A037959
a(n) = n^2*(n+1)*(n+2)!/48.
Original entry on oeis.org
6, 90, 1200, 15750, 211680, 2963520, 43545600, 673596000, 10977120000, 188367379200, 3399953356800, 64457449056000, 1281520880640000, 26676557107200000, 580481882652672000, 13183287756807168000
Offset: 2
- Identity (1.19)/(n+3) in H. W. Gould, Combinatorial Identities, Morgantown, 1972, page 3.
-
[Factorial(n)*StirlingSecond(n+3,n)/(n+3): n in [2..30]]; // G. C. Greubel, Jun 20 2022
-
Table[(n+2)!n^2(n+1)/48,{n,2,20}] (* Harvey P. Dale, Jul 29 2021 *)
-
[factorial(n)*stirling_number2(n+3, n)/(n+3) for n in (2..30)] # G. C. Greubel, Jun 20 2022
A344116
Triangle read by rows: T(n,k) is the number of relations from an n-element set to a k-element set that are not onto functions.
Original entry on oeis.org
1, 3, 14, 7, 58, 506, 15, 242, 4060, 65512, 31, 994, 32618, 1048336, 33554312, 63, 4034, 261604, 16775656, 1073740024, 68719476016, 127, 16258, 2095346, 268427056, 34359721568, 4398046495984, 562949953416272, 255, 65282, 16771420, 4294926472, 1099511501776, 281474976519136, 72057594037786816, 18446744073709511296
Offset: 1
For T(2,2), the number of relations is 2^4 and the number of onto functions is 2, so 2^4 - 2 = 14.
Triangle T(n,k) begins:
1
3 14
7 58 506
15 242 4060 65512
31 994 32618 1048336 33554312
-
TableForm[Table[2^(n*k) - Sum[Binomial[k, k - i] (k - i)^n*(-1)^i, {i, 0, k}], {n, 5}, {k, n}]]
-
T(n,k) = 2^(n*k) - k!*stirling(n, k, 2); \\ Michel Marcus, Jun 26 2021
A371568
Array read by ascending antidiagonals: A(n, k) is the number of paths of length k in Z^n from the origin to points such that x1+x2+...+xn = k with x1,...,xn > 0.
Original entry on oeis.org
1, 0, 1, 0, 2, 1, 0, 0, 6, 1, 0, 0, 6, 14, 1, 0, 0, 0, 36, 30, 1, 0, 0, 0, 24, 150, 62, 1, 0, 0, 0, 0, 240, 540, 126, 1, 0, 0, 0, 0, 120, 1560, 1806, 254, 1, 0, 0, 0, 0, 1800, 8400, 5796, 510, 1
Offset: 1
n\k 1 2 3 4 5 6 7 8 9 10
--------------------------------------------------
1| 1 1 1 1 1 1 1 1 1 1
2| 0 2 6 14 30 62 126 254 510 1022
3| 0 0 6 36 150 540 1806 5796 18150 55980
4| 0 0 0 24 240 1560 8400 40824 186480 818520
5| 0 0 0 0 120 1800 16800 126000 834120 5103000
6| 0 0 0 0 0 720 15120 191520 1905120 16435440
7| 0 0 0 0 0 0 5040 141120 2328480 29635200
8| 0 0 0 0 0 0 0 40320 1451520 30240000
9| 0 0 0 0 0 0 0 0 362880 16329600
10| 0 0 0 0 0 0 0 0 0 3628800
-
A[n_,k_] := Sum[(-1)^(n-i) * i^k * Binomial[n,i], {i,1,n}]
-
# The Akiyama-Tanigawa algorithm for the binomial generates the rows.
# Adds row(0) = 0^k and column(0) = 0^n.
from math import comb as binomial
def ATBinomial(n, len):
A = [0] * len
R = [0] * len
for k in range(len):
R[k] = binomial(k, n)
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(11): print([n], ATBinomial(n, 11)) # Peter Luschny, Apr 19 2024
A372346
a(n) = Sum_{j=0..n} p(n - j, j) where p(n, x) = Sum_{k=0..n} k! * Stirling2(n, k) * x^k. Row sums of A344499.
Original entry on oeis.org
1, 1, 2, 6, 27, 175, 1532, 17276, 243093, 4165261, 85133686, 2039546786, 56447550543, 1783865468187, 63766726231792, 2558290237404920, 114418196763735113, 5670168958036693977, 309630356618418661738, 18536683645526372648446, 1211038603734731649106307, 85983731724631359047504967
Offset: 0
-
p := n -> local k; add(k!*Stirling2(n, k)*x^k, k = 0..n):
a := n -> local j; add(subs(x = j, p(n - j)), j = 0..n):
seq(a(n), n = 0..21);
A105797
"Stirling-Bernoulli transform" of Pell numbers.
