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.

Showing 1-8 of 8 results.

A143494 Triangle read by rows: 2-Stirling numbers of the second kind.

Original entry on oeis.org

1, 2, 1, 4, 5, 1, 8, 19, 9, 1, 16, 65, 55, 14, 1, 32, 211, 285, 125, 20, 1, 64, 665, 1351, 910, 245, 27, 1, 128, 2059, 6069, 5901, 2380, 434, 35, 1, 256, 6305, 26335, 35574, 20181, 5418, 714, 44, 1, 512, 19171, 111645, 204205, 156660, 58107, 11130, 1110, 54, 1
Offset: 2

Views

Author

Peter Bala, Aug 20 2008

Keywords

Comments

This is the case r = 2 of the r-Stirling numbers of the second kind. The 2-Stirling numbers of the second kind give the number of ways of partitioning the set {1,2,...,n} into k nonempty disjoint subsets with the restriction that the elements 1 and 2 belong to distinct subsets.
More generally, the r-Stirling numbers of the second kind give the number of ways of partitioning the set {1,2,...,n} into k nonempty disjoint subsets with the restriction that the numbers 1, 2, ..., r belong to distinct subsets. The case r = 1 gives the usual Stirling numbers of the second kind A008277; for other cases see A143495 (r = 3) and A143496 (r = 4).
The lower unitriangular array of r-Stirling numbers of the second kind equals the matrix product P^(r-1) * S (with suitable offsets in the row and column indexing), where P is Pascal's triangle, A007318 and S is the array of Stirling numbers of the second kind, A008277.
For the definition of and entries relating to the corresponding r-Stirling numbers of the first kind see A143491. For entries on r-Lah numbers refer to A143497. The theory of r-Stirling numbers of both kinds is developed in [Broder].
From Peter Bala, Sep 19 2008: (Start)
Let D be the derivative operator d/dx and E the Euler operator x*d/dx. Then x^(-2)*E^n*x^2 = Sum_{k = 0..n} T(n+2,k+2)*x^k*D^k.
The row generating polynomials R_n(x) := Sum_{k= 2..n} T(n,k)*x^k satisfy the recurrence R_(n+1)(x) = x*R_n(x) + x*d/dx(R_n(x)) with R_2(x) = x^2. It follows that the polynomials R_n(x) have only real zeros (apply Corollary 1.2. of [Liu and Wang]).
Relation with the 2-Eulerian numbers E_2(n,j) := A144696(n,j): T(n,k) = 2!/k!*Sum_ {j = n-k..n-2} E_2(n,j)*binomial(j,n-k) for n >= k >= 2. (End)
From Wolfdieter Lang, Sep 29 2011: (Start)
T(n,k) = S(n,k,2), n>=k>=2, in Mikhailov's first paper, eq.(28) or (A3). E.g.f. column no. k from (A20) with k->2, r->k. Therefore, with offset [0,0], this triangle is the Sheffer triangle (exp(2*x),exp(x)-1) with e.g.f. of column no. m>=0: exp(2*x)*((exp(x)-1)^m)/m!. See one of the formulas given below. For Sheffer matrices see the W. Lang link under A006232 with the S. Roman reference, also found in A132393. (End)

Examples

			Triangle begins
  n\k|...2....3....4....5....6....7
  =================================
  2..|...1
  3..|...2....1
  4..|...4....5....1
  5..|...8...19....9....1
  6..|..16...65...55...14....1
  7..|..32..211..285..125...20....1
  ...
T(4,3) = 5. The set {1,2,3,4} can be partitioned into three subsets such that 1 and 2 belong to different subsets in 5 ways: {{1}{2}{3,4}}, {{1}{3}{2,4}}, {{1}{4}{2,3}}, {{2}{3}{1,4}} and {{2}{4}{1,3}}; the remaining possibility {{1,2}{3}{4}} is not allowed.
From _Peter Bala_, Feb 23 2025: (Start)
The array factorizes as
/ 1               \       /1             \ /1             \ /1            \
| 2    1           |     | 2   1          ||0  1           ||0  1          |
| 4    5   1       |  =  | 4   3   1      ||0  2   1       ||0  0  1       | ...
| 8   19   9   1   |     | 8   7   4  1   ||0  4   3  1    ||0  0  2  1    |
|16   65  55  14  1|     |16  15  11  6  1||0  8   7  4  1 ||0  0  4  3  1 |
|...               |     |...             ||...            ||...           |
where, in the infinite product on the right-hand side, the first array is the Riordan array (1/(1 - 2*x), x/(1 - x)). See A055248. (End)
		

Crossrefs

A001047 (column 3), A005493 (row sums), A008277, A016269 (column 4), A025211 (column 5), A049444 (matrix inverse), A074051 (alt. row sums).

