cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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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

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Author

Peter Bala, Sep 19 2008

Keywords

Comments

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.

Examples

			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).
		

References

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

Crossrefs

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

Programs

  • Magma
    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
    
  • Maple
    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;
  • Mathematica
    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 *)
  • SageMath
    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

Formula

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

A143493 Unsigned 4-Stirling numbers of the first kind.

Original entry on oeis.org

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

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Author

Peter Bala, Aug 20 2008

Keywords

Comments

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

Examples

			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).
		

Crossrefs

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

Programs

  • Maple
    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;

Formula

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

A144829 Partial products of successive terms of A017209; a(0)=1 .

Original entry on oeis.org

1, 4, 52, 1144, 35464, 1418560, 69509440, 4031547520, 270113683840, 20528639971840, 1744934397606400, 164023833375001600, 16894454837625164800, 1892178941814018457600, 228953651959496233369600, 29763974754734510338048000, 4137192490908096936988672000
Offset: 0

Views

Author

Philippe Deléham, Sep 21 2008

Keywords

Examples

			a(0)=1, a(1)=4, a(2)=4*13=52, a(3)=4*13*22=1144, a(4)=4*13*22*31=35464, ...
		

Crossrefs

Programs

  • Magma
    [n le 2 select 4^(n-1) else (9*n-14)*Self(n-1): n in [1..30]]; // G. C. Greubel, May 26 2022
    
  • Mathematica
    Table[4*9^(n-1)*Pochhammer[13/9, n-1], {n, 0, 20}] (* Vaclav Kotesovec, Nov 29 2021 *)
  • PARI
    a(n) = (-5)^n*sum(k=0, n, (9/5)^k*stirling(n+1,n+1-k, 1)); \\ Michel Marcus, Feb 20 2015
    
  • SageMath
    [9^n*rising_factorial(4/9, n) for n in (0..30)] # G. C. Greubel, May 26 2022

Formula

a(n) = Sum_{k=0..n} A132393(n,k)*4^k*9^(n-k).
a(n) = (-5)^n*Sum_{k=0..n} (9/5)^k*s(n+1,n+1-k), where s(n,k) are the Stirling numbers of the first kind, A048994. - Mircea Merca, May 03 2012
a(n) + (5-9*n)*a(n-1) = 0. - R. J. Mathar, Sep 04 2016
From Vaclav Kotesovec, Nov 29 2021: (Start)
a(n) = 9^n * Gamma(n + 4/9) / Gamma(4/9).
a(n) ~ sqrt(2*Pi) * 9^n * n^(n - 1/18) / (Gamma(4/9) * exp(n)). (End)
From G. C. Greubel, May 26 2022: (Start)
G.f.: hypergeometric2F0([1, 4/9], [], 9*x).
E.g.f.: (1-9*x)^(-4/9). (End)
Sum_{n>=0} 1/a(n) = 1 + (e/9^5)^(1/9)*(Gamma(4/9) - Gamma(4/9, 1/9)). - Amiram Eldar, Dec 21 2022

Extensions

a(9) originally given incorrectly as 20520639971840 corrected by Peter Bala, Feb 20 2015

A167569 The lower left triangle of the ED2 array A167560.

Original entry on oeis.org

1, 2, 4, 6, 16, 32, 24, 80, 192, 384, 120, 480, 1344, 3072, 6144, 720, 3360, 10752, 27648, 61440, 122880, 5040, 26880, 96768, 276480, 675840, 1474560, 2949120, 40320, 241920, 967680, 3041280, 8110080, 19169280, 41287680, 82575360
Offset: 1

Views

Author

Johannes W. Meijer, Nov 10 2009

Keywords

Comments

We discovered that the numbers that appear in the lower left triangle of the ED2 array A167560 (m <= n) behave in a regular way, see the formula below. This rather simple regularity doesn't show up in the upper right triangle of the ED2 array (m > n).

Examples

			The first few triangle rows are:
[1]
[2, 4]
[6, 16, 32]
[24, 80, 192, 384]
[120, 480, 1344, 3072, 6144]
[720, 3360, 10752, 27648, 61440, 122880]
		

Crossrefs

A167560 is the ED2 array.
A047053, 2*A034177 and A167570 are the first three right hand triangle columns.
A000142, 4*A001715, 32*A001725, 384* A049388 and 6144* A049398 are the first five left hand triangle columns.
A167571 equals the row sums.

