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 1-10 of 16 results. Next

A371762 Triangle read by rows: the polynomial coefficients of the numerator of the rational solution of the linear recurrence equations of the rows of A371761.

Original entry on oeis.org

1, 0, 1, 0, 1, 2, 0, 1, 7, 6, 0, 1, 19, 46, 24, 0, 1, 46, 251, 326, 120, 0, 1, 104, 1163, 3016, 2556, 720, 0, 1, 225, 4831, 23283, 35848, 22212, 5040, 0, 1, 473, 18523, 158531, 417148, 437228, 212976, 40320, 0, 1, 976, 66886, 976636, 4285549, 7084804, 5586444, 2239344, 362880
Offset: 0

Views

Author

Peter Luschny, Apr 06 2024

Keywords

Comments

Let R(n) = N(n)/D(n) denote the ordinary rational generating function of row n of A371761 as given by its linear recurrence equation. N(n) is the row polynomial Sum_{k=0..n} T(n, k)*x^k and D(n) = Sum_{k=0..n} Stirling1(n+1, n+1-k)*x^k. Thus A371761(n, k) = [x^k] N(n)/D(n).

Examples

			Triangle starts:
  [0] 1;
  [1] 0, 1;
  [2] 0, 1,   2;
  [3] 0, 1,   7,     6;
  [4] 0, 1,  19,    46,     24;
  [5] 0, 1,  46,   251,    326,     120;
  [6] 0, 1, 104,  1163,   3016,    2556,     720;
  [7] 0, 1, 225,  4831,  23283,   35848,   22212,    5040;
  [8] 0, 1, 473, 18523, 158531,  417148,  437228,  212976,   40320;
  [9] 0, 1, 976, 66886, 976636, 4285549, 7084804, 5586444, 2239344, 362880;
.
The rational generating function for row 3 of A371761 is:
gf = (6*x^3 + 7*x^2 + x)/(-6*x^3 + 11*x^2 - 6*x + 1).
		

Crossrefs

Cf. A029767 (row sums), A000142 (main diagonal), A067318 (subdiagonal).
Cf. A371761.

A004248 Array read by ascending antidiagonals: A(n, k) = k^n.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 4, 3, 1, 0, 1, 8, 9, 4, 1, 0, 1, 16, 27, 16, 5, 1, 0, 1, 32, 81, 64, 25, 6, 1, 0, 1, 64, 243, 256, 125, 36, 7, 1, 0, 1, 128, 729, 1024, 625, 216, 49, 8, 1, 0, 1, 256, 2187, 4096, 3125, 1296, 343, 64, 9, 1, 0, 1, 512, 6561, 16384, 15625, 7776, 2401, 512, 81, 10, 1
Offset: 0

Views

Author

Keywords

Comments

This array transforms into A371761 using the Akiyama-Tanigawa algorithm for powers applied to the rows. - Peter Luschny, Apr 16 2024
This array transforms into A344499 using the Akiyama-Tanigawa algorithm for powers applied to the columns. - Peter Luschny, Apr 27 2024

Examples

			Seen as an array that is read by ascending antidiagonals:
[0] 1, 1,   1,    1,     1,     1,      1,      1,       1, ...
[1] 0, 1,   2,    3,     4,     5,      6,      7,       8, ...
[2] 0, 1,   4,    9,    16,    25,     36,     49,      64, ...
[3] 0, 1,   8,   27,    64,   125,    216,    343,     512, ...
[4] 0, 1,  16,   81,   256,   625,   1296,   2401,    4096, ...
[5] 0, 1,  32,  243,  1024,  3125,   7776,  16807,   32768, ...
[6] 0, 1,  64,  729,  4096, 15625,  46656, 117649,  262144, ...
[7] 0, 1, 128, 2187, 16384, 78125, 279936, 823543, 2097152, ...
		