Programs

  • Maple
    with combinat: T := (n, k) -> (1/(k-2)!)*add ((-1)^(k-i)*binomial(k-2,i)*(i+2)^(n-2),i = 0..k-2): for n from 2 to 11 do seq(T(n, k), k = 2..n) end do;
  • Mathematica
    t[n_, k_] := StirlingS2[n, k] - StirlingS2[n-1, k]; Flatten[ Table[ t[n, k], {n, 2, 11}, {k, 2, n}]] (* Jean-François Alcover, Dec 02 2011 *)
  • Sage
    @CachedFunction
    def stirling2r(n, k, r) :
        if n < r: return 0
        if n == r: return 1 if k == r else 0
        return stirling2r(n-1,k-1,r) + k*stirling2r(n-1,k,r)
    A143494 = lambda n,k: stirling2r(n, k, 2)
    for n in (2..6):
        [A143494(n, k) for k in (2..n)] # Peter Luschny, Nov 19 2012

Formula

T(n+2,k+2) = (1/k!)*Sum_{i = 0..k} (-1)^(k-i)*C(k,i)*(i+2)^n, n,k >= 0.
T(n,k) = Stirling2(n,k) - Stirling2(n-1,k) for n, k >= 2.
Recurrence relation: T(n,k) = T(n-1,k-1) + k*T(n-1,k) for n > 2, with boundary conditions T(n,1) = T(1,n) = 0 for all n, T(2,2) = 1 and T(2,k) = 0 for k > 2. Special cases: T(n,2) = 2^(n-2); T(n,3) = 3^(n-2) - 2^(n-2).
As a sum of monomial functions of degree m: T(n+m,n) = Sum_{2 <= i_1 <= ... <= i_m <= n} (i_1*i_2*...*i_m). For example, T(6,4) = Sum_{2 <= i <= j <= 4} (i*j) = 2*2 + 2*3 + 2*4 + 3*3 + 3*4 + 4*4 = 55.
E.g.f. column k+2 (with offset 2): 1/k!*exp(2*x)*(exp(x) - 1)^k.
O.g.f. k-th column: Sum_{n >= k} T(n,k)*x^n = x^k/((1-2*x)*(1-3*x)*...*(1-k*x)).
E.g.f.: exp(2*t + x*(exp(t) - 1)) = Sum_{n >= 0} Sum_{k = 0..n} T(n+2,k+2) *x^k*t^n/n! = Sum_{n >= 0} B_n(2;x)*t^n/n! = 1 + (2 + x)*t/1! + (4 + 5*x + x^2)*t^2/2! + ..., where the row polynomial B_n(2;x) := Sum_{k = 0..n} T(n+2,k+2)*x^k denotes the 2-Bell polynomial.
Dobinski-type identities: Row polynomial B_n(2;x) = exp(-x)*Sum_{i >= 0} (i + 2)^n*x^i/i!. Sum_{k = 0..n} k!*T(n+2,k+2)*x^k = Sum_{i >= 0} (i + 2)^n*x^i/(1 + x)^(i+1).
The T(n,k) are the connection coefficients between falling factorials and the shifted monomials (x + 2)^(n-2). For example, from row 4 we have 4 + 5*x + x*(x - 1) = (x + 2)^2, while from row 5 we have 8 + 19*x + 9*x*(x - 1) + x*(x - 1)*(x - 2) = (x + 2)^3.
The row sums of the array are the 2-Bell numbers, B_n(2;1), equal to A005493(n-2). The alternating row sums are the complementary 2-Bell numbers, B_n(2;-1), equal to (-1)^n*A074051(n-2).
This array is the matrix product P * S, where P denotes the Pascal triangle, A007318 and S denotes the lower triangular array of Stirling numbers of the second kind, A008277 (apply Theorem 10 of [Neuwirth]).
Also, this array equals the transpose of the upper triangular array A126351. The inverse array is A049444, the signed 2-Stirling numbers of the first kind. See A143491 for the unsigned version of the inverse.
Let f(x) = exp(exp(x)). Then for n >= 1, the row polynomials R(n,x) are given by R(n+2,exp(x)) = 1/f(x)*(d/dx)^n(exp(2*x)*f(x)). Similar formulas hold for A008277, A039755, A105794, A111577 and A154537. - Peter Bala, Mar 01 2012

A062137 Coefficient triangle of generalized Laguerre polynomials n!*L(n,3,x) (rising powers of x).

