A008299 -id:A008299 - cp's OEIS frontend

A137375 Triangle read by rows, T(n,k) = (-1)^k*{{n,k}} where {{n,k}} are the second-order Stirling set numbers, n>=0, 0<=k<=n/2.

1, 0, 0, -1, 0, -1, 0, -1, 3, 0, -1, 10, 0, -1, 25, -15, 0, -1, 56, -105, 0, -1, 119, -490, 105, 0, -1, 246, -1918, 1260, 0, -1, 501, -6825, 9450, -945, 0, -1, 1012, -22935, 56980, -17325, 0, -1, 2035, -74316, 302995, -190575, 10395, 0, -1, 4082, -235092

Offset: 0

Roger L. Bagula, Apr 09 2008

Also called "associated Stirling numbers of the second kind" by Riordan in the signless form with different offset A008299.
Old name was: Triangular sequence from coefficients of Mahler polynomials from expansion of: p(x) = exp(x*(1 + t - exp(t))) with weight n!:M(x,n).
M(n,x) = sum(k=0..n, (x)^k*sum(j=0..k, binomial(n,k-j)*stirling2(n-k+j,j)*(-1)^j)). - Vladimir Kruchinin, Jan 13 2012

			[ 0]  1;
[ 1]  0;
[ 2]  0, -1;
[ 3]  0, -1;
[ 4]  0, -1,   3;
[ 5]  0, -1,  10;
[ 6]  0, -1,  25,   -15;
[ 7]  0, -1,  56,  -105;
[ 8]  0, -1, 119,  -490,  105;
[ 9]  0, -1, 246, -1918, 1260;
[10]  0, -1, 501, -6825, 9450, -945;

J. Riordan, Introduction to Combinatory Analysis, Wiley, New York, 1958.

L. Carlitz, The coefficients in an asymptotic expansion, Proc. Amer. Math. Soc. 16 (1965) 248-252.
A. E. Fekete, Apropos two notes on notation, Amer. Math. Monthly, 101 (1994), 771-778.
V. Kruchinin, D. Kruchinin, Application of a composition of generating functions for obtaining explicit formulas of polynomials, arXiv: 1211.0099 [math.NT], 2012.
D. V. Kruchinin and V. V. Kruchinin, Explicit Formulas for Some Generalized Polynomials, Appl. Math. Inf. Sci. 7, No. 5, 2083-2088 (2013).
Andrew Elvey Price, Alan D. Sokal, Phylogenetic trees, augmented perfect matchings, and a Thron-type continued fraction (T-fraction) for the Ward polynomials, arXiv:2001.01468 [math.CO], 2020.
L. M. Smiley, Completion of a rational function sequence of Carlitz, arXiv:math/0006106 [math.CO], 2000.
E. W. Weisstein, Mahler Polynomial.

Row sums are: A000587.

A137375 := proc(n, k) if n = 0 then 1 else
add(binomial(j,n-2*k)* combinat[eulerian2](n-k,n-k-j-1), j=(0..n-k-1))*(-1)^k fi end: for n from 0 to 9 do seq(A137375(n, k), k=(0..n/2)) od; # Peter Luschny, Dec 01 2012

Clear[p, x, t] p[t_] = Exp[x*(1 + t - Exp[t])]; Table[ ExpandAll[n!* SeriesCoefficient[Series[p[t], {t, 0, 30}], n]], {n, 0, 10}] a = Table[ CoefficientList[n!*SeriesCoefficient[Series[p[t], {t, 0, 30}], n], x], {n, 0, 10}]; Flatten[a];
Table[Sum[Binomial[n, k - j] StirlingS2[n - k + j, j] (-1)^j, {j, 0, k}], {n, 0, 15}, {k, 0, n/2}] // Flatten (* Eric W. Weisstein, Nov 13 2018 *)

T(n,k):=sum(binomial(n,k-j)*stirling2(n-k+j,j)*(-1)^(j),j,0,k); /* Vladimir Kruchinin, Jan 13 2012 */

def A137375(n, k): return add(binomial(n,k-j)*(-1)^j*stirling_number2(n-k+j,j) for j in (0..k))
for n in range(11):
    [A137375(n, k) for k in (0..n//2)]  # Peter Luschny, Dec 01 2012

