A131689 -id:A131689 - cp's OEIS frontend

A367392 a(n) = Sum_{k=0..n} (-1)^k * binomial(n, k) * (n - k)^(2*n + 1).

0, 1, 30, 1806, 186480, 29607600, 6711344640, 2060056318320, 823172919528960, 415357755774998400, 258323865658578720000, 194165346649139268480000, 173524374976148227519488000, 181871966450361851863879680000, 220951172052769326900328396800000

Offset: 0

Peter Luschny, Nov 21 2023

nonn

Cf. A000142, A048993, A131689, A247238.

a := n -> n! * Stirling2(2*n + 1, n):
seq(a(n), n = 0..14);

A367392[n_]:=n!StirlingS2[2n+1,n];
Array[A367392,20,0] (* Paolo Xausa, Nov 24 2023 *)

a(n) = n! * Stirling2(2*n + 1, n) = A000142(n) * A247238(n).

A380977 Triangle read by rows: T(n,m) (1<=m<=n) = number of surjections f:[n]->[m] with f(n) != f(j), j

Original entry on oeis.org

1, 0, 2, 0, 2, 6, 0, 2, 18, 24, 0, 2, 42, 144, 120, 0, 2, 90, 600, 1200, 720, 0, 2, 186, 2160, 7800, 10800, 5040, 0, 2, 378, 7224, 42000, 100800, 105840, 40320, 0, 2, 762, 23184, 204120, 756000, 1340640, 1128960, 362880, 0, 2, 1530, 72600, 932400, 5004720, 13335840, 18627840, 13063680, 3628800

Offset: 1

Manfred Boergens, Feb 10 2025

Number of n-tuples containing all elements of [m] with a unique last element.
Consider an urn with m balls of pairwise different colors. T(n,m) is motivated by the probability p(n,m) for exactly n draws with replacement needed to obtain all colors; p(n,m)=T(n,m)/m^n. - With m fixed and n running, p(n,m) is a probability distribution. The expected number of draws needed to obtain all colors is Sum_{j=1..m} m/j.
In the Laplace link, an asymptotic formula is given for i(n,k), the number of draws required to have probability 1/k that all balls have been drawn. However, Laplace did not consider the distribution of the draw-count.
Feller's Introduction, section 'Waiting times in sampling' (p. 211 in 2nd edition, 224 in 3rd) seems to be the first to discuss the expectation, then gives the variance as a problem (p. 224 in 2nd, p. 239 in 3rd).

			The triangle T(n,m) begins:
  n\m  1 2    3     4      5       6        7        8        9      10 ...
   1:  1
   2:  0 2
   3:  0 2    6
   4:  0 2   18    24
   5:  0 2   42   144    120
   6:  0 2   90   600   1200     720
   7:  0 2  186  2160   7800   10800     5040
   8:  0 2  378  7224  42000  100800   105840    40320
   9:  0 2  762 23184 204120  756000  1340640  1128960   362880
  10:  0 2 1530 72600 932400 5004720 13335840 18627840 13063680 3628800
  ...
T(4,3)=18 is the number of 4-sequences of draws from [3] completing the covering of [3] with the last draw; these sequences are (without brackets and commas):
   1123 1213 1223 2113 2123 2213 1132 1312 1332
   3112 3132 3312 2231 2321 2331 3221 3231 3321

William Feller, Introduction to Probability Theory and Its Applications, 2nd ed., 1957.

Pierre-Simon Laplace, Théorie analytique des probabilités, 1812.
Michael Shackleford, Problem 74. Free gift in the cereal box problem #2, Mathproblems.info.
Wikipedia, Coupon collector's problem.

Cf. A048993, A068293, A131689.

Row sums give A005649(n-1) for n>=1.

Table[m! StirlingS2[n - 1, m - 1], {n, 10}, {m, n}]//Flatten

T(n,m) = m!*S2(n-1,m-1) = m!*A048993(n-1,m-1).
T(n,m) = m*A131689(n-1,m-1).
T(n,3) = A068293(n-1), n>1.
From Natalia L. Skirrow, Aug 29 2025: (Start)
E.g.f.: (1/(1-(e^x-1)*y) + (x-log(y*(e^x-1)-1))/(1+y)) * y/(1+y)
E.g.f. for T(n+1,k+1): 1/(1-y*(e^x-1))^2.
In general, the e.g.f. of sequence T(n,k) = S2(n,k)*(o)_k is 1/(1-y(e^x-1))^o, where (o)_k = (o+k-1)!/(o-1)! is the Pochhammer symbol. (End)

A156820 T(n,m) = Sum_{j=0..m} (-1)^(j + m)(j + 1)^nbinomial(m, j) + Sum_{j=0..(n-m)} (-1)^(j - m + n )(1 + j)^nbinomial(n-m, j).