Original entry on oeis.org

1, 4, -1, 20, -10, 1, 120, -90, 18, -1, 840, -840, 252, -28, 1, 6720, -8400, 3360, -560, 40, -1, 60480, -90720, 45360, -10080, 1080, -54, 1, 604800, -1058400, 635040, -176400, 25200, -1890, 70, -1, 6652800, -13305600, 9313920
Offset: 0

Views

Author

Wolfdieter Lang, Jun 19 2001

Keywords

Comments

The row polynomials s(n,x) := n!*L(n,3,x) = Sum_{m=0..n} a(n,m)*x^m have e.g.f. exp(-z*x/(1-z))/(1-z)^4. They are Sheffer polynomials satisfying the binomial convolution identity s(n,x+y) = Sum_{k=0..n} binomial(n,k)*s(k,x)*p(n-k,y), with polynomials p(n,x) = Sum_{m=1..n} |A008297(n,m)|*(-x)^m, n >= 1 and p(0,x)=1 (for Sheffer polynomials see A048854 for S. Roman reference).
These polynomials appear in the radial part of the l=1 (p-wave) eigen functions for the discrete energy levels of the H-atom. See Messiah reference.
The unsigned version of this triangle is the triangle of unsigned 2-Lah numbers A143497. - Peter Bala, Aug 25 2008
This matrix (unsigned) is embedded in that for n!*L(n,-3,-x). Introduce 0,0,0 to each unsigned row and then add 1,-2,1,4,2,1 to the beginning of the array as the first three rows to generate n!*L(n,-3,-x). - Tom Copeland, Apr 21 2014
From Wolfdieter Lang, Jul 07 2014: (Start)
The standard Rodrigues formula for the monic generalized Laguerre polynomials (also called Laguerre-Sonin polynomials) is Lm(n,alpha,x) := (-1)^n*n!*L(n,3,x) is x^(-alpha)*exp(x)*(d/dx)^n(exp(-x)*x^(n+alpha)).
Another Rodrigues type formula is Lm(n,alpha,x) = exp(x)*x^(-alpha+n+1)*(-x^2*d/dx)^n*(exp(-x)*x^(alpha+1)). This is derivable from the differential - difference relation of Lm(n,alpha,x) which yields first the creation operator formula -(x*d/dx + (-x + alpha + n + 1))*Lm(n,alpha,x) = Lm(n+1,alpha,x) or in the variable q = log(x) the operator -(d/dq + alpha + n + 1 - exp(q)).
The identity (due to Christoph Mayer) (d/dq - (d/dq)W(q))*f(q) = exp(W(q)*d/dq(exp(-W(q)*f(q)) is then iterated with W(q) = W(alpha,n,q) = exp(q) - (alpha + n + 1)*q and a further change of variables leads then to the given result. (End)

Examples

			The triangle a(n,m) begins:
n\m       0        1       2     3    4   5 ...
0:        1
1:        4       -1
2:       20      -10      1
3:      120      -90     18     -1
4:      840     -840    252    -28    1
5:     6720    -8400   3360   -560   40  -1
... Formatted by _Wolfdieter Lang_, Jul 07 2014
For more rows see the link.
n = 2: 2!*L(2,3,x) = 20 - 10*x + x^2.
		

References

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

Crossrefs

For m=0..5 the (unsigned) columns give A001715, A061206, A062141-A062144. The row sums (signed) give A062146, the row sums (unsigned) give A062147.
Cf. A143497. - Peter Bala, Aug 25 2008
Cf. A062139, A105278. - Wolfdieter Lang, Jul 07 2014

Programs

  • Mathematica
    Flatten[Table[((-1)^m)*n!*Binomial[n+3,n-m]/m!,{n,0,9},{m,0,n}]] (* Indranil Ghosh, Feb 23 2017 *)
  • PARI
    row(n) = Vecrev(n!*pollaguerre(n, 3)); \\ Michel Marcus, Feb 06 2021

Formula

a(n, m) = ((-1)^m)*n!*binomial(n+3, n-m)/m!.
E.g.f. for m-th column sequence: ((-x/(1-x))^m)/(m!*(1-x)^4), m >= 0.

