A060540
Square array read by antidiagonals downwards: T(n,k) = (n*k)!/(k!^n*n!), (n>=1, k>=1), the number of ways of dividing nk labeled items into n unlabeled boxes with k items in each box.
Original entry on oeis.org
1, 1, 1, 1, 3, 1, 1, 10, 15, 1, 1, 35, 280, 105, 1, 1, 126, 5775, 15400, 945, 1, 1, 462, 126126, 2627625, 1401400, 10395, 1, 1, 1716, 2858856, 488864376, 2546168625, 190590400, 135135, 1, 1, 6435, 66512160, 96197645544, 5194672859376, 4509264634875, 36212176000, 2027025, 1
Offset: 1
Array begins:
1, 1, 1, 1, 1, 1, ...
1, 3, 10, 35, 126, 462, ...
1, 15, 280, 5775, 126126, 2858856, ...
1, 105, 15400, 2627625, 488864376, 96197645544, ...
1, 945, 1401400, 2546168625, 5194672859376, 11423951396577720, ...
...
- Seiichi Manyama, Antidiagonals n = 1..50, flattened (first 20 antidiagonals from Harry J. Smith)
- Tom Copeland, Calculus, Combinatorics, and Geometry Underlying OEIS A060540, and the Exponential Formula, 2021.
- Nattawut Phetmak and Jittat Fakcharoenphol, Uniformly Generating Derangements with Fixed Number of Cycles in Polynomial Time, Thai J. Math. (2023) Vol. 21, No. 4, 899-915. See pp. 901, 914.
- Elena L. Wang and Guoce Xin, On Ward Numbers and Increasing Schröder Trees, arXiv:2507.15654 [math.CO], 2025. See p. 13.
Cf.
A000217,
A000292,
A000332,
A000389,
A000579,
A000580,
A007318,
A036040,
A099174,
A133314,
A132440,
A135278 (associations in Copeland link).
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T[n_, k_] := (n*k)!/(k!^n*n!);
Table[T[n-k+1, k], {n, 1, 10}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Jun 29 2018 *)
-
{ i=0; for (m=1, 20, for (n=1, m, k=m - n + 1; write("b060540.txt", i++, " ", (n*k)!/(k!^n*n!))); ) } \\ Harry J. Smith, Jul 06 2009
A361948
Array read by ascending antidiagonals. A(n, k) = Product_{j=0..k-1} binomial((j + 1)*n - 1, n - 1) if n >= 1, and A(0, k) = 1 for all k.
Original entry on oeis.org
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 10, 15, 1, 1, 1, 1, 35, 280, 105, 1, 1, 1, 1, 126, 5775, 15400, 945, 1, 1, 1, 1, 462, 126126, 2627625, 1401400, 10395, 1, 1, 1, 1, 1716, 2858856, 488864376, 2546168625, 190590400, 135135, 1, 1
Offset: 0
Array A(n, k) starts:
[0] 1, 1, 1, 1, 1, 1, ...
[1] 1, 1, 1, 1, 1, 1, ...
[2] 1, 1, 3, 15, 105, 945, ... A001147
[3] 1, 1, 10, 280, 15400, 1401400, ... A025035
[4] 1, 1, 35, 5775, 2627625, 2546168625, ... A025036
[5] 1, 1, 126, 126126, 488864376, 5194672859376, ... A025037
[6] 1, 1, 462, 2858856, 96197645544, 11423951396577720, ... A025038
.
Triangle A(n-k, k) starts:
[0] 1;
[1] 1, 1;
[2] 1, 1, 1;
[3] 1, 1, 1, 1;
[4] 1, 1, 3, 1, 1;
[5] 1, 1, 10, 15, 1, 1;
[6] 1, 1, 35, 280, 105, 1, 1;
- Cyril Banderier, Philippe Marchal, and Michael Wallner, Rectangular Young tableaux with local decreases and the density method for uniform random generation (short version), arXiv:1805.09017 [cs.DM], 2018.
- Antoine Chambert-Loir, Combinatorics of partitions, blog post 2024.
- Alexander Karpov, Generalized knockout tournament seedings, International Journal of Computer Science in Sport, vol. 17(2), 2018.
-
A := (n, k) -> mul(binomial((j + 1)*n - 1, n - 1), j = 0..k-1):
seq(seq(A(n-k, k), k = 0..n), n = 0..9);
# Alternative, using recursion:
A := proc(n, k) local P; P := proc(n, k) option remember;
if n = 0 then return x^k*k! fi; if k = 0 then 1 else add(binomial(n*k, n*j)*
P(n,k-j)*x, j=1..k) fi end: coeff(P(n, k), x, k) / k! end:
seq(print(seq(A(n, k), k = 0..5)), n = 0..6);
# Alternative, using exponential generating function:
egf := n -> ifelse(n=0, 1, exp(x^n/n!)): ser := n -> series(egf(n), x, 8*n):
row := n -> local k; seq((n*k)!*coeff(ser(n), x, n*k), k = 0..6):
for n from 0 to 6 do [n], row(n) od; # Peter Luschny, Aug 15 2024
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A[n_, k_] := Product[Binomial[n (j + 1) - 1, n - 1], {j, 0, k - 1}]; Table[A[n - k, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Michael De Vlieger, Apr 13 2023 *)
-
def Arow(n, size):
if n == 0: return [1] * size
return [prod(binomial((j + 1)*n - 1, n - 1) for j in range(k)) for k in range(size)]
for n in range(7): print(Arow(n, 7))
# Alternative, using exponential generating function:
def SetPolyLeadCoeff(m, n):
x, z = var("x, z")
if m == 0: return 1
w = exp(2 * pi * I / m)
o = sum(exp(z * w ** k) for k in range(m)) / m
t = exp(x * (o - 1)).taylor(z, 0, m*n)
p = factorial(m*n) * t.coefficient(z, m*n)
return p.leading_coefficient(x)
for m in range(7):
print([SetPolyLeadCoeff(m, k) for k in range(6)])
Showing 1-2 of 2 results.
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