Crossrefs

For other versions see A051129 and A009998.
Row sums are A026898, diagonal sums are A104872. [Paul Barry, Mar 28 2005]

Programs

  • Mathematica
    T[x_, y_] := If[y == 0, 1, (x - y)^y];
    Table[T[x, y], {x, 0, 11}, {y, x, 0, -1}] // Flatten (* Jean-François Alcover, Dec 15 2017 *)
  • PARI
    T(x, y) = x^y \\ Charles R Greathouse IV, Feb 07 2017
    
  • SageMath
    def Arow(n, len): return [k**n for k in range(len)]
    for n in range(8): print([n], Arow(n, 9))  # Peter Luschny, Apr 16 2024

Formula

Table of x^y, where (x,y) = (0,0), (0,1), (1,0), (0,2), (1,1), (2,0), ...
As a number triangle, columns have g.f. x^k/(1 - kx). - Paul Barry, Mar 28 2005
From Paul Barry, Jul 13 2005: (Start)
T(n, k) = if(k <= n, k^(n - k), 0).
T(n, k) = Sum_{j=0..floor((n-k)/2)} (-1)^j*C(n-k, j)*C(n-k-j, n-k)*k^(n-k-2j).
(End)

Extensions

New name by Peter Luschny, Apr 16 2024.

A048144 a(n) = Sum_{k=0..n} (k!)^2 * Stirling_2(n,k)^2.

Original entry on oeis.org

1, 1, 5, 73, 2069, 95401, 6487445, 610093513, 75796724309, 12020754177001, 2369364111428885, 568128719132038153, 162835627057766030549, 54975855375379966645801, 21593185551426744571090325, 9762238510837560633366673993, 5033241437347149354018370856789
Offset: 0

Views

Author

Keywords

Comments

Number of digraphs with loops, with labeled vertices and labeled arcs, with n arcs and with no vertex of indegree 0 or outdegree 0, cf. A121936, A122418, A122399. - Vladeta Jovovic, Sep 06 2006
Chromatic invariant of the complete bipartite graph K_{n+1,n+1}. - Eric W. Weisstein, Jul 11 2011
Generally, for p >= 1, Sum_{k=0..n} (k!*StirlingS2(n,k))^p is asymptotic to n^(p*n+1/2) * sqrt(Pi/(2*p*(1-log(2))^(p-1))) / (exp(p*n) * log(2)^(p*n+1)). - Vaclav Kotesovec, May 10 2014

Crossrefs

Programs

  • Maple
    a := proc(n) local A, j; A := proc(n, k) option remember; if n = 0 then n^k else add(binomial(k + `if`(j>0, 1, 0), j+1) * A(n-1, k-j), j = 0..k) fi end: A(n,n) end:
    seq(a(n), n = 0..16);  # Peter Luschny, Nov 20 2024
  • Mathematica
    Table[Sum[(k!)^2*StirlingS2[n,k]^2,{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, May 07 2014 *)
  • PARI
    a(n) = sum(k=0, n, k!^2*stirling(n, k, 2)^2); \\ Michel Marcus, Mar 07 2020
    
  • Python
    from functools import cache
    from math import comb as binomial
    @cache
    def A(n, k): return int(k == 0) if n == 0 else sum(binomial(k + int(j > 0), j + 1) * A(n - 1, k - j) for j in range(k + 1))
    a = lambda n: A(n, n)
    print([a(n) for n in range(17)])  # Peter Luschny, Nov 20 2024

Formula

E.g.f.: Sum_{n>=0} Sum_{j=0..n} (-1)^(n-j)*binomial(n,j)*(exp(j*x)-1)^n. a(n) = Sum_{k=0..n} Stirling2(n,k)*k!*A104602(k). - Vladeta Jovovic, Mar 25 2006
a(n) ~ sqrt(Pi/(1-log(2))) * n^(2*n+1/2) / (2*exp(2*n) * (log(2))^(2*n+1)). - Vaclav Kotesovec, May 09 2014
E.g.f.: Sum_{n>=0} (1 - exp(-n*x))^n * exp(-n*x). - Paul D. Hanna, Mar 26 2018
E.g.f.: Sum_{n>=0} (exp(n*x) - 1)^n * exp(-n*(n+1)*x). - Paul D. Hanna, Mar 26 2018
a(n) = A272644(2n,n). - Alois P. Heinz, Oct 17 2024
a(n) = A371761(n, n). - Peter Luschny, Nov 20 2024
a(n) = (n!)^2 * [(x*y)^n] 1 / (exp(x) + exp(y) - exp(x + y)). - Ilya Gutkovskiy, Apr 24 2025