A143491 Unsigned 2-Stirling numbers of the first kind.

Original entry on oeis.org

1, 2, 1, 6, 5, 1, 24, 26, 9, 1, 120, 154, 71, 14, 1, 720, 1044, 580, 155, 20, 1, 5040, 8028, 5104, 1665, 295, 27, 1, 40320, 69264, 48860, 18424, 4025, 511, 35, 1, 362880, 663696, 509004, 214676, 54649, 8624, 826, 44, 1, 3628800, 6999840, 5753736, 2655764
Offset: 2

Views

Author

Peter Bala, Aug 20 2008

Keywords

Comments

Essentially the same as A136124 but with column numbers differing by one. See A049444 for a signed version of this array. The unsigned 2-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 and 2 belong to distinct cycles. This is the particular case r = 2 of the unsigned r-Stirling numbers of the first kind, which count the permutations of the set {1,2,...,n} into k disjoint cycles, with the restriction that the numbers 1, 2, ..., r belong to distinct cycles. The case r = 1 gives the usual unsigned Stirling numbers of the first kind, abs(A008275); for other cases see A143492 (r = 3) and A143493 (r = 4). The corresponding 2-Stirling numbers of the second kind can be found in A143494.
In general, the lower unitriangular array of unsigned r-Stirling numbers of the first kind (with suitable offsets in the row and column indexing) equals the matrix product St1 * P^(r-1), where St1 is the array of unsigned Stirling numbers of the first kind, abs(A008275) and P is Pascal's triangle, A007318. The theory of r-Stirling numbers of both kinds is developed in [Broder]. For details of the related r-Lah numbers see A143497.
This sequence also represents the number of permutations in the alternating group An of length k, where the length is taken with respect to the generators set {(12)(ij)}. For a bijective proof of the relation between these numbers and the 2-Stirling numbers of the first kind see the Rotbart link. - Aviv Rotbart, May 05 2011
With offset n=0,k=0 : triangle T(n,k), read by rows, given by [2,1,3,2,4,3,5,4,6,5,...] DELTA [1,0,1,0,1,0,1,0,1,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Sep 29 2011
With offset n=0 and k=0, this is the Sheffer triangle (1/(1-x)^2,-log(1-x)) (in the umbral notation of S. Roman's book this would be called Sheffer for (exp(-2*t),1-exp(-t))). See the e.g.f. given below. Compare also with the e.g.f. for the signed version A049444. - Wolfdieter Lang, Oct 10 2011
Reversed rows correspond to the Betti numbers of the moduli space M(0,n) of smooth Riemann surfaces (see Murri link). - Tom Copeland, Sep 19 2012

Examples

			Triangle begins
  n\k|.....2.....3.....4.....5.....6.....7
  ========================================
  2..|.....1
  3..|.....2.....1
  4..|.....6.....5.....1
  5..|....24....26.....9.....1
  6..|...120...154....71....14.....1
  7..|...720..1044...580...155....20.....1
  ...
T(4,3) = 5. The permutations of {1,2,3,4} with 3 cycles such that 1 and 2 belong to different cycles are: (1)(2)(3 4), (1)(3)(24), (1)(4)(23), (2)(3)(14) and (2)(4)(13). The remaining possibility (3)(4)(12) is not allowed.
From _Aviv Rotbart_, May 05 2011: (Start)
Example of the alternating group permutations numbers:
Triangle begins
  n\k|.....0.....1.....2.....3.....4.....5.....6.....7
  ====================================================
  2..|.....1
  3..|.....1.....2
  4..|.....1.....5.....6
  5..|.....1.....9....26....24
  6..|.....1....14....71...154...120
  7..|.....1....20...155...580..1044..720
A(n,k) = number of permutations in An of length k, with respect to the generators set {(12)(ij)}. For example, A(2,0)=1 (only the identity is there), for A4, the generators are {(12)(13),(12)(14),(12,23),(12)(24),(12)(34)}, thus we have A(4,1)=5 (exactly 5 generators), the permutations of length 2 are:
   (12)(13)(12)(13) = (312)
   (12)(13)(12)(14) = (41)(23)
   (12)(13)(12)(24) = (432)(1)
   (12)(13)(12)(34) = (342)(1)
   (12)(23)(12)(24) = (13)(24)
   (12)(14)(12)(14) = (412)(3)
Namely, A(4,2)=6. Together with the identity [=(12)(12), of length 0. therefore A(4,0)=1] we have 12 permutations, comprising all A4 (4!/2=12). (End)
		

Crossrefs

Cf. A001705 - A001709 (column 3..7), A001710 (row sums), A008275, A049444 (signed version), A136124, A143492, A143493, A143494, A143497.
Cf. A094638.