T(n,k) = Sum_{j=0..k} C(n,k-j)*stirling2(n-k+j,j)*(-1)^(j). - Vladimir Kruchinin, Jan 13 2012
T(n,k) = (-1)^k*Sum_{j=0..n-k} C(j,n-2*k)*E2(n-k,n-k-j-1) for n>0, T(0,0) = 1, where E2(n,k) are the second-order Eulerian numbers A201637. - Peter Luschny, Nov 27 2012
Let p(x,t) = exp(x*(1+t-exp(t))) then T(n,k) = [x^k](n!*[t^n] series(p(x,t))) where [s^m] denotes the coefficient of s^m. - Peter Luschny, Dec 01 2012

Edited and simpler definition by Peter Luschny, Nov 27 2012

A059025 Triangle of Stirling numbers of order 6.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 462, 1, 1716, 1, 4719, 1, 11440, 1, 25883, 1, 56134, 1, 118456, 2858856, 1, 245480, 23279256, 1, 502588, 124710300, 1, 1020680, 551496660, 1, 2061709, 2181183147, 1, 4149752, 8021782197, 1, 8333153, 28051272535

Offset: 6

Barbara Haas Margolius (margolius(AT)math.csuohio.edu), Dec 14 2000

The number of partitions of the set N, |N|=n, into k blocks, all of cardinality greater than or equal to 6. This is the 6-associated Stirling number of the second kind.
This is entered as a triangular array. The entries S_6(n,k) are zero for 6k>n, so these values are omitted. Initial entry in sequence is S_6(6,1).
Rows are of lengths 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, ...

			There are 462 ways of partitioning a set N of cardinality 12 into 2 blocks each of cardinality at least 6, so S_6(12,2)=462.

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 222.
J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 76.

Michael De Vlieger, Table of n, a(n) for n = 6..13205 (rows n = 6..400, flattened).
Bishal Deb and Alan D. Sokal, Higher-order Stirling cycle and subset triangles: Total positivity, continued fractions and real-rootedness, arXiv:2507.18959 [math.CO], 2025. See p. 5.
A. E. Fekete, Apropos two notes on notation, Amer. Math. Monthly, 101 (1994), 771-778.

Cf. A008299, A059022, A059023, A059024.

S6[n_ /; 6 <= n <= 11, 1] = 1; S6[n_, k_] /; 1 <= k <= Floor[n/6] := S6[n, k] = k*S6[n-1, k] + Binomial[n-1, 5]*S6[n-6, k-1]; S6[, ] = 0; Flatten[ Table[ S6[n, k], {n, 6, 24}, {k, 1, Floor[n/6]}]] (* Jean-François Alcover, Feb 21 2012 *)

S_r(n+1, k)=k S_r(n, k)+binomial(n, r-1)S_r(n-r+1, k-1) for this sequence, r=6.

G.f.: Sum_{n>=0, k>=0} S_r(n,k)*u^k*t^n/n! = exp(u(e^t - Sum_{i=0..r-1} t^i/i!)).

A139541 There are 4n players who wish to play bridge at n tables. Each player must have another player as partner and each pair of partners must have another pair as opponents. The choice of partners and opponents can be made in exactly a(n)=(4n)!/(n!*8^n) different ways.

Original entry on oeis.org

1, 3, 315, 155925, 212837625, 618718975875, 3287253918823875, 28845653137679503125, 388983632561608099640625, 7637693625347175036443671875, 209402646126143497974176151796875, 7752714167528210725497923667975703125, 377130780679409810741846496828678078515625

Offset: 0

Reinhard Zumkeller, Apr 25 2008

nonn

From Karol A. Penson, Oct 05 2009: (Start)
Integral representation as n-th moment of a positive function on a positive semi-axis (solution of the Stieltjes moment problem), in Maple notation:
a(n)=int(x^n*((1/4)*sqrt(2)*(Pi^(3/2)*2^(1/4)*hypergeom([], [1/2, 3/4], -(1/32)*x)*sqrt(x)-2*Pi*hypergeom([], [3/4, 5/4], -(1/32)*x)*GAMMA(3/4)*x^(3/4)+sqrt(Pi)*GAMMA(3/4)^2*2^(1/4)*hypergeom([], [5/4, 3/2],-(1/32)*x)*x)/(Pi^(3/2)*GAMMA(3/4)*x^(5/4))), x=0..infinity), n=0,1... .
This solution may not be unique. (End)

G. Pólya and G. Szegő, Problems and Theorems in Analysis II (Springer 1924, reprinted 1976), Appendix: Problem 203.1, p164.