Original entry on oeis.org

2, 2, 2, 3, 6, 3, 7, 19, 19, 7, 25, 75, 100, 75, 25, 121, 391, 570, 570, 391, 121, 721, 2583, 3962, 4200, 3962, 2583, 721, 5041, 20287, 33852, 35406, 35406, 33852, 20287, 5041, 40321, 181695, 338690, 364140, 333648, 364140, 338690, 181695, 40321, 362881

Offset: 0

Roger L. Bagula, Feb 16 2009

Row sums are 2, 4, 12, 52, 300, 2164, 18732, 189172, 2183340, 28349044, ... = 2 * A000629.

			Triangle begins:
2
2, 2
3, 6, 3
7, 19, 19, 7
25, 75, 100, 75, 25
121, 391, 570, 570, 391, 121
721, 2583, 3962, 4200, 3962, 2583, 721
5041, 20287, 33852, 35406, 35406, 33852, 20287, 5041
40321, 181695, 338690, 364140, 333648, 364140, 338690, 181695, 40321

Cf. A000629, A028246, A131689, A123125, A007318.

w[n_, m_] = Sum[(-1)^(j + m)*(j + 1)^n*Binomial[m, j], {j, 0, m}] +
Sum[(-1)^( j - m + n )*(1 + j)^n Binomial[ -m + n, j], {j, 0, n - m}];
Table[Table[w[n, m], {m, 0, n}], {n, 0, 10}];
Flatten[%]

T(n,m) = Sum_{j=0..m}[(-1)^(j + m)*(j + 1)^n*binomial(m, j)] + Sum_{j=0..(n-m)}[(-1)^(j - m + n )*(1 + j)^n*binomial(n-m, j)].

Apparently, T = A007318 * A131689 + A123125 * A007318 = A028246 + row reversed A028246. - Tom Copeland, Jan 26 2015

A321434 Triangle read by rows; T(n,k) is the number of achiral rows of n colors using exactly k colors.

Original entry on oeis.org

1, 0, 1, 0, 1, 0, 1, 2, 0, 1, 2, 0, 1, 6, 6, 0, 1, 6, 6, 0, 1, 14, 36, 24, 0, 1, 14, 36, 24, 0, 1, 30, 150, 240, 120, 0, 1, 30, 150, 240, 120, 0, 1, 62, 540, 1560, 1800, 720, 0, 1, 62, 540, 1560, 1800, 720, 0, 1, 126, 1806, 8400, 16800, 15120, 5040, 0, 1, 126, 1806, 8400, 16800, 15120, 5040

Offset: 0

Robert A. Russell, Nov 09 2018

Each zero in the data is the beginning of a new row.

Same as A131689, with rows (except for the first) repeated. - Joerg Arndt, Sep 08 2019

			The triangle begins with T(0,0):
1
0 1
0 1
0 1   2
0 1   2
0 1   6     6
0 1   6     6
0 1  14    36     24
0 1  14    36     24
0 1  30   150    240    120
0 1  30   150    240    120
0 1  62   540   1560   1800    720
0 1  62   540   1560   1800    720
0 1 126  1806   8400  16800   15120    5040
0 1 126  1806   8400  16800   15120    5040
0 1 254  5796  40824 126000  191520  141120   40320
0 1 254  5796  40824 126000  191520  141120   40320
0 1 510 18150 186480 834120 1905120 2328480 1451520 362880
For T(7,2)=14, the rows are AAABAAA, AABABAA, AABBBAA, ABAAABA, ABABABA, ABBABBA, ABBBBBA, BAAAAAB, BAABAAB, BABABAB, BABBBAB, BBAAABB, BBABABB, and BBBABBB.

Columns 0-6 are A000007, A057427, A056453, A056454, A056455, A056456, A056457.
Cf. A019538 (oriented), A305621 (unoriented), A305622 (chiral).
Cf. A131689.

Table[k! StirlingS2[Ceiling[n/2], k], {n, 0, 18}, {k, 0, (n+1)/2}] // Flatten

T(n,k) = k!*S2(ceiling(n/2),k), where S2 is the Stirling subset number A008277.

A326717 Coefficients of polynomials related to ordered set partitions. Triangle read by rows, T_{m}(n, k) for m = 5 and 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 127, 126, 0, 255256, 381381, 126126, 0, 2979852651, 5447453786, 2956465512, 488864376, 0, 127156445503275, 264284637872750, 184292523727620, 52359004217520, 5194672859376

Offset: 0

Peter Luschny, Jul 21 2019

			Triangle starts:
[0] [1]
[1] [0, 1]
[2] [0, 127, 126]
[3] [0, 255256, 381381, 126126]
[4] [0, 2979852651, 5447453786, 2956465512, 488864376]
[5] [0, 127156445503275, 264284637872750, 184292523727620, 52359004217520, 5194672859376]
[6] [0, 15160169962750251082, 34544220081315967665, 28276496764200664980, 10634436034307385300, 1865368063755476280, 123378675083039376]

Row sums A243666. Main diagonal A025037.