Programs

  • Maple
    with combinat: T := (n, k) -> (n-2)! * add((n-j-1)*abs(stirling1(j,k-2))/j!,j = k-2..n-2): for n from 2 to 10 do seq(T(n, k), k = 2..n) end do;
  • Mathematica
    t[n_, k_] := (n-2)!*Sum[(n-j-1)*Abs[StirlingS1[j, k-2]]/j!, {j, k-2, n-2}]; Table[t[n, k], {n, 2, 11}, {k, 2, n}] // Flatten (* Jean-François Alcover, Apr 16 2013, after Maple *)

Formula

T(n,k) = (n-2)! * Sum_{j = k-2 .. n-2} (n-j-1)*|stirling1(j,k-2)|/j!.
Recurrence relation: T(n,k) = T(n-1,k-1) + (n-1)*T(n-1,k) for n > 2, with boundary conditions: T(n,1) = T(1,n) = 0, for all n; T(2,2) = 1; T(2,k) = 0 for k > 2.
Special cases: T(n,2) = (n-1)!; T(n,3) = (n-1)!*(1/2 + 1/3 + ... + 1/(n-1)).
T(n,k) = Sum_{2 <= i_1 < ... < i_(n-k) < n} (i_1*i_2*...*i_(n-k)). For example, T(6,4) = Sum_{2 <= i < j < 6} (i*j) = 2*3 + 2*4 + 2*5 + 3*4 + 3*5 + 4*5 = 71.
Row g.f.: Sum_{k = 2..n} T(n,k)*x^k = x^2*(x+2)*(x+3)*...*(x+n-1).
E.g.f. for column (k+2): Sum_{n>=k} T(n+2,k+2)*x^n/n! = (1/k!)*(1/(1-x)^2)*(log(1/(1-x)))^k.
E.g.f.: (1/(1-t))^(x+2) = Sum_{n>=0} Sum_{k = 0..n} T(n+2,k+2)*x^k*t^n/n! = 1 + (2+x)*t/1! + (6+5*x+x^2)*t^2/2! + ... .
This array is the matrix product St1 * P, 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!/2 ( A001710 ). The alternating row sums are (n-2)!.
If we define f(n,i,a) = Sum_{k=0..n-i} binomial(n,k)*Stirling1(n-k,i)*Product_{j=0..k-1} (-a - j), then T(n-1,i) = |f(n,i,2)|, for n=1,2,...; i=0..n. - Milan Janjic, Dec 21 2008
From Gary W. Adamson, Jul 19 2011: (Start)
n-th row of the triangle = top row of M^(n-2), M = a reversed variant of the (1,2) Pascal triangle (Cf. A029635); as follows:
2, 1, 0, 0, 0, 0, ...
2, 3, 1, 0, 0, 0, ...
2, 5, 4, 1, 0, 0, ...
2, 7, 9, 5, 1, 0, ...
... (End)
The reversed, row polynomials of this entry multiplied by (1+x) are the row polynomials of A094638. E.g., (1+x)(1+5x+6x^2) = (1+6x+11x^2+6x^3). - Tom Copeland, Dec 11 2016

A144696 Triangle of 2-Eulerian numbers.