Programs

  • Maple
    a := proc(n, m): 4^(m-1)*(m-1)!*(n+m-1)!/(2*m-1)! end: seq(seq(a(n, m), m=1..n), n=1..8); # Johannes W. Meijer, revised Nov 23 2012
  • Mathematica
    Flatten[Table[4^(m - 1)*(m - 1)!*(n + m - 1)!/(2*m - 1)!, {n, 1, 50}, {m, n}]] (* G. C. Greubel, Jun 16 2016 *)

Formula

a(n,m) = 4^(m-1)*(m-1)!*(n+m-1)!/(2*m-1)!.

A144886 Lower triangular array called S1hat(4) related to partition number array A144885.

Original entry on oeis.org

1, 4, 1, 20, 4, 1, 120, 36, 4, 1, 840, 200, 36, 4, 1, 6720, 1720, 264, 36, 4, 1, 60480, 12480, 2040, 264, 36, 4, 1, 604800, 118560, 16000, 2296, 264, 36, 4, 1, 6652800, 1081920, 149600, 17280, 2296, 264, 36, 4, 1, 79833600, 11793600, 1362240, 163680, 18304, 2296, 264
Offset: 1

Views

Author

Wolfdieter Lang Oct 09 2008

Keywords

Comments

If in the partition array M31hat(4):=A144885 entries with the same parts number m are summed one obtains this triangle of numbers S1hat(4). In the same way the signless Stirling1 triangle |A008275| is obtained from the partition array M_2 = A036039.
The first columns are A001715(n+2), A144888, A144889,...

Examples

			[1];[4,1];[20,4,1];[120,36,4,1];[840,200,36,4,1];...
		

Crossrefs

A144887 (row sums).

Formula

a(n,m)=sum(product(|S1(4;j,1)|^e(n,m,q,j),j=1..n),q=1..p(n,m)) if n>=m>=1, else 0. Here p(n,m)=A008284(n,m), the number of m parts partitions of n and e(n,m,q,j) is the exponent of j in the q-th m part partition of n. |S1(4,n,1)|= A049352(n,1) = A001715(n+2) = (n+2)!/3!.

A161742 Third left hand column of the RSEG2 triangle A161739.

Original entry on oeis.org

1, 4, 13, 30, -14, -504, 736, 44640, -104544, -10644480, 33246720, 5425056000, -20843695872, -5185511654400, 23457840537600, 8506857655296000, -44092609863966720, -22430879475779174400, 130748316971139072000
Offset: 2

Views

Author

Johannes W. Meijer & Nico Baken (n.h.g.baken(AT)tudelft.nl), Jun 18 2009

Keywords

Crossrefs

Equals third left hand column of A161739 (RSEG2 triangle).
Other left hand columns are A129825 and A161743.
A008955 is a central factorial number triangle.
A028246 is Worpitzky's triangle.
A001710 (n!/2!), A001715 (n!/3!), A001720 (n!/4!), A001725 (n!/5!), A001730 (n!/6!), A049388 (n!/7!), A049389 (n!/8!), A049398 (n!/9!), A051431 (n!/10!) appear in Maple program.

Programs

  • Maple
    nmax:=21; for n from 0 to nmax do A008955(n,0):=1 end do: for n from 0 to nmax do A008955(n,n):=(n!)^2 end do: for n from 1 to nmax do for m from 1 to n-1 do A008955(n,m):= A008955(n-1,m-1)*n^2+A008955(n-1,m) end do: end do: for n from 1 to nmax do A028246(n,1):=1 od: for n from 1 to nmax do A028246(n,n):=(n-1)! od: for n from 3 to nmax do for m from 2 to n-1 do A028246(n,m):=m*A028246(n-1,m)+(m-1)*A028246(n-1,m-1) od: od: for n from 2 to nmax do a(n):=sum(((-1)^k/((k+1)!*(k+2)!)) *(n!)*A028246(n,k+2)* A008955(k+1,k),k=0..n-2) od: seq(a(n),n=2..nmax);

Formula

a(n) = sum(((-1)^k/((k+1)!*(k+2)!))*(n!)*A028246(n, k+2)*A008955(k+1, k), k=0..n-2)

A161743 Fourth left hand column of the RSEG2 triangle A161739.