A129062 (m=1, associated with A131689), A326477 (m=2, associated with A241171), A326587 (m=3, associated with A278073), A326585 (m=4, associated with A278074), this sequence (m=5).

Maple
```
# See A326477.
```

T(n, k) = T_{5}(n, k) where T_{m}(n, k) is defined in A326477.

A344392 T(n, k) = k!*Stirling2(n - k, k), for n >= 0 and 0 <= k <= floor(n/2). Triangle read by rows.

Original entry on oeis.org

1, 0, 0, 1, 0, 1, 0, 1, 2, 0, 1, 6, 0, 1, 14, 6, 0, 1, 30, 36, 0, 1, 62, 150, 24, 0, 1, 126, 540, 240, 0, 1, 254, 1806, 1560, 120, 0, 1, 510, 5796, 8400, 1800, 0, 1, 1022, 18150, 40824, 16800, 720, 0, 1, 2046, 55980, 186480, 126000, 15120

Offset: 0

Peter Luschny, May 17 2021

The antidiagonal representation of the Fubini numbers (A131689).

			Triangle starts:
[ 0] [1]
[ 1] [0]
[ 2] [0, 1]
[ 3] [0, 1]
[ 4] [0, 1, 2]
[ 5] [0, 1, 6]
[ 6] [0, 1, 14, 6]
[ 7] [0, 1, 30, 36]
[ 8] [0, 1, 62, 150, 24]
[ 9] [0, 1, 126, 540, 240]
[10] [0, 1, 254, 1806, 1560, 120]
[11] [0, 1, 510, 5796, 8400, 1800]

Cf. A105795 (row sums).

Cf. A019538, A131689, A048993.

T := (n, k) -> k!*Stirling2(n - k, k):
seq(seq(T(n, k), k=0..n/2), n = 0..11);

A344398 a(n) = (-1)^n * F_{n}((-1)^n * n), where F_{n}(x) is the Fubini polynomial.

Original entry on oeis.org

1, 1, 10, 111, 8676, 243005, 49729758, 2634606331, 1026912225160, 88276603008249, 55954905981282210, 7103694104486331671, 6655958151527584785900, 1171100778886715057133493, 1521436331153097968932487206, 354408430829377435361459172915, 609729139653483641913607434550800

Offset: 0

Peter Luschny, May 21 2021

nonn

The coefficients of the Fubini polynomials are A131689.

Cf. A094420.

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:
a := n -> (-1)^n*subs(x = (-1)^n*n, F(n)):
seq(a(n), n = 0..17);

F[n_][x_] := If[n == 0, 1, Sum[k! StirlingS2[n, k] x^k, {k, 0, n}]];
a[n_] := (-1)^n F[n][(-1)^n*n];
Table[a[n], {n, 0, 16}] (* Jean-François Alcover, May 09 2024 *)

@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 a(n):
    return (-1)^n*F(n).substitute(x = (-1)^n*n)
print([a(n) for n in range(17)])

cp's OEIS Frontend

A367392 a(n) = Sum_{k=0..n} (-1)^k * binomial(n, k) * (n - k)^(2*n + 1).

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A380977 Triangle read by rows: T(n,m) (1<=m<=n) = number of surjections f:[n]->[m] with f(n) != f(j), j

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A156820 T(n,m) = Sum_{j=0..m} (-1)^(j + m)*(j + 1)^n*binomial(m, j) + Sum_{j=0..(n-m)} (-1)^(j - m + n )*(1 + j)^n*binomial(n-m, j).

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A321434 Triangle read by rows; T(n,k) is the number of achiral rows of n colors using exactly k colors.

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A326717 Coefficients of polynomials related to ordered set partitions. Triangle read by rows, T_{m}(n, k) for m = 5 and 0 <= k <= n.

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A344392 T(n, k) = k!*Stirling2(n - k, k), for n >= 0 and 0 <= k <= floor(n/2). Triangle read by rows.

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A344398 a(n) = (-1)^n * F_{n}((-1)^n * n), where F_{n}(x) is the Fubini polynomial.

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A156820 T(n,m) = Sum_{j=0..m} (-1)^(j + m)(j + 1)^nbinomial(m, j) + Sum_{j=0..(n-m)} (-1)^(j - m + n )(1 + j)^nbinomial(n-m, j).