Original entry on oeis.org

1, 1, 2, 1, 7, 4, 1, 18, 33, 8, 1, 41, 171, 131, 16, 1, 88, 718, 1208, 473, 32, 1, 183, 2682, 8422, 7197, 1611, 64, 1, 374, 9327, 49780, 78095, 38454, 5281, 128, 1, 757, 30973, 264409, 689155, 621199, 190783, 16867, 256
Offset: 2

Views

Author

Peter Bala, Sep 19 2008

Keywords

Comments

Let [n] denote the ordered set {1,2,...,n}. The symmetric group S_n consists of the injective mappings p:[n] -> [n]. Such a permutation p has an excedance at position i, 1 <= i < n, if p(i) > i. One well-known interpretation of the Eulerian numbers A(n,k) is that they count the permutations in the symmetric group S_n with k excedances. The triangle of Eulerian numbers is A008292 (but with an offset of 1 in the column numbering). We generalize this definition to restricted permutations as follows.
Let r be a nonnegative integer and 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 p has an excedance at position i, 1 <= i <= n-r, if p(i) > i. Then the r-Eulerian number, denoted by A(r;n,k), is defined as the number of permutations in Permute(n,n-r) having k excedances. Thus the current array of 2-Eulerian numbers gives the number of permutations in Permute(n,n-2) with k excedances. See the example section below for some numerical examples.
Clearly A(0;n,k) = A(n,k). The case r = 1 also produces the ordinary Eulerian numbers A(n,k). There is an obvious bijection from Permute(n,n) to Permute(n,n-1) that preserves the number of excedances of a permutation. Consequently, the 1-Eulerian numbers are equal to the 0-Eulerian numbers: A(1;n,k) = A(0;n,k) = A(n,k).
For other cases of r-Eulerian numbers see A144697 (r = 3), A144698 (r = 4) and A144699 (r = 5). There is also a concept of r-Stirling numbers of the first and second kinds - see A143491 and A143494. If we multiply the entries of the current array by a factor of 2 and then reverse the rows we obtain A120434.
An alternative interpretation of the current array due to [Strosser] involves the 2-excedance statistic of a permutation (see also [Foata & Schutzenberger, Chapitre 4, Section 3]). We define a permutation p in Permute(n,n-2) to have a 2-excedance at position i (1 <=i <= n-2) if p(i) >= i + 2.
Given a permutation p in Permute(n,n-2), define ~p to be the permutation in Permute(n,n-2) that takes i to n+1 - p(n-i-1). The map ~ is a bijection of Permute(n,n-2) with the property that if p has (resp. does not have) an excedance in position i then ~p does not have (resp. has) a 2-excedance at position n-i-1. Hence ~ gives a bijection between the set of permutations with k excedances and the set of permutations with (n-k) 2-excedances. Thus reading the rows of this array in reverse order gives a triangle whose entries count the permutations in Permute(n,n-2) with k 2-excedances.
Example: Represent a permutation p:[n-2] -> [n] in Permute(n,n-2) by its image vector (p(1),...,p(n-2)). In Permute(10,8) the permutation p = (1,2,4,7,10,6,5,8) does not have an excedance in the first two positions (i = 1 and 2) or in the final three positions (i = 6, 7 and 8). The permutation ~p = (3,6,5,1,4,7,9,10) has 2-excedances only in the first three positions and the final two positions.
From Peter Bala, Dec 27 2019: (Start)
This is the array A(1,1,3) in the notation of Hwang et al. (p. 25), where the authors remark that the r-Eulerian numbers were first studied by Shanlan Li (Duoji Bilei, Ch. 4), who gave the summation formulas
Sum_{i = 2..n+1} (i-1)*C(i,2) = C(n+3,4) + 2*C(n+2,4)
Sum_{i = 2..n+1} (i-1)^2*C(i,2) = C(n+4,5) + 7*C(n+3,5) + 4*C(n+2,5)
Sum_{i = 2..n+1} (i-1)^3*C(i,2) = C(n+5,6) + 18*C(n+4,6) + 33*C(n+3,6) + 8*C(n+2,6). (End)

Examples

			The triangle begins
===========================================
n\k|..0.....1.....2.....3.....4.....5.....6
===========================================
2..|..1
3..|..1.....2
4..|..1.....7.....4
5..|..1....18....33.....8
6..|..1....41...171...131....16
7..|..1....88...718..1208...473....32
8..|..1...183..2682..8422..7197..1611....64
...
Row 4 = [1,7,4]: We represent a permutation p:[n-2] -> [n] in Permute(n,n-2) by its image vector (p(1),...,p(n-2)). Here n = 4. The permutation (1,2) has no excedances; 7 permutations have a single excedance, namely, (1,3), (1,4), (2,1), (3,1), (3,2), (4,1) and (4,2); the remaining 4 permutations, (2,3), (2,4), (3,4) and (4,3) each have two excedances.
		