Original entry on oeis.org

1, 10, 73, 425, 1561, -2856, -73520, 380160, 15376416, -117209664, -7506967104, 72162155520, 7045087741056, -80246202992640, -11448278791372800, 149576169325363200, 30017051616972275712, -440857664887810867200
Offset: 3

Views

Author

Johannes W. Meijer & Nico Baken (n.h.g.baken(AT)tudelft.nl), Jun 18 2009

Keywords

Crossrefs

Equals fourth left hand column of A161739 (RSEG2 triangle).
Other left hand columns are A129825 and A161742.
A008955 is a central factorial number triangle.
A028246 is Worpitzky's triangle.
A001710 (n!/2!), A001715 (n!/3!), A001720 (n!/4!), A001725 (n!/5!), A001730 (n!/6!), A049388 (n!/7!), A049389 (n!/8!), A049398 (n!/9!), A051431 (n!/10!) appear in Maple program.

Programs

  • Maple
    nmax:=21; for n from 0 to nmax do A008955(n,0):=1 end do: for n from 0 to nmax do A008955(n,n):=(n!)^2 end do: for n from 1 to nmax do for m from 1 to n-1 do A008955(n,m):= A008955(n-1,m-1)*n^2+A008955(n-1,m) end do: end do: for n from 1 to nmax do A028246(n,1):=1 od: for n from 1 to nmax do A028246(n,n):=(n-1)! od: for n from 3 to nmax do for m from 2 to n-1 do A028246(n,m):=m*A028246(n-1,m)+(m-1)*A028246(n-1,m-1) od: od: for n from 3 to nmax do a(n) := sum(((-1)^k/((k+2)!*(k+3)!))*(n!)*A028246(n,k+3)* A008955(k+2,k), k=0..n-3) od: seq(a(n),n=3..nmax);

Formula

a(n) = sum(((-1)^k/((k+2)!*(k+3)!))*(n!)*A028246(n, k+3)*A008955(k+2, k), k = 0..n-3).

A162995 A scaled version of triangle A162990.

Original entry on oeis.org

1, 3, 1, 12, 4, 1, 60, 20, 5, 1, 360, 120, 30, 6, 1, 2520, 840, 210, 42, 7, 1, 20160, 6720, 1680, 336, 56, 8, 1, 181440, 60480, 15120, 3024, 504, 72, 9, 1, 1814400, 604800, 151200, 30240, 5040, 720, 90, 10, 1
Offset: 1

Views

Author

Johannes W. Meijer, Jul 27 2009

Keywords

Comments

We get this scaled version of triangle A162990 by dividing the coefficients in the left hand columns by their 'top-values' and then taking the square root.
T(n,k) = A173333(n+1,k+1), 1 <= k <= n. - Reinhard Zumkeller, Feb 19 2010
T(n,k) = A094587(n+1,k+1), 1 <= k <= n. - Reinhard Zumkeller, Jul 05 2012

Examples

			The first few rows of the triangle are:
[1]
[3, 1]
[12, 4, 1]
[60, 20, 5, 1]
		

Crossrefs

Cf. A094587.
A056542(n) equals the row sums for n>=1.
A001710, A001715, A001720, A001725, A001730, A049388, A049389, A049398, A051431 are related to the left hand columns.
A000012, A009056, A002378, A007531, A052762, A052787, A053625 and A159083 are related to the right hand columns.

Programs

  • Haskell
    a162995 n k = a162995_tabl !! (n-1) !! (k-1)
    a162995_row n = a162995_tabl !! (n-1)
    a162995_tabl = map fst $ iterate f ([1], 3)
       where f (row, i) = (map (* i) row ++ [1], i + 1)
    -- Reinhard Zumkeller, Jul 04 2012
  • Maple
    a := proc(n, m): (n+1)!/(m+1)! end: seq(seq(a(n, m), m=1..n), n=1..9); # Johannes W. Meijer, revised Nov 23 2012
  • Mathematica
    Table[(n+1)!/(m+1)!, {n, 10}, {m, n}] (* Paolo Xausa, Mar 31 2024 *)

Formula

a(n,m) = (n+1)!/(m+1)! for n = 1, 2, 3, ..., and m = 1, 2, ..., n.

