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 1-3 of 3 results.

A143495 Triangle read by rows: 3-Stirling numbers of the second kind.

Original entry on oeis.org

1, 3, 1, 9, 7, 1, 27, 37, 12, 1, 81, 175, 97, 18, 1, 243, 781, 660, 205, 25, 1, 729, 3367, 4081, 1890, 380, 33, 1, 2187, 14197, 23772, 15421, 4550, 644, 42, 1, 6561, 58975, 133057, 116298, 47271, 9702, 1022, 52, 1, 19683, 242461, 724260, 830845, 447195
Offset: 3

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Author

Peter Bala, Aug 20 2008

Keywords

Comments

This is the case r = 3 of the r-Stirling numbers of the second kind. The 3-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, 2 and 3 belong to distinct subsets. For remarks on the general case see A143494 (r = 2). The corresponding array of 3-Stirling numbers of the first kind is A143492. The theory of r-Stirling numbers of both kinds is developed in [Broder]. For 3-Lah numbers refer to A143498.
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^(-3)*E^n*x^3 = Sum_{k = 0..n} T(n+3,k+3)*x^k*D^k.
The row generating polynomials R_n(x) := Sum_{k= 3..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_3(x) = x^3. It follows that the polynomials R_n(x) have only real zeros (apply Corollary 1.2. of [Liu and Wang]).
Relation with the 3-Eulerian numbers E_3(n,j) := A144697(n,j): T(n,k) = (3!/k!)*Sum_{j = n-k..n-3} E_3(n,j)*binomial(j,n-k) for n >= k >= 3.
(End)
T(n,k) = S(n,k,3), n>=k>=3, in Mikhailov's first paper, eq.(28) or (A3). E.g.f. column k from (A20) with k->3, r->k. Therefore, with offset [0,0], this triangle is the Sheffer triangle (exp(3*x),exp(x)-1) with e.g.f. of column no. m>=0: exp(3*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. - Wolfdieter Lang, Sep 29 2011

Examples

			Triangle begins
  n\k|....3....4....5....6....7....8
  ==================================
  3..|....1
  4..|....3....1
  5..|....9....7....1
  6..|...27...37...12....1
  7..|...81..175...97...18....1
  8..|..243..781..660..205...25....1
  ...
T(5,4) = 7. The set {1,2,3,4,5} can be partitioned into four subsets such that 1, 2 and 3 belong to different subsets in 7 ways: {{1}{2}{3}{4,5}}, {{1}{2}{5}{3,4}}, {{1}{2}{4}{3,5}}, {{1}{3}{4}{2,5}}, {{1}{3}{5}{2,4}}, {{2}{3}{4}{1,5}} and {{2}{3}{5}{1,4}}.
From _Peter Bala_, Feb 23 2025: (Start)
The array factorizes as
/ 1               \       /1             \ /1             \ /1            \
| 3    1           |     | 3   1          ||0  1           ||0  1          |
| 9    7   1       |  =  | 9   4   1      ||0  3   1       ||0  0  1       | ...
|27   37  12   1   |     |27  13   5  1   ||0  9   4  1    ||0  0  3  1    |
|81  175  97  18  1|     |81  40  18  6  1||0 27  13  5  1 ||0  0  9  4  1 |
|...               |     |...             ||...            ||...           |
where, in the infinite product on the right-hand side, the first array is the Riordan array (1/(1 - 3*x), x/(1 - x)). See A106516. (End)
		

Crossrefs

Cf. A005061 (column 4), A005494 (row sums), A008277, A016753 (column 5), A028025 (column 6), A049458 (matrix inverse), A106516, A143492, A143494, A143496, A143498.