References

  • J. Riordan. An introduction to combinatorial analysis. New York, J. Wiley, 1958.
  • R. Strosser. Séminaire de théorie combinatoire, I.R.M.A., Université de Strasbourg, 1969-1970.
  • Li, Shanlan (1867). Duoji bilei (Series summation by analogies), 4 scrolls. In Zeguxizhai suanxue (Mathematics from the Studio Devoted to the Imitation of the Ancient Chinese Tradition) (Jinling ed.), Volume 4.
  • Li, Shanlan (2019). Catégories analogues d’accumulations discrètes (Duoji bilei), traduit et commenté par Andrea Bréard. La Bibliothèque Chinoise. Paris: Les Belles Lettres.

Crossrefs

Cf. A000079 (right diagonal), A001710 (row sums).
Cf. A000182 (related to alt. row sums).

Programs

  • Magma
    m:=2; [(&+[(-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
    
  • Maple
    with(combinat):
    T:= (n,k) -> 1/2!*add((-1)^(k-j)*binomial(n+1,k-j)*(j+1)^(n-1)*(j+2), j = 0..k):
    for n from 2 to 10 do
    seq(T(n,k),k = 0..n-2)
    end do;
  • Mathematica
    T[n_, k_]:= 1/2!*Sum[(-1)^(k-j)*Binomial[n+1, k-j]*(j+1)^(n-1)*(j+2), {j, 0, k}];
    Table[T[n, k], {n,2,10}, {k,0,n-2}]//Flatten (* Jean-François Alcover, Oct 15 2019 *)
  • SageMath
    m=2 # A144696
    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

Formula

T(n,k) = (1/2!)*Sum_{j = 0..k} (-1)^(k-j)*binomial(n+1,k-j)*(j+1)^(n-1)*(j+2);
T(n,n-k) = (1/2!)*Sum_{j = 2..k} (-1)^(k-j)*binomial(n+1,k-j)*j^(n-1)*(j-1).
Recurrence relations:
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 >= 2, T(2,k) = 0 for k >= 1. Special cases: T(n,n-2) = 2^(n-2); T(n,n-3) = A066810(n-1).
E.g.f. (with suitable offsets): (1/2)*[(1 - x)/(1 - x*exp(t - t*x))]^2 = 1/2 + x*t + (x + 2*x^2)*t^2/2! + (x + 7*x^2 + 4*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_2(x) = 1. It follows that the polynomials R_n(x) for n >= 3 have only real zeros (apply Corollary 1.2. of [Liu and Wang]).
The (n+1)-th row generating polynomial = (1/2!)*Sum_{k = 1..n} (k+1)!*Stirling2(n,k) *x^(k-1)*(1-x)^(n-k).
For n >= 2,
(1/2)*(x*d/dx)^(n-1) (1/(1-x)^2) = x/(1-x)^(n+1) * Sum_{k = 0..n-2} T(n,k)*x^k,
(1/2)*(x*d/dx)^(n-1) (x^2/(1-x)^2) = 1/(1-x)^(n+1) * Sum_{k = 2..n} T(n,n-k)*x^k,
1/(1-x)^(n+1)*Sum_{k = 0..n-2} T(n,k)*x^k = (1/2!) * Sum_{m = 0..inf} (m+1)^(n-1)*(m+2)*x^m,
1/(1-x)^(n+1)*Sum_{k = 2..n} T(n,n-k)*x^k = (1/2!) * Sum_{m = 2..inf} m^(n-1)*(m-1)*x^m.
Worpitzky-type identities:
Sum_{k = 0..n-2} T(n,k)*binomial(x+k,n) = (1/2!)*x^(n-1)*(x - 1);
Sum_{k = 2..n} T(n,n-k)*binomial(x+k,n) = (1/2!)*(x + 1)^(n-1)*(x + 2).
Relation with Stirling numbers (Frobenius-type identities):
T(n+1,k-1) = (1/2!) * Sum_{j = 0..k} (-1)^(k-j)*(j+1)!* binomial(n-j,k-j)*Stirling2(n,j) for n,k >= 1;
T(n+1,k-1) = (1/2!) * Sum_{j = 0..n-k} (-1)^(n-k-j)*(j+1)!* binomial(n-j,k)*S(2;n+2,j+2) for n,k >= 1 and
T(n+2,k) = (1/2!) * Sum_{j = 0..n-k} (-1)^(n-k-j)*(j+2)!* binomial(n-j,k)*S(2;n+2,j+2) for n,k >= 0, where S(2;n,k) denotes the 2-Stirling numbers A143494(n,k).
The row polynomials of this array are related to the Eulerian polynomials. For example, 1/2*x*d/dx [x*(x + 4*x^2 + x^3)/(1-x)^4] = x^2*(1 + 7*x + 4*x^2)/(1-x)^5 and 1/2*x*d/dx [x*(x + 11*x^2 + 11*x^3 + x^4)/(1-x)^5] = x^2*(1 + 18*x + 33*x^2 + 8*x^3)/(1-x)^6.
Row sums A001710. Alternating row sums [1, -1, -2, 8, 16, -136, -272, 3968, 7936, ... ] are alternately (signed) tangent numbers and half tangent numbers - see A000182.
Sum_{k = 0..n-2} 2^k*T(n,k) = A069321(n-1). Sum_{k = 0..n-2} 2^(n-k)*T(n,k) = 4*A083410(n-1).
For n >=2, the shifted row polynomial t*R(n,t) = (1/2)*D^(n-1)(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))^(-2). - Peter Bala, Apr 22 2012