A371081 Triangle read by rows, (2, 2)-Lah numbers.

Original entry on oeis.org

1, 16, 1, 400, 52, 1, 14400, 2948, 116, 1, 705600, 203072, 12344, 216, 1, 45158400, 17154432, 1437472, 38480, 360, 1, 3657830400, 1760601600, 191088544, 6978592, 99320, 556, 1, 365783040000, 216690624000, 29277351936, 1370470592, 26445312, 224420, 812, 1
Offset: 2

Views

Author

Aleks Zigon Tankosic, Mar 10 2024

Keywords

Comments

The (2, 2)-Lah numbers T(n, k) count ordered 2-tuples (pi(1), pi(2)) of partitions of the set {1, ..., n} into k linearly ordered blocks (lists, for short) such that the numbers 1, 2 are in distinct lists, and bl(pi(1)) = bl(pi(2)) where for i = {1, 2} and pi(i) = b(1)^i, b(2)^i, ..., b(k)^i, where b(1)^i, b(2)^i, ..., b(k)^i are the blocks of partition pi(i), bl(pi(i)) = {min(b(1))^i, min(b(2))^i, ..., min(b(k))^i} is the set of block leaders, i.e., of minima of the lists in partition pi(i).
The (2, 2)-Lah numbers T(n, k) are the (m, r)-Lah numbers for m=r=2. More generally, the (m, r)-Lah numbers count ordered m-tuples (pi(1), pi(2), ..., pi(m)) of partitions of the set {1, 2, ..., n} into k linearly ordered blocks (lists, for short) such that the numbers 1, 2, ..., r are in distinct lists, and bl(pi(1)) = bl(pi(2)) = ... = bl(pi(m)) where for i = {1, 2, ..., m} and pi(i) = {b(1)^i, b(2)^i, ..., b(k)^i}, where b(1)^i, b(2)^i,..., b(k)^i are the blocks of partition pi(i), bl(pi(i)) = {min(b(1))^i, min(b(2))^i, ..., min (b(k))^i} is the set of block leaders, i.e., of minima of the lists in partition pi(i).

Examples

			Triangle begins:
         1;
        16,          1;
       400,         52,         1;
     14400,       2984,       116,       1;
    705600,     203072,     12344,     216,     1;
  45158400,    1715443,   1437472,   38480,   360,    1;
3657830400, 1760601600, 191088544,  6978592, 99320, 556, 1.
  ...
An example for T(4, 3). The corresponding partitions are
pi(1) = {(1),(2),(3,4)},
pi(2) = {(1),(2),(4,3)},
pi(3) = {(1),(2,3),(4)},
pi(4) = {(1),(3,2),(4)},
pi(5) = {(1),(2,4),(3)},
pi(6) = {(1),(4,2),(3)},
pi(7) = {(1,3),(2),(4)},
pi(8) = {(3,1),(2),(4)},
pi(9) = {(1,4),(2),(3)},
pi(10) = {(4,1),(2),(3)}, since A143497 for n=4, k=3 equals 10. Sets of their block leaders are bl(pi(1)) = bl(pi(2)) = bl(pi(5)) = bl(pi(6)) = bl(pi(9)) = bl(pi(10)) = {1,2,3} and
 bl(pi(3)) = bl(pi(4)) = bl(pi(7)) = bl(pi(8)) = {1,2,4}.
Compute the number of ordered 2-tuples (i.e., ordered pairs) of partitions pi(1), pi(2), ..., pi(10) such that partitions in the same pair share the same set of block leaders. As there are six partitions with the set of block leaders equal to {1,2,3}, and four partitions with the set of block leaders equal to {1,2,4}, T(4, 3) = 6^2 + 4^2 = 52.
		

Crossrefs

Column k=2 gives A001715(n+1)^2.
Cf. A143497, A371259, A371277, A371488 (row sums), A372208.