A136126 Triangle read by rows: T(n,k) is the number of permutations of {1,2,...,k+n} having excedance set {1,2,...,k} (the empty set for k=0), 0 <= k <= n-1.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 7, 7, 1, 1, 15, 31, 15, 1, 1, 31, 115, 115, 31, 1, 1, 63, 391, 675, 391, 63, 1, 1, 127, 1267, 3451, 3451, 1267, 127, 1, 1, 255, 3991, 16275, 25231, 16275, 3991, 255, 1, 1, 511, 12355, 72955, 164731, 164731, 72955, 12355, 511, 1
Offset: 1

Views

Author

Emeric Deutsch, Jan 17 2008

Keywords

Comments

The excedance set of a permutation p in S_n is the set of indices i such that p(i) > i.
Columns 1,2,3,4 yield A000225, A091344, A091347, A091348, respectively. Row sums yield A136127.
T(a+b-1,b-1)*(-1)^(a+b-1) = Sum_{k=0..} F(a,b,k)*(-1)^k where F(a,b,k) is the number of connected subgraphs of K(a,b) (the complete bipartite graph) with k edges. F(n,n,k) is A255192(n,k). - Thomas Dybdahl Ahle, Feb 18 2015 [The sum starts with k=0, and F(n,n,k) is A255192(n,k), but there seems to be no A255192(n,0). Is there no upper k-summation limit? - Wolfdieter Lang, Mar 15 2015]
Comment from Don Knuth, Aug 25 2020, added by N. J. A. Sloane, Sep 07 2020: (Start)
This array also arises from a problem about {0,1}-matrices. Symmetric array read by antidiagonals: A(n,k) (n >= 1, k >= 0) = number of n X k matrices of 0's and 1's satisfying two conditions: (i) no column is entirely 0; (ii) no 0 has simultaneously a 1 above it and another 1 to its left.
Equivalently (see the Steingrímsson-Williams reference) A(n,k) is the number of permutations p_1...p_{n+k} on {1,...,n+k} for which p_1 >= 1, ..., p_n >= n, p_{n+1} < n+1,..., p_{n+k} < n+k. Then A(n,k) = A(k+1,n-1), for n >= 1 and k >= 0.
For example, the seven 2 X 2 matrices satisfying (i) and (ii) are
00 01 10 10 11 11 11
11 11 01 11 00 01 11
and the seven permutations of {1, 2, 3, 4} satisfying the other definition are
1423, 2413, 3412, 3421, 4213, 4312, 4321.
(End)

Examples

			T(4,2) = 7 because 3412, 4312, 2413, 2314, 2431, 3421 and 4321 are the only permutations of {1,2,3,4} with excedance set {1,2}.
Triangle starts:
  1;
  1,   1;
  1,   3,    1;
  1,   7,    7,     1;
  1,  15,   31,    15,     1;
  1,  31,  115,   115,    31,     1;
  1,  63,  391,   675,   391,    63,    1;
  1, 127, 1267,  3451,  3451,  1267,  127,   1;
  1, 255, 3991, 16275, 25231, 16275, 3991, 255, 1;
  ...
Formatted as a square array A(n,k) with 0 <= k <= n:
  1,   1,    1,     1,      1,        1,         1,          1, ... [A000012]
  1,   3,    7,    15,     31,       63,       127,        255, ... [A000225]
  1,   7,   31,   115,    391,     1267,      3991,      12355, ... [A091344]
  1,  15,  115,   675,   3451,    16275,     72955,     316275, ... [A091347]
  1,  31,  391,  3451,  25231,   164731,    999391,    5767051, ... [A091348]
  1,  63, 1267, 16275, 164731,  1441923,  11467387,   85314915, ...
  1, 127, 3991, 72955, 999391, 11467387, 116914351, 1096832395, ...
		