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

Views

Author

Peter Bala, Aug 25 2008

Keywords

Comments

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.

Examples

			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}.
		

Crossrefs

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.

Programs

  • Magma
    /* 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
  • Maple
    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;
  • Mathematica
    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 *)

Formula

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.

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

Views

Author

Peter Bala, Aug 25 2008

Keywords

Comments

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.

Examples

			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}.
		

Crossrefs

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

Programs

  • Maple
    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;
  • Mathematica
    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 *)

Formula

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.

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.

A371277 Triangle read by rows, (3, 2)-Lah numbers.

Original entry on oeis.org

1, 64, 1, 8000, 280, 1, 1728000, 104040, 792, 1, 592704000, 54996480, 681408, 1792, 1, 303464448000, 40685137920, 736404480, 3066560, 3520, 1, 221225582592000, 40988602368000, 1020839500800, 6035420160, 10800000, 6264, 1, 221225582592000000, 54777055334400000, 1804999259750400, 14280657592320, 35670620160, 31941000, 10360, 1
Offset: 2

Views

Author

Aleks Zigon Tankosic, Mar 17 2024

Keywords

Comments

The (3, 2)-Lah numbers T(n, k) count ordered 3-tuples (pi(1), pi(2), pi(3)) 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))= bl(pi(3)) where for i = {1, 2, 3} 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 (3, 2)-Lah numbers T(n, k) are the (m, r)-Lah numbers for m=3 and 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;
             64,              1;
           8000,            280,             1;
        1728000,         104040,           792,          1;
      592704000,       54996480,        681408,       1792,        1;
   303464448000,    40685137920,     736404480,    3066560,     3520,    1;
221225582592000, 40988602368000, 1020839500800, 6035420160, 10800000, 6264,  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 3-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) = 6^3 + 4^3 = 280.
		

Crossrefs

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)^3*T(n-1, k)))
        end:
    seq(seq(T(n, k), k=2..n), n=2..10);
  • Mathematica
    A371277[n_, k_] := A371277[n, k] = Which[n < k || k < 2, 0, n == k, 1, True, A371277[n-1, k-1] + (n+k-1)^3*A371277[n-1, k]];
    Table[A371277[n, k], {n, 2, 10}, {k, 2, n}] (* Paolo Xausa, Jun 12 2024 *)
  • Python
    A371277 = lambda n, k: 0 if (k < 2 or k > n) else (1 if (n == 2 and k == 2) else (A371277(n-1, k-1) + ((n + k - 1)**3) * A371277(n-1, k)))
    print([A371277(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)^3*T(n-1, k).
Explicit formula: T(n, k) = Sum_{3 <= j(1) < j(2) < ... < j(n-k) <= n} (2j(1)-2)^3 * (2j(2)-3)^3 * ... * (2j(n-k)-(n-k+1))^3.
Special cases:
T(n, k) = 0 for n < k or k < 2.
T(n, n) = 1.
T(n, 2) = (A143497(n,2))^3 = ((n+1)!)^3/216.
T(n, n-1) = 2^3 * Sum_{j=2..n-1} j^3.
Showing 1-8 of 8 results.