Original entry on oeis.org
0, 1, 3, 19, 135, 1291, 14343, 188539, 2815095, 47412811, 886239783, 18231365659, 409053408855, 9943622273131, 260300948527623, 7300927107288379, 218426614502831415, 6943261704033434251, 233692323131307301863
Offset: 0
-
CoefficientList[Series[E^x*(1-E^x)/(1-4*E^x+2*E^(2*x)), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Sep 26 2013 *)
A225476
Triangle read by rows, k!*2^k*S_2(n, k) where S_m(n, k) are the Stirling-Frobenius subset numbers of order m; n >= 0, k >= 0.
Original entry on oeis.org
1, 1, 1, 1, 4, 2, 1, 13, 18, 6, 1, 40, 116, 96, 24, 1, 121, 660, 1020, 600, 120, 1, 364, 3542, 9120, 9480, 4320, 720, 1, 1093, 18438, 74466, 121800, 94920, 35280, 5040, 1, 3280, 94376, 576576, 1394064, 1653120, 1028160, 322560, 40320, 1, 9841, 478440, 4319160
Offset: 0
[n\k][0, 1, 2, 3, 4, 5 ]
[0] 1,
[1] 1, 1,
[2] 1, 4, 2,
[3] 1, 13, 18, 6,
[4] 1, 40, 116, 96, 24,
[5] 1, 121, 660, 1020, 600, 120.
- Vincenzo Librandi, Rows n = 0..50, flattened
- Peter Luschny, Generalized Eulerian polynomials.
- Peter Luschny, The Stirling-Frobenius numbers.
- Shi-Mei Ma, Toufik Mansour, Matthias Schork, Normal ordering problem and the extensions of the Stirling grammar, Russian Journal of Mathematical Physics, 2014, 21(2), arXiv 1308.0169 p. 12.
Alternating row sum ~
A000364 (Euler secant numbers).
-
SF_SSO := proc(n, k, m) option remember;
if n = 0 and k = 0 then return(1) fi;
if k > n or k < 0 then return(0) fi;
k*SF_SSO(n-1, k-1, m) + (m*(k+1)-1)*SF_SSO(n-1, k, m) end:
seq(print(seq(SF_SSO(n, k, 2), k=0..n)), n = 0..5);
-
EulerianNumber[n_, k_, m_] := EulerianNumber[n, k, m] = (If[ n == 0, Return[If[k == 0, 1, 0]]]; Return[(m*(n - k) + m - 1)*EulerianNumber[n - 1, k - 1, m] + (m*k + 1)*EulerianNumber[n - 1, k, m]]); SFSSO[n_, k_, m_] := Sum[ EulerianNumber[n, j, m]*Binomial[j, n - k], {j, 0, n}]/m^k; Table[ SFSSO[n, k, 2], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 29 2013, translated from Sage *)
-
@CachedFunction
def EulerianNumber(n, k, m) :
if n == 0: return 1 if k == 0 else 0
return (m*(n-k)+m-1)*EulerianNumber(n-1, k-1, m)+(m*k+1)*EulerianNumber(n-1, k, m)
def SF_SSO(n, k, m):
return add(EulerianNumber(n, j, m)*binomial(j, n - k) for j in (0..n))/m^k
for n in (0..6): [SF_SSO(n, k, 2) for k in (0..n)]
A285867
Triangle T(n, k) read by rows: T(n, k) = S2(n, k)*k! + S2(n, k-1)*(k-1)! with the Stirling2 triangle S2 = A048993.
Original entry on oeis.org
1, 0, 1, 0, 1, 3, 0, 1, 7, 12, 0, 1, 15, 50, 60, 0, 1, 31, 180, 390, 360, 0, 1, 63, 602, 2100, 3360, 2520, 0, 1, 127, 1932, 10206, 25200, 31920, 20160, 0, 1, 255, 6050, 46620, 166824, 317520, 332640, 181440, 0, 1, 511, 18660, 204630, 1020600, 2739240, 4233600, 3780000, 1814400, 0, 1, 1023, 57002, 874500, 5921520, 21538440, 46070640, 59875200, 46569600, 19958400
Offset: 0
The triangle T(n, k) begins:
n\k 0 1 2 3 4 5 6 7 8 9 ...
0: 1
1: 0 1
2: 0 1 3
3: 0 1 7 12
4: 0 1 15 50 60
5: 0 1 31 180 390 360
6: 0 1 63 602 2100 3360 2520
7: 0 1 127 1932 10206 25200 31920 20160
8: 0 1 255 6050 46620 166824 317520 332640 181440
9: 0 1 511 18660 204630 1020600 2739240 4233600 3780000 1814400
...