Andrew Howroyd, Table of n, a(n) for n = 0..100
Eric Weisstein's World of Mathematics, Tournament
Index entries for sequences related to tornaments.

Cf. A008299, A000142, A100733, A001018.

a(n)={(4*n)!/(n!*8^n)} \\ Andrew Howroyd, Jan 07 2020

a(n) = (4*n)!/(n!*8^n).
a(n) = A001147(n)*A001147(2*n).
a(n) = A008977(n)*(A049606(n)/A001316(n))^3. - Reinhard Zumkeller, Apr 28 2008

Terms a(11) and beyond from Andrew Howroyd, Jan 07 2020

A290586 Number of irredundant sets in the n X n rook graph.

Original entry on oeis.org

2, 11, 94, 1185, 20106, 453271, 13169346, 476777153, 20869990066, 1076251513071, 64077661097418, 4337014196039377, 329768528011095642, 27905789218764082151, 2608140451597365915346, 267506385903592339178241, 29943760423790270319833826

Offset: 1

Eric W. Weisstein, Aug 07 2017

nonn

Andrew Howroyd, Table of n, a(n) for n = 1..100
Andrew Howroyd, Irredundant Sets in Rook Graphs
Eric Weisstein's World of Mathematics, Irredundant Set
Eric Weisstein's World of Mathematics, Rook Graph

Main diagonal of A290818.
Row sums of A290823.
Cf. A008299, A248744.

s[n_, k_]:=Sum[(-1)^i*Binomial[n, i] StirlingS2[n - i, k - i], {i, 0, Min[n, k]}];
c[m_, n_, x_]:=Sum[Binomial[m, i] (n^i - n!*StirlingS2[i, n])*x^i, {i, 0, m - 1}];
p[m_, n_, x_]:=Sum[Sum[Binomial[m, k] Binomial[n, r]* k!*s[r, k]*x^r*c[m - k, n - r, x], {r, 2k, n - 1}], {k,0, m - 1}];
Table[2*n^n - n! + p[n, n, 1], {n, 30}]
(* Indranil Ghosh, Aug 12 2017, after PARI code *)

\\ here s(n,k) is A008299, 2*n^n - n! is A248744.
s(n,k)=sum(i=0, min(n, k), (-1)^i * binomial(n, i) * stirling(n-i, k-i, 2) );
c(m,n,x)=sum(i=0, m-1, binomial(m, i) * (n^i - n!*stirling(i, n, 2))*x^i);
p(m,n,x)={sum(k=0, m-1, sum(r=2*k, n-1, binomial(m,k) * binomial(n,r) * k! * s(r,k) * x^r * c(m-k,n-r,x) ))}
a(n) = 2*n^n - n! + p(n,n,1); \\ Andrew Howroyd, Aug 11 2017

a(n) = 2*n^n - n! + Sum_{k=0..n-1} Sum_{r=2*k..n-1} binomial(n,k) * binomial(n,r) * k! * A008299(r,k) * c(n-k,n-r) where c(m,n) = Sum_{i=0..m-1} binomial(n,i) * (n^i - n!*stirling2(i, n)). - Andrew Howroyd, Aug 11 2017

a(4) corrected and a(5) from Andrew Howroyd, Aug 07 2017

Terms a(6) and beyond from Andrew Howroyd, Aug 11 2017

A131106 Rectangular array read by antidiagonals: k objects are each put into one of n boxes, independently with equal probability. a(n, k) is the expected number of boxes with exactly one object (n, k >= 1). Sequence gives the numerators.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 4, 3, 0, 1, 3, 4, 1, 0, 1, 8, 27, 32, 5, 0, 1, 5, 48, 27, 80, 3, 0, 1, 12, 25, 256, 405, 64, 7, 0, 1, 7, 108, 125, 256, 729, 448, 1, 0, 1, 16, 147, 864, 3125, 6144, 5103, 1024, 9, 0, 1, 9, 64, 343, 6480, 3125, 28672, 2187, 256, 5, 0, 1, 20, 243, 2048, 12005

Offset: 1

David Wasserman, Jun 15 2007

Problem suggested by Brandon Zeidler. To motivate this sequence, suppose that when objects are placed in the same box, they mix and the information they contain is lost. The sequence tells us how much information we can expect to recover.