Crossrefs

Programs

  • Maple
    with(combinat): T:=proc(n,k) if k < n then add((-1)^(k+1-i)*factorial(i)*i^(n-1-k)* stirling2(k+1,i),i=1..k+1) else 0 end if end proc: for n to 10 do seq(T(n,k),k=0..n-1) end do; # yields sequence in triangular form
    # Alternatively as a square array:
    A := (n, k) -> add((-1)^(k-j)*j!*Stirling2(k+1,j+1)*(j+1)^(n+1), j=0..k);
    seq(print(seq(A(n, k), k=0..7)), n=0..6); # Peter Luschny, Mar 14 2018
    # Using the exponential generating function as given by Arakawa & Kaneko:
    gf := polylog(-t, 1-exp(-x))/(exp(x)-1):
    ser := series(gf, x, 12): c := n -> n!*coeff(ser, x, n):
    seq(lprint(seq(subs(t=k, c(n)), n=0..8)), k=0..8); # Peter Luschny, Apr 29 2021
    # Using recurrence relations:
    A := proc(n, k) option remember; local j; if n = 0 then return k^n fi;
    add(binomial(k+1, j+1)*A(n-1, k-j), j = 0..k) end:
    for n from 0 to 7 do lprint(seq(A(n, k), k=0..8)) od;  # Peter Luschny, Apr 19 2024
  • Mathematica
    T[n_, k_] := Sum[(-1)^(k + 1 - i)*i!*i^(n - 1 - k)*StirlingS2[k + 1, i], {i, 1, k + 1}];
    Table[T[n, k], {n, 1, 10}, {k, 0, n - 1}] // Flatten (* Jean-François Alcover, Nov 16 2017 *)
  • PARI
    {T(n,k)=polcoeff(polcoeff( x*y*sum(m=0, n, m!*x^m*prod(k=1, m, (1+y+k*x*y)/(1+(1+y)*k*x+k^2*x^2*y +x*O(x^n))) ), n,x),k,y)} \\ Paul D. Hanna, Feb 01 2013
    for(n=1, 10,for(k=1,n, print1(T(n,k), ", "));print(""))
    
  • PARI
    tabl(nn) = {default(seriesprecision, nn+1); pol = log(1/(1-(exp(x)-1)*(exp(y)-1))) + O(x^nn); for (n=1, nn-1, poly = n!*polcoeff(pol, n, x); for (k=1, n, print1(k!*polcoeff(poly, k, y), ", ");); print(););} \\ Michel Marcus, Apr 17 2015

Formula

T(n,k) = Sum_{i=1..k+1} (-1)^(k+1-i)*i!*i^(n-1-k)*Stirling2(k+1,i) (0 <= k <= n-1).
G.f.: A(x,y) = x*y*Sum_{n>=1} n! * x^n*Product_{k=1..n} (1 + y + k*x*y) / (1 + (1+y)*k*x + k^2*x^2*y). - Paul D. Hanna, Feb 01 2013
Central terms of triangle equals A092552. - Paul D. Hanna, Feb 01 2013
T(n,k-1) = Sum_{i=0..k, m=0..i} binomial(i,m)*(-1)^(k-m)*i^(n-k)*m^k (1 <= k <= n). - Thomas Dybdahl Ahle, Feb 18 2015
E.g.f.: log(1/(1-(exp(x)-1)*(exp(y)-1))). - Vladimir Kruchinin, Apr 17 2015
Let W(n,k) = k!*Stirling2(n+1, k+1) denote the Worpitzky numbers, then A(n,k) = Sum_{j=0..k} (-1)^(k-j)*W(k,j)*(j+1)^(n+1) enumerates the square array. - Peter Luschny, Mar 14 2018
Assume the missing first row (1,0,0,...) of the array which Ayyer and Bényi call the 'poly-Bernoulli numbers of type C'. Then T(n, k) = p_{n}(k) where p_{n}(x) = Sum_{k=0..n} (-1)^(n-k)*(k+1)^x*Sum_{j=0..n} E1(n,j)*binomial(n-j, n-k), and E1(n, k) are the Eulerian numbers of first order. This reflects the Worpitzky approach to the Bernoulli numbers. This formula can alternatively be written as: T(n, k) = Sum_{j=0..k} (-1)^(k-j)*(j+1)^n*A028246(k+1, j+1). - Peter Luschny, Apr 29 2021

