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 1-5 of 5 results.

A143496 4-Stirling numbers of the second kind.

Original entry on oeis.org

1, 4, 1, 16, 9, 1, 64, 61, 15, 1, 256, 369, 151, 22, 1, 1024, 2101, 1275, 305, 30, 1, 4096, 11529, 9751, 3410, 545, 39, 1, 16384, 61741, 70035, 33621, 7770, 896, 49, 1, 65536, 325089, 481951, 305382, 95781, 15834, 1386, 60, 1, 262144, 1690981, 3216795
Offset: 4

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Author

Peter Bala, Aug 20 2008

Keywords

Comments

This is the case r = 4 of the r-Stirling numbers of the second kind. The 4-Stirling numbers of the second kind count the ways of partitioning the set {1,2,...,n} into k nonempty disjoint subsets with the restriction that the elements 1, 2, 3 and 4 belong to distinct subsets. For remarks on the general case see A143494 (r = 2). The corresponding array of 4-Stirling numbers of the first kind is A143493. The theory of r-Stirling numbers of both kinds is developed in [Broder]. For 4-Lah numbers refer to A143499.
From Wolfdieter Lang, Sep 29 2011: (Start)
T(n,k) = S(n,k,4), n >= k >= 4, in Mikhailov's first paper, eq.(28) or (A3). E.g.f. column k from (A20) with k->4, r->k. Therefore, with offset [0,0], this triangle is the Sheffer triangle (exp(4*x),exp(x)-1) with e.g.f. of column no. m >= 0: exp(4*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|.....4.....5.....6.....7.....8.....9
========================================
4..|.....1
5..|.....4.....1
6..|....16.....9.....1
7..|....64....61....15.....1
8..|...256...369...151....22.....1
9..|..1024..2101..1275...305....30.....1
...
T(6,5) = 9. The set {1,2,3,4,5,6} can be partitioned into five subsets such that 1, 2, 3 and 4 belong to different subsets in 9 ways: {{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. A003468 (column 7), A005060 (column 5), A008277, A016103 (column 6), A045379 (row sums), A049459 (matrix inverse), A143493, A143494, A143495, A143499.

Programs

  • Maple
    with combinat: T := (n, k) -> 1/(k-4)!*add ((-1)^(k-i)*binomial(k-4,i)*(i+4)^(n-4),i = 0..k-4): for n from 4 to 13 do seq(T(n, k), k = 4..n) end do;
  • Mathematica
    t[n_, k_] := StirlingS2[n, k] - 6*StirlingS2[n-1, k] + 11*StirlingS2[n-2, k] - 6*StirlingS2[n-3, k]; Flatten[ Table[ t[n, k], {n, 4, 13}, {k, 4, n}]] (* Jean-François Alcover, Dec 02 2011 *)