			Array begins:
  1 0 0 0 0 0 ...
  1 1 3/4 1/2 5/16 3/16 ...
  1 4/3 4/3 32/27 80/81 64/81 ...
  ...

Cf. A131107 gives the denominators. A131103, A131104 and A131105 give f(n, k, 0), f(n, k, 1) and f(n, k, 2).

a(n, k) = k*(1 - 1/n)^(k - 1). Let f(n, k, i) be the number of assignments such that exactly i boxes have exactly one object. For i > n, f(n, k, i) = 0. For i = k <= n, f(n, k, i) = n!/(n-k)!. Otherwise, f(n, k, i) = Sum_{j = 1..min(floor((k-i)/2), n-i)} A008299(k-i, j)*n!*binomial(k, i)/(n-i-j)!. Then a(n, k) = Sum_{i=1..min(n, k)} i*f(n, k, i)/n^k.

A278987 Array read by antidiagonals downwards: T(b,n) = number of words of length n over an alphabet of size b that are in standard order and which have the property that every letter that appears in the word is repeated.

Original entry on oeis.org

0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 4, 1, 1, 0, 1, 11, 4, 1, 1, 0, 1, 26, 11, 4, 1, 1, 0, 1, 57, 41, 11, 4, 1, 1, 0, 1, 120, 162, 41, 11, 4, 1, 1, 0, 1, 247, 610, 162, 41, 11, 4, 1, 1, 0, 1, 502, 2165, 715, 162, 41, 11, 4, 1, 1, 0, 1, 1013, 7327, 3425, 715, 162, 41, 11, 4, 1, 1, 0

Offset: 1

Joerg Arndt and N. J. A. Sloane, Dec 06 2016

We study words made of letters from an alphabet of size b, where b >= 1. We assume the letters are labeled {1,2,3,...,b}. There are b^n possible words of length n.

We say that a word is in "standard order" if it has the property that whenever a letter i appears, the letter i-1 has already appeared in the word. This implies that all words begin with the letter 1.

			The array begins:
0,.1,..1,...1,...1,...1,...1,....1..; b=1,
0,.1,..4,...8,..16,..32,..64,..128..; b=2,
0,.1,..4,..14,..41,.122,.365,.1094..; b=3,
0,.1,..4,..14,..51,.187,.715,.2795..; b=4,
0,.1,..4,..14,..51,.202,.855,.3845..; b=5,
0,.1,..4,..14,..51,.202,.876,.4111..; b=6,
...
Rows b=1 through b=4 of the array are A000012, A000295 (or A130103), A278988, A278989.

Joerg Arndt and N. J. A. Sloane, Counting Words that are in "Standard Order"

The words for b=9 are listed in A273978.

with(combinat);
A008299 := proc(n,k) local i,j,t1;
if k<1 or k>floor(n/2) then t1:=0; else
t1 := add( (-1)^i*binomial(n, i)*add( (-1)^j*(k - i - j)^(n - i)/(j!*(k - i - j)!), j = 0..k - i), i = 0..k); fi; t1; end;
f3:=proc(L,b) global A008299; local i; add(A008299(L,i),i=1..b); end;
Q3:=b->[seq(f3(L,b),L=1..40)];
for b from 1 to 6 do lprint(Q3(b)); od:

The number of words of length n over an alphabet of size b that are in standard order and in which every symbol that appears in a word is repeated is Sum_{j = 1..b} A008299(n,j).

A290818 Array read by antidiagonals: T(m,n) = number of irredundant sets in the lattice (rook) graph K_m X K_n.