Extensions

Definition corrected. Changed "T(n,k) is the number of permutations of {1,2,...,n}..." to "T(n,k) is the number of permutations of {1,2,...,k+n}..." - Karel Casteels (kcasteel(AT)sfu.ca), Feb 17 2010

A272644 Triangle read by rows: T(n,m) = Sum_{i=0..m} Stirling2(m+1,i+1)*(-1)^(m-i)*i^(n-m)*i!, for n >= 2, m = 1..n-1.

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 13, 13, 1, 1, 29, 73, 29, 1, 1, 61, 301, 301, 61, 1, 1, 125, 1081, 2069, 1081, 125, 1, 1, 253, 3613, 11581, 11581, 3613, 253, 1, 1, 509, 11593, 57749, 95401, 57749, 11593, 509, 1, 1, 1021, 36301, 268381, 673261, 673261, 268381, 36301, 1021, 1
Offset: 2

Views

Author

N. J. A. Sloane, May 07 2016

Keywords

Comments

Gives number of bitriangular permutations. Could be prefixed with row 0 containing a single 1. - N. J. A. Sloane, Jan 10 2018

Examples

			Triangle begins:
n\m  [1]     [2]     [3]     [4]     [5]     [6]     [7]     [8]
[2]  1;
[3]  1,      1;
[4]  1,      5,      1;
[5]  1,     13,     13,      1;
[6]  1,     29,     73,     29,      1;
[7]  1,     61,    301,    301,     61,      1;
[8]  1,    125,   1081,   2069,   1081,    125,      1;
[9]  1,    253,   3613,  11581,  11581,   3613,    253,      1;
...
		

Crossrefs

Column 2 is A036563.
Largest term in each row gives A272645.
Second diagonal from the right is 2^i - 3.
Third diagonal from the right edge is A006230.
T(2n,n) gives A048144.
For row sums see A297195.

Programs

  • Maple
    A272644 := proc(n,m)
        add(combinat[stirling2](m+1,i+1)*(-1)^(m-i)*i^(n-m)*i!,i=0..m) ;
    end proc:
    seq(seq(A272644(n,m),m=1..n-1),n=2..10) ; # R. J. Mathar, Mar 04 2018
  • Mathematica
    Table[Sum[StirlingS2[m + 1, i + 1] (-1)^(m - i) i^(n - m) i!, {i, 0, m} ], {n, 11}, {m, n - 1}] /. {} -> {0} // Flatten  (* Michael De Vlieger, May 19 2016 *)
  • PARI
    A(n,m) = sum(i=0, m, stirling(m+1, i+1, 2) * (-1)^((m-i)%2) * i^(n - m) * i!);
    concat(vector(10, n, vector(n, m, A(n+1, m))))  \\ Gheorghe Coserea, May 16 2016

Formula

T(n,m) = Sum_{i=0..m} Stirling2(m+1, i+1)*(-1)^(m-i)*i^(n-m)*i!, for n>=2, m=1..n-1, where Stirling2(n,k) is defined by A008277.
A001469(n+1) = Sum_{m=1..2*n-1} (-1)^(m-1)*T(2*n,m). - Gheorghe Coserea, May 18 2016

Extensions

More terms from Gheorghe Coserea, May 16 2016

A382735 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where A(n,k) = n! * k! * [x^n * y^k] 1 / (exp(x) + exp(y) - exp(x+y))^3.

