := Stirr[n, k, r] = Which[n < r, 0, n == r, If[k == r, 1, 0], True, Stirr[n-1, k-1, r] + k*Stirr[n-1, k, r]]; a[n_, r_] := Sum[ k!*Stirr[n, k, r], {k, 0, n}]; Table[Table[a[n, r], {r, 1, n}], {n, 1, 12}] // Flatten (* Jean-François Alcover, Jan 10 2014, translated from Maple *)
Fubini[n_, r_] := Sum[k!*Sum[(-1)^(i+k+r)*(i+r)^(n-r)/(i!*(k-i-r)!), {i, 0, k-r}], {k, r, n}]; Table[Fubini[n, r], {n, 1, 10}, {r, 1, n}] // Flatten (* Jean-François Alcover, Mar 30 2016 *)

@CachedFunction
def stirling_number2r(n, k, r) :
    if n < r: return 0
    if n == r: return 1 if k == r else 0
    return stirling_number2r(n-1,k-1,r)+ k*stirling_number2r(n-1,k,r)
def A219374(n, r):
    return add(factorial(k)*stirling_number2r(n, k, r) for k in (0..n))
for n in (1..6):
    print([A219374(n, r) for r in (1..n)]) # Peter Luschny, Nov 19 2012

F(n,r) = Sum_{k=0..n} {n over k}_r *k!.

A338282 a(n) = (1/e^n) * Sum_{j>=3} j^n * n^j / (j-3)!.

Original entry on oeis.org

0, 4, 216, 7371, 239424, 8127875, 296315496, 11685617608, 498593804800, 22959117809685, 1137033860419000, 60338078785131785, 3418430599382500800, 206053517402599981504, 13172124530670958537160, 890361160360138336174875, 63463906792476058870550528, 4758276450884470061869230823

Offset: 0

Pedro Caceres, Oct 20 2020

nonn

			a(3) = 7371 = (1/e^3) * Sum_{j>=3} j^3 * 3^j / factorial(j-3).

Cf. A005494, A143495, A242817, A299824.

seq(add(n^(k+3)*A143495(n+3, k+3), k = 0..n), n = 0..17); # Peter Luschny, Oct 21 2020

a[n_] := Exp[-n] * Sum[j^n * n^j/(j - 3)!, {j, 3, Infinity}]; Array[a, 17, 0] (* Amiram Eldar, Oct 20 2020 *)

a(n)={sum(k=0, n+3, n^k*(stirling(n+3,k,2) - 3*stirling(n+2,k,2) + 2*stirling(n+1,k,2)))} \\ Andrew Howroyd, Oct 20 2020

# Increase precision for larger n!
R = RealField(100)
t = 3
sol = [0]*18
for n in range(0, 18):
    suma = R(0)
    for j in range(t, 1000):
        suma += (j^n * n^j) / factorial(j - t)
    suma *= exp(-n)
    sol[n] = round(suma)
print(sol) # Peter Luschny, Oct 20 2020

a(n) = Sum_{k=0..n+3} n^k*(Stirling2(n+3,k) - 3*Stirling2(n+2,k) + 2*Stirling2(n+1,k)). - Andrew Howroyd, Oct 20 2020

a(n) = Sum_{k=0..n} n^(k+3)*A143495(n+3, k+3). - Peter Luschny, Oct 21 2020

cp's OEIS Frontend

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A016753 Expansion of 1/((1-3*x)*(1-4*x)*(1-5*x)).

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A028025 Expansion of 1/((1-3x)(1-4x)(1-5x)*(1-6x)).

A016753 Expansion of 1/((1-3x)(1-4x)(1-5*x)).