Formula

T(n+4,k+4) = (1/k!)*Sum_{i = 0..k} (-1)^(k-i)*C(k,i)*(i+4)^n, n,k >= 0.
T(n,k) = Stirling2(n,k) - 6*Stirling2(n-1,k) + 11*Stirling2(n-2,k) - 6*Stirling2(n-3,k) for n,k >= 4.
Recurrence relation: T(n,k) = T(n-1,k-1) + k*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) = 4^(n-4); T(n,5) = 5^(n-4) - 4^(n-4).
E.g.f. (k+4)-th column (with offset 4): (1/k!)*exp(4*x)*(exp(x)-1)^k.
O.g.f. k-th column: Sum_{n>=k} T(n,k)*x^n = x^k/((1-4*x)*(1-5*x)*...*(1-k*x)).
E.g.f.: exp(4*t + x*(exp(t)-1)) = Sum_{n = 0..infinity} Sum_(k = 0..n) T(n+4,k+4)*x^k*t^n/n! = Sum_{n = 0..infinity} B_n(4;x)*t^n/n! = 1 + (4+x)*t/1! + (16+9*x+x^2)*t^2/2! + ..., where the row polynomials, B_n(4;x) := Sum_{k = 0..n} T(n+4,k+4)*x^k, may be called the 4-Bell polynomials.
Dobinski-type identities: Row polynomial B_n(4;x) = exp(-x)*Sum_{i = 0..infinity} (i+4)^n*x^i/i!; Sum_{k = 0..n} k!*T(n+4,k+4)*x^k = Sum_{i = 0..infinity} (i+4)^n*x^i/(1+x)^(i+1).
The T(n,k) are the connection coefficients between the falling factorials and the shifted monomials (x+4)^(n-4). For example, 16 + 9*x + x*(x-1) = (x+4)^2; 64 + 61*x + 15*x*(x-1) + x*(x-1)*(x-2) = (x+4)^3.
This array is the matrix product P^3 * S, where P denotes Pascal's triangle, A007318 and S denotes the lower triangular array of Stirling numbers of the second kind, A008277 (apply Theorem 10 of [Neuwirth]).
The inverse array is A049459, the signed 4-Stirling numbers of the first kind.
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^(-4)*E^n*x^4 = Sum_{k = 0..n} T(n+4,k+4)*x^k*D^k.
The row generating polynomials R_n(x) := Sum_{k=4..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_4(x) = x^4. It follows that the polynomials R_n(x) have only real zeros (apply Corollary 1.2. of [Liu and Wang]).
Relation with the 4-Eulerian numbers E_4(n,j) := A144698(n,j): T(n,k) = 4!/k!*Sum_{j = n-k..n-4} E_4(n,j)*binomial(j,n-k) for n >= k >= 4.
(End)

A143497 Triangle of unsigned 2-Lah numbers.

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, 3104640, 554400, 55440, 3080, 88, 1
Offset: 2

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Author

Peter Bala, Aug 25 2008

Keywords

Comments

For a signed version of this triangle see A062137. The unsigned 2-Lah number L(2; n,k) gives the number of partitions of the set {1, 2, ..., n} into k ordered lists with the restriction that the elements 1 and 2 must belong to different lists. More generally, the unsigned r-Lah number L(r; n, k) gives the number of partitions of the set {1, 2, ..., n} into k ordered lists with the restriction that the elements 1, 2, ..., r belong to different lists. If r = 1 there is no restriction and we obtain the unsigned Lah numbers A105278. For other cases see A143498 (r=3) and A143499 (r=4). We make some remarks on the general case.
The unsigned r-Lah numbers occur as connection constants in the generalized Lah identity (x + 2*r - 1)*(x + 2*r)*...*(x + 2*r + n - r - 2) = Sum_{k=r..n} L(r; n, k)*(x - 1)*(x - 2)*...*(x - k + r) for n >= r and where any empty products are taken equal to 1 (for a bijective proof of the identity, follow the proof of [Petkovsek and Pisanski] but restrict the first r of the Argonauts to different paths).
The unsigned r-Lah numbers satisfy the same recurrence as the unsigned Lah numbers, namely, L(r; n, k) = (n + k - 1)*L(r; n - 1,k) + L(r; n - 1,k - 1), but with the boundary conditions: L(r; n, k) = 0 if n < r or if k < r; L(r; r, r) = 1. The recurrence has the explicit solution L(r; n, k) = ((n - r)!/(k - r)!)*binomial(n + r - 1, k + r - 1) for n, k >= r. It follows that the unsigned r-Lah numbers have 'vertical' generating functions for k >= r of the form Sum_{n>=k} L(r; n, k)*t^n/(n -r)! = 1/(k - r)!*t^k/(1 - t)^(k + r). This yields the e.g.f. for the array of unsigned r-restricted Lah numbers in the form: Sum_{n,k>=r} L(r; n, k)*x^k*t^n/(n-r)! = (x*t)^r * 1/(1 - t)^(2*r) * exp(x*t/(1 - t)) = (x*t)^r (1 + (2*r + x)*t + (2r*(2*r + 1) + 2*(2*r + 1)*x + x^2)*t^2/2! + ...).
The array of unsigned r-Lah numbers begins
1
2r 1
2r*(2r+1) 2*(2r+1) 1
2r*(2r+1)*(2r+2) 3*(2r+1)*(2r+2) 3*(2r+2) 1
...
and equals exp(D(r)), where D(r) is the array with the sequence (2*r, 2*(2*r + 1), 3*(2*r + 2), 4*(2*r + 3), ...) on the main subdiagonal and zeros everywhere else.
The unsigned r-Lah numbers are related to the r-Stirling numbers: the lower triangular array of unsigned r-Lah numbers may be expressed as the matrix product St1(r) * St2(r), where St1(r) and St2(r) denote the arrays of r-Stirling numbers of the first and second kind respectively. The theory of r-Stirling numbers is developed in [Broder]. See A143491 - A143496 for tables of r-Stirling numbers. An alternative factorization for the array is as St1 * P^(2r - 2) * 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 (apply Theorem 10 of [Neuwirth]).
The array of unsigned r-Lah numbers is an example of the fundamental matrices sketched in A133314. So redefining the offset as n=0, given matrices A and B with A(n, k) = T(n, k)*a(n - k) and B(n, k) = T(n, k)*b(n - k), then A*B = C where C(n, k) = T(n,k)*[a(.) + b(.)]^(n - k), umbrally. An e.g.f. for the row polynomials of A is exp(x*t) exp{-x*t*[a*t/(a*t - 1)]}/(1 - a*t)^4 = exp(x*t) exp[(.)!*Laguerre(., 3, -x*t)* a(.)*t)], umbrally. - Tom Copeland, Sep 19 2008