Original entry on oeis.org

1, 0, 0, 0, 3, 0, 0, 3, 3, 0, 0, 3, 27, 3, 0, 0, 3, 75, 75, 3, 0, 0, 3, 171, 579, 171, 3, 0, 0, 3, 363, 2667, 2667, 363, 3, 0, 0, 3, 747, 10083, 22779, 10083, 747, 3, 0, 0, 3, 1515, 34635, 142923, 142923, 34635, 1515, 3, 0, 0, 3, 3051, 112899, 761211, 1396803, 761211, 112899, 3051, 3, 0
Offset: 0

Views

Author

Seiichi Manyama, Apr 04 2025

Keywords

Examples

			Square array begins:
  1, 0,   0,     0,      0,       0, ...
  0, 3,   3,     3,      3,       3, ...
  0, 3,  27,    75,    171,     363, ...
  0, 3,  75,   579,   2667,   10083, ...
  0, 3, 171,  2667,  22779,  142923, ...
  0, 3, 363, 10083, 142923, 1396803, ...
		

Crossrefs

Main diagonal gives A382738.

Programs

  • PARI
    a(n, k) = sum(j=0, min(n, k), j!^2*binomial(j+2, 2)*stirling(n, j, 2)*stirling(k, j, 2));

Formula

E.g.f.: 1 / (exp(x) + exp(y) - exp(x+y))^3.
A(n,k) = A(k,n).
A(n,k) = Sum_{j=0..min(n,k)} (j!)^2 * binomial(j+2,2) * Stirling2(n,j) * Stirling2(k,j).

A344499 T(n, k) = F(n - k, k), where F(n, x) is the Fubini polynomial. Triangle read by rows, T(n, k) for 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 3, 2, 1, 0, 13, 10, 3, 1, 0, 75, 74, 21, 4, 1, 0, 541, 730, 219, 36, 5, 1, 0, 4683, 9002, 3045, 484, 55, 6, 1, 0, 47293, 133210, 52923, 8676, 905, 78, 7, 1, 0, 545835, 2299754, 1103781, 194404, 19855, 1518, 105, 8, 1, 0, 7087261, 45375130, 26857659, 5227236, 544505, 39390, 2359, 136, 9, 1
Offset: 0

Views

Author

Peter Luschny, May 21 2021

Keywords

Comments

The array rows are recursively generated by applying the Akiyama-Tanigawa algorithm to the powers (see the Python implementation below). In this way the array becomes the image of A004248 under the AT-transformation when applied to the columns of A004248. This makes the array closely linked to A371761, which is generated in the same way, but applied to the rows of A004248. - Peter Luschny, Apr 27 2024

Examples

			Triangle starts:
[0] 1;
[1] 0, 1;
[2] 0, 1,      1;
[3] 0, 3,      2,       1;
[4] 0, 13,     10,      3,       1;
[5] 0, 75,     74,      21,      4,      1;
[6] 0, 541,    730,     219,     36,     5,     1;
[7] 0, 4683,   9002,    3045,    484,    55,    6,    1;
[8] 0, 47293,  133210,  52923,   8676,   905,   78,   7,   1;
[9] 0, 545835, 2299754, 1103781, 194404, 19855, 1518, 105, 8, 1;
.
Seen as an array A(n, k) = T(n + k, n):
[0] [1, 0,   0,    0,     0,       0,         0, ...  A000007
[1] [1, 1,   3,   13,    75,     541,      4683, ...  A000670
[2] [1, 2,  10,   74,   730,    9002,    133210, ...  A004123
[3] [1, 3,  21,  219,  3045,   52923,   1103781, ...  A032033
[4] [1, 4,  36,  484,  8676,  194404,   5227236, ...  A094417
[5] [1, 5,  55,  905, 19855,  544505,  17919055, ...  A094418
[6] [1, 6,  78, 1518, 39390, 1277646,  49729758, ...  A094419
[7] [1, 7, 105, 2359, 70665, 2646007, 118893705, ...  A238464
		

Crossrefs

Variant of the array is A094416 (which has column 0 and row 0 missing).
The coefficients of the Fubini polynomials are A131689.
Cf. A094420 (main diagonal of array), A372346 (row sums), A004248, A371761.