-
Table[If[k == 0, Boole[n == 0], StirlingS2[n, k] k! + StirlingS2[n, k - 1] (k - 1)!], {n, 0, 10}, {k, 0, n}] (* Michael De Vlieger, May 08 2017 *)
A344053
a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*Stirling2(n, k)*k!.
Original entry on oeis.org
1, 1, 0, -9, -40, 125, 3444, 18571, -241872, -5796711, -24387220, 1132278191, 25132445832, 8850583573, -10681029498972, -214099676807085, 1643397436986464, 176719161389104817, 2976468247699317468, -71662294521163070153, -4638920054290748840520, -55645074852328083377619
Offset: 0
-
a[n_] := Sum[(-1)^(n - k) * Binomial[n, k] * StirlingS2[n, k] * k!, {k, 0, n}]; Array[a, 22, 0] (* Amiram Eldar, May 10 2021 *)
-
a(n) = sum(k=0, n, (-1)^(n-k)*binomial(n, k)*stirling(n, k, 2)*k!); \\ Michel Marcus, May 10 2021
A344397
a(n) = Stirling2(n, floor(n/2)) * floor(n/2)!.
Original entry on oeis.org
1, 0, 1, 1, 14, 30, 540, 1806, 40824, 186480, 5103000, 29607600, 953029440, 6711344640, 248619571200, 2060056318320, 86355926616960, 823172919528960, 38528927611574400, 415357755774998400, 21473732319740064000, 258323865658578720000, 14620825330739032204800
Offset: 0
-
a := n -> add((-1)^k*binomial((2*n-1)/4 + (-1)^n/4,k)*((2*n-1)/4 + (-1)^n/4 - k)^n, k = 0..n/2):
# Alternative, via Fubini recurrence:
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 -> coeff(F(n), x, iquo(n, 2));
seq(a(n), n = 0..22);
-
a[n_] := StirlingS2[n, Floor[n/2]] * Floor[n/2]!; Array[a, 23, 0] (* Amiram Eldar, May 22 2021 *)
-
def a(n): return stirling_number2(n, n//2) * factorial(n//2)
print([a(n) for n in range(23)])
A348576
Triangle read by rows: T(n,k) is the number of ordered partitions of [n] into k nonempty subsets, in which the first subset has size at least 2, n >= 1 and 1 <= k <= n.
Original entry on oeis.org
0, 1, 0, 1, 3, 0, 1, 10, 12, 0, 1, 25, 80, 60, 0, 1, 56, 360, 660, 360, 0, 1, 119, 1372, 4620, 5880, 2520, 0, 1, 246, 4788, 26376, 58800, 57120, 20160, 0, 1, 501, 15864, 134316, 466704, 771120, 604800, 181440, 0, 1, 1012, 50880, 637020, 3238200, 8094240, 10584000, 6955200, 1814400, 0
Offset: 1
For n=3, the ordered partitions of {1,2,3} in which the first block has size at least 2 are 123, 12/3, 13/2 and 23/1, so T(3,1)=1, T(3,2)=3 and T(3,3)=0.
Triangle begins:
0;
1, 0;
1, 3, 0;
1, 10, 12, 0;
1, 25, 80, 60, 0;
1, 56, 360, 660, 360, 0;
1, 119, 1372. 4620, 5880, 2520, 0;
1, 246, 4788, 26376, 58800, 57120, 20160, 0;
1, 501, 15864, 134316, 466704, 771120, 604800, 181440, 0;
1, 1012, 50880, 637020, 3238200, 8094240, 10584000, 6955200, 1814400, 0;
...
-
b:= proc(n, t) option remember; expand(`if`(n=0, 1,
add(x*b(n-j, 1)*binomial(n, j), j=t..n)))
end:
T:= n-> (p-> seq(coeff(p, x, i), i=1..n))(b(n, 2)):
seq(T(n), n=1..10); # Alois P. Heinz, Oct 24 2021
-
eulerian[n_,m_] := eulerian[n,m] =
Sum[((-1)^k)*Binomial[n+1,k]*((m+1-k)^n), {k,0,m+1}] (* eulerian[n, m] is an Eulerian number, counting the permutations of [n] with m descents *);
op2[n_,k_] := op2[n,k] =
Sum[(n-j)*eulerian[n-1,j-1]*Binomial[j-1,n-k-1], {j,1,n-1}] (* op2[n,k] counts ordered partitions on [n] with k parts, with first part having size at least 2 *); Table[op2[n, k],{n,1,12},{k,1,n}]
-
TE(n, k) = sum(j=0, k, (-1)^j * (k-j)^n * binomial( n+1, j)); \\ A008292
T(n,k) = sum(j=1, n-1, (n-j)*TE(n-1,j)*binomial(j-1,n-k-1)); \\ Michel Marcus, Oct 24 2021
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