Examples

			Triangle begins:
=========================================
n\k |     2     3     4     5     6     7
----+------------------------------------
  2 |     1
  3 |     4     1
  4 |    20    10     1
  5 |   120    90    18     1
  6 |   840   840   252    28     1
  7 |  6720  8400  3360   560    40     1
 ...
T(4,3) = 10. The ten partitions of {1,2,3,4} into 3 ordered lists such that the elements 1 and 2 lie in different lists are: {1}{2}{3,4} and {1}{2}{4,3}, {1}{3}{2,4} and {1}{3}{4,2}, {1}{4}{2,3} and {1}{4}{3,2}, {2}{3}{1,4} and {2}{3}{4,1}, {2}{4}{1,3} and {2}{4}{3,1}. The remaining two partitions {3}{4}{1,2} and {3}{4}{2,1} are not allowed because the elements 1 and 2 belong to the same block.
		

Crossrefs

Cf. A001715 (column 2), A007318, A008275, A008277, A061206 (column 3), A062137, A062141 - A062144 ( column 4 to column 7), A062146 (alt. row sums), A062147 (row sums), A105278 (unsigned Lah numbers), A143491, A143494, A143498, A143499.

Programs

  • GAP
    T:=Flat(List([2..10],n->List([2..n],k->(Factorial(n-2)/Factorial(k-2))*Binomial(n+1,k+1)))); # Muniru A Asiru, Nov 27 2018
  • Maple
    T := (n, k) -> ((n-2)!/(k-2)!)*binomial(n+1, k+1):
    for n from 2 to 11 do seq(T(n, k), k = 2..n) od;
  • Mathematica
    T[n_, k_] := (n-2)!/(k-2)!*Binomial[n+1, k+1]; Table[T[n, k], {n,2,10}, {k,2,n}] // Flatten (* Amiram Eldar, Nov 27 2018 *)
  • Maxima
    create_list((n - 2)!/(k - 2)!*binomial(n + 1, k + 1), n, 2, 12, k, 2, n); /* Franck Maminirina Ramaharo, Nov 27 2018 */
    