Programs

  • Maple
    F := proc(n) option remember; if n = 0 then return 1 fi:
    expand(add(binomial(n, k)*F(n - k)*x, k = 1..n)) end:
    seq(seq(subs(x = k, F(n - k)), k = 0..n), n = 0..10);
  • Mathematica
    F[n_] := F[n] = If[n == 0, 1,
       Expand[Sum[Binomial[n, k]*F[n - k]*x, {k, 1, n}]]];
    Table[Table[F[n - k] /. x -> k, {k, 0, n}], {n, 0, 10}] // Flatten (* Jean-François Alcover, Jun 06 2024, after Peter Luschny *)
  • SageMath
    # Computes the triangle.
    @cached_function
    def F(n):
        R. = PolynomialRing(ZZ)
        if n == 0: return R(1)
        return R(sum(binomial(n, k)*F(n - k)*x for k in (1..n)))
    def Fval(n): return [F(n - k).substitute(x = k) for k in (0..n)]
    for n in range(10): print(Fval(n))
    
  • SageMath
    # Computes the square array using the Akiyama-Tanigawa algorithm.
    def ATFubini(n, len):
        A = [0] * len
        R = [0] * len
        for k in range(len):
            R[k] = (n + 1)**k  # Chancing this to R[k] = k**n generates A371761.
            for j in range(k, 0, -1):
                R[j - 1] = j * (R[j] - R[j - 1])
            A[k] = R[0]
        return A
    for n in range(8): print([n], ATFubini(n, 7))  # Peter Luschny, Apr 27 2024

Formula

T(n, k) = (n - k)! * [x^(n - k)] (1 / (1 + k * (1 - exp(x)))).
T(2*n, n) = A094420(n).

A382734 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where A(n,k) = n! * k! * [x^n * y^k] 1 / (exp(x) + exp(y) - exp(x+y))^2.

Original entry on oeis.org

1, 0, 0, 0, 2, 0, 0, 2, 2, 0, 0, 2, 14, 2, 0, 0, 2, 38, 38, 2, 0, 0, 2, 86, 254, 86, 2, 0, 0, 2, 182, 1118, 1118, 182, 2, 0, 0, 2, 374, 4142, 8654, 4142, 374, 2, 0, 0, 2, 758, 14078, 51662, 51662, 14078, 758, 2, 0, 0, 2, 1526, 45614, 267566, 467102, 267566, 45614, 1526, 2, 0
Offset: 0

Views

Author

Seiichi Manyama, Apr 04 2025

Keywords

Examples

			Square array begins:
  1, 0,   0,    0,     0,      0, ...
  0, 2,   2,    2,     2,      2, ...
  0, 2,  14,   38,    86,    182, ...
  0, 2,  38,  254,  1118,   4142, ...
  0, 2,  86, 1118,  8654,  51662, ...
  0, 2, 182, 4142, 51662, 467102, ...
		

Crossrefs

Main diagonal gives A382737.

Programs

  • PARI
    a(n, k) = sum(j=0, min(n, k), j!*(j+1)!*stirling(n, j, 2)*stirling(k, j, 2));

Formula

E.g.f.: 1 / (exp(x) + exp(y) - exp(x+y))^2.
A(n,k) = A(k,n).
A(n,k) = Sum_{j=0..min(n,k)} j! * (j+1)! * Stirling2(n,j) * Stirling2(k,j).

A382736 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where A(n,k) = n! * k! * [x^n * y^k] 1 / (exp(x) + exp(y) - exp(x+y))^4.