Original entry on oeis.org

2, 3, 3, 4, 11, 4, 5, 24, 24, 5, 6, 47, 94, 47, 6, 7, 88, 272, 272, 88, 7, 8, 163, 774, 1185, 774, 163, 8, 9, 304, 2230, 4280, 4280, 2230, 304, 9, 10, 575, 6542, 15781, 20106, 15781, 6542, 575, 10, 11, 1104, 19452, 60604, 88512, 88512, 60604, 19452, 1104, 11

Offset: 1

Andrew Howroyd, Aug 11 2017

			Array begins:
===============================================================
m\n| 1   2     3      4       5        6        7         8
---+-----------------------------------------------------------
1  | 2   3     4      5       6        7        8         9 ...
2  | 3  11    24     47      88      163      304       575 ...
3  | 4  24    94    272     774     2230     6542     19452 ...
4  | 5  47   272   1185    4280    15781    60604    240073 ...
5  | 6  88   774   4280   20106    88512   400728   1879744 ...
6  | 7 163  2230  15781   88512   453271  2326534  12363513 ...
7  | 8 304  6542  60604  400728  2326534 13169346  76446456 ...
8  | 9 575 19452 240073 1879744 12363513 76446456 476777153 ...
...

Andrew Howroyd, Table of n, a(n) for n = 1..1275
Eric Weisstein's World of Mathematics, Irredundant Set
Eric Weisstein's World of Mathematics, Rook Graph

Row 2 is A290707 for n > 1.
Main diagonal is A290586.
Cf. A287274, A290632.

s[n_, k_]:=Sum[(-1)^i*Binomial[n, i] StirlingS2[n - i, k - i], {i, 0, Min[n, k]}];
c[m_, n_, x_]:=Sum[Binomial[m, i] (n^i - n!*StirlingS2[i, n])*x^i, {i, 0, m - 1}];
p[m_, n_, x_]:=Sum[Sum[Binomial[m, k] Binomial[n, r]* k!*s[r, k]*x^r*c[m - k, n - r, x], {r, 2k, n - 1}], {k,0, m - 1}];
b[m_, n_, x_]:=m^n*x^n + n^m*x^m - If[n<=m, n!*x^m*StirlingS2[m, n], m!*x^n*StirlingS2[n, m]];
T[m_, n_]:= b[m, n, 1] + p[m, n, 1];
Table[T[n, m -n + 1], {m, 10}, {n, m}]//Flatten
(* Indranil Ghosh, Aug 12 2017, after PARI code *)

\\ See A. Howroyd note in A290586 for explanation.
s(n,k)=sum(i=0, min(n, k), (-1)^i * binomial(n, i) * stirling(n-i, k-i, 2) );
c(m,n,x)=sum(i=0, m-1, binomial(m, i) * (n^i - n!*stirling(i, n, 2))*x^i);
p(m,n,x)={sum(k=0, m-1, sum(r=2*k, n-1, binomial(m, k) * binomial(n, r) * k! * s(r, k) * x^r * c(m-k, n-r, x) ))}
b(m,n,x) = m^n*x^n + n^m*x^m - if(n<=m, n!*x^m*stirling(m, n, 2), m!*x^n*stirling(n, m, 2));
T(m,n) = b(m,n,1) + p(m,n,1);
for(m=1,10,for(n=1,m,print1(T(n,m-n+1),", ")));

T(m,n) = A290632(m, n) + Sum_{k=0..m-1} Sum_{r=2*k..n-1} binomial(m,k) * binomial(n,r) * k! * A008299(r,k) * c(m-k,n-r) where c(m,n) = Sum_{i=0..m-1} binomial(n,i) * (n^i - n!*stirling2(i, n)).

A131103 Rectangular array read by antidiagonals: a(n, k) is the number of ways to put k labeled objects into n labeled boxes so that there are no boxes with exactly one object (n, k >= 1).

Original entry on oeis.org

0, 0, 1, 0, 2, 1, 0, 3, 2, 1, 0, 4, 3, 8, 1, 0, 5, 4, 21, 22, 1, 0, 6, 5, 40, 63, 52, 1, 0, 7, 6, 65, 124, 243, 114, 1, 0, 8, 7, 96, 205, 664, 969, 240, 1, 0, 9, 8, 133, 306, 1405, 3196, 3657, 494, 1, 0, 10, 9, 176, 427, 2556, 7425, 15712, 12987, 1004, 1, 0, 11, 10, 225, 568, 4207

Offset: 1

David Wasserman, Jun 14 2007, Jun 15 2007

Problem suggested by Brandon Zeidler. Columns four and five are A000567 and A051874. Second row is A130102.