Formula

T(n, k) = ((n - 2)!/(k - 2)!)*C(n+1, k+1), for n, k >= 2.
Recurrence: T(n, k) = (n + k - 1)*T(n-1, k) + T(n-1, k-1) for n, k >= 2, with the boundary conditions: T(n, k) = 0 if n < 2 or k < 2; T(2, 2) = 1.
E.g.f. for column k: Sum_{n>=k} T(n, k)*t^n/(n - 2)! = 1/(k - 2)!*t^k/(1 - t)^(k+2) for k >= 2.
E.g.f: Sum_{n=2..inf} Sum_{k=2..n} T(n, k)*x^k*t^n/(n - 2)! = (x*t)^2/(1 - t)^4* exp(x*t/(1 - t)) = (x*t)^2*(1 + (4 + x)*t + (20 + 10*x + x^2)*t^2/2! + ... ).
Generalized Lah identity: (x + 3)*(x + 4)*...*(x + n) = Sum_{k = 2..n} T(n, k)*(x - 1)*(x - 2)*...*(x - k + 2).
The polynomials 1/n!*Sum_{k=2..n+2} T(n+2, k)*(-x)^(k - 2) for n >= 0 are the generalized Laguerre polynomials Laguerre(n,3,x). See A062137.
Array = A143491 * A143494 = abs(A008275) * (A007318)^2 * A008277 (apply Theorem 10 of [Neuwirth]). Array equals exp(D), where D is the array with the quadratic sequence (4, 10, 18, 28, ...) on the main subdiagonal and zeros elsewhere.
After adding 1 to the head of the main diagonal and a zero to each of the subdiagonals, the n-th diagonal may be generated as coefficients of (1/n!) [D^(-1) tDt t^(-3)D t^3]^n exp(x*t), where D is the derivative w.r.t. t and D^(-1) t^j/j! = t^(j + 1)/(j + 1)!. E.g., n = 2 generates 20*x*t^3/3! + 90*x^2*t^4/4! + 252*x^3* t^5/5! + ... . For the general unsigned r-Lah number array, replace the threes by (2*r - 1) in the operator. The e.g.f. of the row polynomials is then exp[D^(-1) tDt t^(-(2*r-1))D t^(2*r - 1)] exp(x*t), with offset n = 0. - Tom Copeland, Sep 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

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.

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

Views

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

A144698 Triangle of 4-Eulerian numbers.

Original entry on oeis.org

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

Views

Author

Peter Bala, Sep 19 2008

Keywords

Comments

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

Examples

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

References

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

Crossrefs

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

Programs

  • Magma
    m:=4; [(&+[(-1)^(k-j)*Binomial(n+1,k-j)*Binomial(j+m,m-1)*(j+1)^(n-m+1): j in [0..k]])/m: k in [0..n-m], n in [m..m+10]]; // G. C. Greubel, Jun 04 2022
    
  • Maple
    with(combinat):
    T:= (n,k) -> 1/4!*add((-1)^(k-j)*binomial(n+1,k-j)*(j+1)^(n-3)*(j+2)*(j+3)*(j+4),j = 0..k):
    for n from 4 to 12 do
    seq(T(n,k),k = 0..n-4)
    end do;
  • Mathematica
    T[n_, k_] /; 0 < k <= n-4 := T[n, k] = (k+1) T[n-1, k] + (n-k) T[n-1, k-1];
    T[, 0] = 1; T[, _] = 0;
    Table[T[n, k], {n, 4, 12}, {k, 0, n-4}] // Flatten (* Jean-François Alcover, Nov 11 2019 *)
  • SageMath
    m=4 # A144698
    def T(n,k): return (1/m)*sum( (-1)^(k-j)*binomial(n+1,k-j)*binomial(j+m,m-1)*(j+1)^(n-m+1) for j in (0..k) )
    flatten([[T(n,k) for k in (0..n-m)] for n in (m..m+10)]) # G. C. Greubel, Jun 04 2022

Formula

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