Programs

  • Maple
    T:= proc(n, k) option remember; `if`(k<2 or k>n, 0,
          `if`(n=k, 1, T(n-1, k-1)+(n+k-1)^2*T(n-1, k)))
        end:
    seq(seq(T(n, k), k=2..n), n=2..10);  # Alois P. Heinz, Mar 11 2024
  • Mathematica
    A371081[n_, k_] := A371081[n, k] = Which[n < k || k < 2, 0, n == k, 1, True, A371081[n-1, k-1] + (n+k-1)^2*A371081[n-1, k]];
    Table[A371081[n, k], {n, 2, 10}, {k, 2, n}] (* Paolo Xausa, Jun 12 2024 *)
  • Python
    A371081 = lambda n, k: 0 if (k < 2 or k > n) else (1 if (n == 2 and k == 2) else (A371081(n-1, k-1) + ((n + k - 1)**2) * A371081(n-1, k)))
    print([A371081(n, k) for n in range(2, 10) for k in range(2, n+1)])

Formula

Recurrence relation: T(n, k) = T(n-1, k-1) + (n+k-1)^2*T(n-1, k).
Explicit formula: T(n, k) = Sum_{3 <= j(1) < j(2) < ... < j(n-k) <= n} (2j(1)-2)^2 * (2j(2)-3)^2 * ... * (2j(n-k)-(n-k+1))^2.
Special cases:
T(n, k) = 0 for n < k or k < 2,
T(n, n) = 1,
T(n, 2) = (A143497(n,2))^2 = A001715(n+1)^2 = ((n+1)!)^2/36,
T(n, n-1) = 2^2 * Sum_{j=2..n-1} j^2.

A092580 Triangle read by rows: T(n,k) is the number of permutations p of [n] in which exactly the first k terms satisfy the up-down property, i.e., p(1)p(3), p(3)

Original entry on oeis.org

1, 1, 1, 3, 1, 2, 12, 4, 3, 5, 60, 20, 15, 9, 16, 360, 120, 90, 54, 35, 61, 2520, 840, 630, 378, 245, 155, 272, 20160, 6720, 5040, 3024, 1960, 1240, 791, 1385, 181440, 60480, 45360, 27216, 17640, 11160, 7119, 4529, 7936, 1814400, 604800, 453600, 272160, 176400, 111600, 71190, 45290, 28839, 50521
Offset: 1

Views

Author

Emeric Deutsch and Warren P. Johnson (wjohnson(AT)bates.edu), Apr 10 2004

Keywords

Comments

Row sums are the factorial numbers (A000142). First column is A001710. Second column is A001715. Diagonal is A000111.

Examples

			T(4,3)=3 because 1432, 2431, 3421 are the only permutations of [4] in which exactly the first 3 entries satisfy the up-down property.
Triangle starts:
    1;
    1,   1;
    3,   1,  2;
   12,   4,  3,  5;
   60,  20, 15,  9, 16;
  360, 120, 90, 54, 35, 61;
  ...
		

Crossrefs

Programs

  • Maple
    b:= proc(u, o) option remember;
          `if`(u+o=0, 1, add(b(o-1+j, u-j), j=1..u))
        end:
    E:= n-> b(n, 0):
    T:= (n, k)-> `if`(n=k, E(n), n!*((k+1)*E(k)-E(k+1))/(k+1)!):
    seq(seq(T(n, k), k=1..n), n=1..10);  # Alois P. Heinz, Aug 12 2016
  • Mathematica
    b[u_, o_] := b[u, o] = If[u + o == 0, 1, Sum[b[o - 1 + j, u - j], {j, 1, u}]]; e[n_] := b[n, 0]; T[n_, k_] := If[n == k, e[n], n!*((k + 1)*e[k] - e[k + 1])/(k + 1)!]; Table[T[n, k], {n, 1, 10}, {k, 1, n}] // Flatten (* Jean-François Alcover, Dec 21 2016, after Alois P. Heinz *)

Formula

T(n, k) = n!*[(k+1)*E(k)-E(k+1)]/(k+1)! for k=0} [E(n)x^n/n!] (i.e., E(n) = A000111(n)).
Sum_{k=0..n} (k+1) * T(n,k) = A230960(n). - Alois P. Heinz, Apr 27 2023
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