Programs

  • Maple
    A143495 := (n, k) -> (1/(k-3)!)*add((-1)^(k-i-1)*binomial(k-3,i)*(i+3)^(n-3), i = 0..k-3): for n from 3 to 12 do seq(A143495(n, k), k = 3..n) end do;
  • Mathematica
    nmax = 12; t[n_, k_] := 1/(k-3)!* Sum[ (-1)^(k-j-1)*Binomial[k-3, j]*(j+3)^(n-3), {j, 0, k-3}]; Flatten[ Table[ t[n, k], {n, 3, nmax}, {k, 3, n}]] (* Jean-François Alcover, Dec 07 2011, after Maple *)
  • 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)
    A143495 = lambda n, k: stirling2r(n, k, 3)
    for n in (3..8): [A143495(n, k) for k in (3..n)] # Peter Luschny, Nov 19 2012

Formula

T(n+3,k+3) = (1/k!)*Sum_{i = 0..k} (-1)^(k-i)*binomial(k,i)*(i+3)^n, n,k >= 0.
T(n,k) = Stirling2(n,k) - 3*Stirling2(n-1,k) + 2*Stirling2(n-2,k), n,k >= 3.
Recurrence relation: T(n,k) = T(n-1,k-1) + k*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) = 3^(n-3); T(n,4) = 4^(n-3) - 3^(n-3).
E.g.f. (k+3) column (with offset 3): (1/k!)*exp(3x)*(exp(x)-1)^k.
O.g.f. k-th column: Sum_{n >= k} T(n,k)*x^n = x^k/((1-3*x)*(1-4*x)*...*(1-k*x)).
E.g.f.: exp(3*t + x*(exp(t)-1)) = Sum_{n >= 0} Sum_{k = 0..n} T(n+3,k+3)*x^k*t^n/n! = Sum_{n >= 0} B_n(3;x)*t^n/n! = 1 + (3+x)*t/1! + (9+7*x+x^2)*t^2/2! + ..., where the row polynomials, B_n(3;x) := Sum_{k = 0..n} T(n+3,k+3)*x^k, may be called the 3-Bell polynomials.
Dobinski-type identities: Row polynomial B_n(3;x) = exp(-x)*Sum_{i >= 0} (i+3)^n*x^i/i!; Sum_{k = 0..n} k!*T(n+3,k+3)*x^k = Sum_{i >= 0} (i+3)^n*x^i/(1+x)^(i+1).
The T(n,k) are the connection coefficients between the falling factorials and the shifted monomials (x+3)^(n-3). For example, 9 + 7*x + x*(x-1) = (x+3)^2 and 27 + 37*x + 12x*(x-1) + x*(x-1)*(x-2) = (x+3)^3.
This array is the matrix product P^2 * 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 A049458, the signed 3-Stirling numbers of the first kind.

A245019 Number of ordered n-tuples of positive integers, whose minimum is 0 and maximum is 4.

Original entry on oeis.org

0, 2, 24, 194, 1320, 8162, 47544, 266114, 1448520, 7727522, 40616664, 211117634, 1088079720, 5571427682, 28384443384, 144041002754, 728708854920, 3677645732642, 18524892775704, 93171895169474, 468051525534120, 2349032799986402
Offset: 1

Views

Author

Ovidiu Bagdasar, Sep 17 2014

Keywords

Comments

For given k and n positive integers, let T(k,n) represent the number of n-tuples of positive integers, whose minimum is zero and maximum is k. In this notation, the sequence corresponds to a(n) = T(4,n).

Examples

			For n=2 the a(2)=2 solutions are (0,4) and (4,0).
		

Crossrefs

T(1,n) gives A000918, T(2,n-1) gives A028243, T(n,3) gives A008588, T(n,4) gives A005914.

Programs

  • PARI
    concat(0, Vec(-2*x^2/((3*x-1)*(4*x-1)*(5*x-1)) + O(x^100))) \\ Colin Barker, Sep 17 2014

Formula

a(n) = 5^n-2*4^n+3^n.
a(n) = 12*a(n-1)-47*a(n-2)+60*a(n-3). G.f.: -2*x^2 / ((3*x-1)*(4*x-1)*(5*x-1)). - Colin Barker, Sep 17 2014
a(n) = 2*A016753(n-2) for n>1. - Colin Barker, Sep 17 2014

A372118 Square array A(n, k) = ((k+2)^(n+2) - 2 * (k+1)^(n+2) + k^(n+2))/2 for k, n >= 0 read by ascending antidiagonals.