Original entry on oeis.org

1, 0, 0, 0, 4, 0, 0, 4, 4, 0, 0, 4, 44, 4, 0, 0, 4, 124, 124, 4, 0, 0, 4, 284, 1084, 284, 4, 0, 0, 4, 604, 5164, 5164, 604, 4, 0, 0, 4, 1244, 19804, 48044, 19804, 1244, 4, 0, 0, 4, 2524, 68524, 313804, 313804, 68524, 2524, 4, 0, 0, 4, 5084, 224284, 1707884, 3281404, 1707884, 224284, 5084, 4, 0
Offset: 0

Views

Author

Seiichi Manyama, Apr 04 2025

Keywords

Examples

			Square array begins:
  1, 0,   0,     0,      0,       0, ...
  0, 4,   4,     4,      4,       4, ...
  0, 4,  44,   124,    284,     604, ...
  0, 4, 124,  1084,   5164,   19804, ...
  0, 4, 284,  5164,  48044,  313804, ...
  0, 4, 604, 19804, 313804, 3281404, ...
		

Crossrefs

Main diagonal gives A382739.

Programs

  • PARI
    a(n, k) = sum(j=0, min(n, k), j!^2*binomial(j+3, 3)*stirling(n, j, 2)*stirling(k, j, 2));

Formula

E.g.f.: 1 / (exp(x) + exp(y) - exp(x+y))^4.
A(n,k) = A(k,n).
A(n,k) = Sum_{j=0..min(n,k)} (j!)^2 * binomial(j+3,3) * Stirling2(n,j) * Stirling2(k,j).

A006230 Bitriangular permutations.

Original entry on oeis.org

1, 13, 73, 301, 1081, 3613, 11593, 36301, 111961, 342013, 1038313, 3139501, 9467641, 28501213, 85700233, 257493901, 773268121, 2321377213, 6967277353, 20908123501, 62736953401, 188236026013, 564758409673, 1694375892301, 5083329003481, 15250389663613
Offset: 4

Views

Author

Keywords

Comments

Prepending the term 0 and setting the offset to 0 makes this sequence row 3 of A371761. In this form it can be generated by the Akiyama-Tanigawa algorithm for powers (see the Python script). - Peter Luschny, Apr 12 2024

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A136301 (row 4), A371761 (row 3).

Programs

  • Maple
    A006230:=-(z+1)*(6*z+1)/(z-1)/(3*z-1)/(2*z-1); # Conjectured by Simon Plouffe in his 1992 dissertation.
  • Mathematica
    12*StirlingS2[n+1, 3]+1; (* Brian Parsonnet, Feb 25 2011 *)
    Sum[ StirlingS2[n,i] * StirlingS2[ 3,i ] * i!^2, {i,3} ]; (* alternative, Brian Parsonnet, Feb 25 2011 *)
  • PARI
    Vec(x^4*(1 + x)*(1 + 6*x) / ((1 - x)*(1 - 2*x)*(1 - 3*x)) + O(x^40))
    \\ Colin Barker, Dec 27 2017
    
  • Python
    # Using the Akiyama-Tanigawa algorithm for powers from A371761.
    print(ATPowList(3, 27))  # Peter Luschny, Apr 12 2024

Formula

a(n) = 12*S(n-2) + 1, with S(n)=A000392(n) the Stirling numbers of second kind, 3rd column. - Ralf Stephan, Jul 07 2003
a(n+3) = Sum_{i=1..3} A008277(n,i) * A008277(3,i) * i!^2. - Brian Parsonnet, Feb 25 2011
From Colin Barker, Dec 27 2017: (Start)
G.f.: x^4*(1 + x)*(1 + 6*x) / ((1 - x)*(1 - 2*x)*(1 - 3*x)).
a(n) = 12*(3 - 3*2^(n-2) + 3^(n-2))/6 + 1.
a(n) = 6*a(n-1) - 11*a(n-2) + 6*a(n-3) for n>6. (End)
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