			Array begins:
0 1 1 1 1 1 1
0 2 2 8 22 52 114
0 3 3 21 63 243 969

Cf. A131104, A131105, A131106, A131107.

a(n, k) = sum_{j=1..min(floor(k/2), n)} A008299(k, j)*n!/(n-j)!.

A131104 Rectangular array read by antidiagonals: a(n, k) is the number of ways to put k labeled objects into n labeled boxes so that there is one box with exactly one object (n, k >= 1).

Original entry on oeis.org

1, 2, 0, 3, 0, 0, 4, 0, 6, 0, 5, 0, 18, 8, 0, 6, 0, 36, 24, 10, 0, 7, 0, 60, 48, 120, 12, 0, 8, 0, 90, 80, 420, 396, 14, 0, 9, 0, 126, 120, 1000, 1512, 1092, 16, 0, 10, 0, 168, 168, 1950, 3720, 6804, 2736, 18, 0, 11, 0, 216, 224, 3360, 7380, 23240, 31008, 6480, 20, 0, 12, 0

Offset: 1

David Wasserman, Jun 14 2007, Jun 15 2007

Problem suggested by Brandon Zeidler. Columns 3 through 5 are A028896, A033996, 10*A007586.

			Array begins:
1 0 0 0 0 0 0
2 0 6 8 10 12 14
3 0 18 24 120 396 1092

Cf. A131103, A131105, A131106, A131107.

a(n, 1) = n. For k > 1, a(n, k) = sum_{j=1..min(floor((k-1)/2), n-1)} A008299(k-1, j)*n!*k*/(n-j-1)!.

A131105 Rectangular array read by antidiagonals: a(n, k) is the number of ways to put k labeled objects into n labeled boxes so that there are exactly two boxes with exactly one object (n, k >= 2).

Original entry on oeis.org

2, 6, 0, 12, 0, 0, 20, 0, 36, 0, 30, 0, 144, 60, 0, 42, 0, 360, 240, 90, 0, 56, 0, 720, 600, 1440, 126, 0, 72, 0, 1260, 1200, 6300, 5544, 168, 0, 90, 0, 2016, 2100, 18000, 26460, 17472, 216, 0, 110, 0, 3024, 3360, 40950, 78120, 136080, 49248, 270, 0, 132, 0, 4320

Offset: 2

David Wasserman, Jun 15 2007

Problem suggested by Brandon Zeidler. Columns 2, 4 and 5 are A002378, 36*A000292 and 60*A000292.

			Array begins:
2 0 0 0 0 0
6 0 36 60 90 126
12 0 144 240 1440 5544

Cf. A131103, A131104, A131106, A131107.

a(n, 2) = n^2-n. For k > 2, a(n, k) = sum_{j=1..min(floor(k/2)-1, n-2)} A008299(k-2, j)*n!*(k^2-k)/(2*(n-j-2)!).

cp's OEIS Frontend

A137375 Triangle read by rows, T(n,k) = (-1)^k*{{n,k}} where {{n,k}} are the second-order Stirling set numbers, n>=0, 0<=k<=n/2.

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A059025 Triangle of Stirling numbers of order 6.

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A139541 There are 4*n players who wish to play bridge at n tables. Each player must have another player as partner and each pair of partners must have another pair as opponents. The choice of partners and opponents can be made in exactly a(n)=(4*n)!/(n!*8^n) different ways.

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PARI

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A290586 Number of irredundant sets in the n X n rook graph.

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A131106 Rectangular array read by antidiagonals: k objects are each put into one of n boxes, independently with equal probability. a(n, k) is the expected number of boxes with exactly one object (n, k >= 1). Sequence gives the numerators.

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A278987 Array read by antidiagonals downwards: T(b,n) = number of words of length n over an alphabet of size b that are in standard order and which have the property that every letter that appears in the word is repeated.

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Maple

Formula

A290818 Array read by antidiagonals: T(m,n) = number of irredundant sets in the lattice (rook) graph K_m X K_n.

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Mathematica

A139541 There are 4n players who wish to play bridge at n tables. Each player must have another player as partner and each pair of partners must have another pair as opponents. The choice of partners and opponents can be made in exactly a(n)=(4n)!/(n!*8^n) different ways.