Original entry on oeis.org

1, 3, 1, 7, 6, 1, 15, 25, 9, 1, 31, 90, 55, 12, 1, 63, 301, 285, 97, 15, 1, 127, 966, 1351, 660, 151, 18, 1, 255, 3025, 6069, 4081, 1275, 217, 21, 1, 511, 9330, 26335, 23772, 9751, 2190, 295, 24, 1, 1023, 28501, 111645, 133057, 70035, 19981, 3465, 385, 27, 1
Offset: 0

Views

Author

Werner Schulte, Apr 19 2024

Keywords

Comments

Depending on some fixed integer m >= 0 we define a family of square arrays A(m; n, k) = (Sum_{i=0..m} (-1)^i * binomial(m, i) * (k + m - i)^(n+m)) / m! for k, n >= 0. Special cases are: A004248 (m=0), A343237 (m=1) and this array (m=2). The A(m; n, k) satisfy: A(m; n, k) = (k+m) * A(m; n-1, k) + A(m-1; n, k) with initial values A(0; n, k) = k^n and A(m; 0, k) = 1.
Further properties are conjectures:
(1) O.g.f. of column k is Prod_{i=k..k+m} 1 / (1 - i * t);
(2) E.g.f. of row n is exp(x) * (Sum_{k=0..n} binomial(k+m, m) * A048993(n+m, k+m) * x^k);
(3) The LU decompositions of these arrays are given by the upper triangular matrix U which is the transpose of A007318 and the lower triangular matrix L, where L is defined L(m; n, k) = A048993(n+m, k+m) * (k+m)! / m!, i.e., A(m; n, k) = Sum_{i=0..k} L(m; n, i) * binomial(k, i).
The three conjectures are true, see links. - Sela Fried, Jul 07 2024

Examples

			Square array A(n, k) starts:
n\k :    0     1       2       3        4         5         6         7
=======================================================================
  0 :    1     1       1       1        1         1         1         1
  1 :    3     6       9      12       15        18        21        24
  2 :    7    25      55      97      151       217       295       385
  3 :   15    90     285     660     1275      2190      3465      5160
  4 :   31   301    1351    4081     9751     19981     36751     62401
  5 :   63   966    6069   23772    70035    170898    365001    706104
  6 :  127  3025   26335  133057   481951   1398097   3463615   7628545
  7 :  255  9330  111645  724260  3216795  11075670  31794105  79669320
  etc.
		

Crossrefs

Rows: A000012 (n=0), A008585 (n=1), A227776 (n=2).
Columns: A000225 (k=0), A000392 (k=1), A016269 (k=2), A016753 (k=3), A016103 (k=4), A019757 (k=5), A020570 (k=6), A020782 (k=7).
Main diagonal: A281596(n+2).

Programs

  • Mathematica
    A372118[n_, k_] := ((k+2)^(n+2) - 2*(k+1)^(n+2) + k^(n+2))/2;
    Table[A372118[n-k, k], {n, 0, 10}, {k, 0, n}] (* Paolo Xausa, Jul 10 2024 *)
  • PARI
    A(n, k) = ((k+2)^(n+2) - 2 * (k+1)^(n+2) + k^(n+2))/2

Formula

A(n, k) = (k+2) * A(n-1, k) + (k+1)^(n+1) - k^(n+1) for n > 0.
Conjectures:
(1) O.g.f. of column k is Prod_{i=k..k+2} 1 / (1 - i * t);
(2) E.g.f. of row n is exp(x) * (Sum_{k=0..n} binomial(k+2, 2) * A048993(n+2, k+2) * x^k);
(3) The LU decomposition of this array is given by the upper triangular matrix U which is the transpose of A007318 and the lower triangular matrix L, where L is defined L(n, k) = A048993(n+2, k+2) * (k+2)! / 2!, i.e., A(n, k) = Sum_{i=0..k} L(n, i) * binomial(k, i).
The three conjectures are true. See comments. - Sela Fried, Jul 09 2024
Showing 1-3 of 3 results.