A032011
Partition n labeled elements into sets of different sizes and order the sets.
Original entry on oeis.org
1, 1, 1, 7, 9, 31, 403, 757, 2873, 12607, 333051, 761377, 3699435, 16383121, 108710085, 4855474267, 13594184793, 76375572751, 388660153867, 2504206435681, 20148774553859, 1556349601444477, 5050276538344665, 33326552998257031, 186169293932977115, 1305062351972825281, 9600936552132048553, 106019265737746665727, 12708226588208611056333, 47376365554715905155127
Offset: 0
-
b:= proc(n, i, p) option remember;
`if`(i*(i+1)/2n, 0, b(n-i, i-1, p+1)*binomial(n,i))))
end:
a:= n-> b(n$2, 0):
seq(a(n), n=0..30); # Alois P. Heinz, Sep 02 2015
-
f[list_]:=Apply[Multinomial,list]*Length[list]!; Table[Total[Map[f, Select[IntegerPartitions[n], Sort[#] == Union[#] &]]], {n, 1, 30}]
b[n_, i_, p_] := b[n, i, p] = If[i*(i+1)/2n, 0, b[n-i, i-1, p+1]*Binomial[n, i]]]]; a[n_] := b[n, n, 0]; Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Nov 16 2015, after Alois P. Heinz *)
-
seq(n)=[subst(serlaplace(y^0*p),y,1) | p <- Vec(serlaplace(prod(k=1, n, 1 + x^k*y/k! + O(x*x^n))))] \\ Andrew Howroyd, Sep 13 2018
A285824
Number T(n,k) of ordered set partitions of [n] into k blocks such that equal-sized blocks are ordered with increasing least elements; triangle T(n,k), n>=0, 0<=k<=n, read by rows.
Original entry on oeis.org
1, 0, 1, 0, 1, 1, 0, 1, 6, 1, 0, 1, 11, 18, 1, 0, 1, 30, 75, 40, 1, 0, 1, 52, 420, 350, 75, 1, 0, 1, 126, 1218, 3080, 1225, 126, 1, 0, 1, 219, 4242, 17129, 15750, 3486, 196, 1, 0, 1, 510, 14563, 82488, 152355, 63756, 8526, 288, 1, 0, 1, 896, 42930, 464650, 1049895, 954387, 217560, 18600, 405, 1
Offset: 0
T(3,1) = 1: 123.
T(3,2) = 6: 1|23, 23|1, 2|13, 13|2, 3|12, 12|3.
T(3,3) = 1: 1|2|3.
Triangle T(n,k) begins:
1;
0, 1;
0, 1, 1;
0, 1, 6, 1;
0, 1, 11, 18, 1;
0, 1, 30, 75, 40, 1;
0, 1, 52, 420, 350, 75, 1;
0, 1, 126, 1218, 3080, 1225, 126, 1;
0, 1, 219, 4242, 17129, 15750, 3486, 196, 1;
...
Columns k=0-10 give:
A000007,
A057427,
A285917,
A285918,
A285919,
A285920,
A285921,
A285922,
A285923,
A285924,
A285925.
-
b:= proc(n, i, p) option remember; expand(`if`(n=0 or i=1,
(p+n)!/n!*x^n, add(b(n-i*j, i-1, p+j)*x^j*combinat
[multinomial](n, n-i*j, i$j)/j!^2, j=0..n/i)))
end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b(n$2, 0)):
seq(T(n), n=0..12);
-
multinomial[n_, k_List] := n!/Times @@ (k!);
b[n_, i_, p_] := b[n, i, p] = Expand[If[n == 0 || i == 1, (p + n)!/n!*x^n, Sum[b[n-i*j, i-1, p+j]*x^j*multinomial[n, Join[{n-i*j}, Table[i, j]]]/ j!^2, {j, 0, n/i}]]];
T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 0, n}]][b[n, n, 0]];
Table[T[n], {n, 0, 12}] // Flatten (* Jean-François Alcover, Apr 28 2018, after Alois P. Heinz *)
A327244
Number T(n,k) of colored compositions of n using all colors of a k-set such that all parts have different color patterns and the patterns for parts i are sorted and have i colors in (weakly) increasing order; triangle T(n,k), n>=0, 0<=k<=n, read by rows.
Original entry on oeis.org
1, 0, 1, 0, 1, 2, 0, 3, 10, 8, 0, 3, 27, 54, 31, 0, 5, 70, 255, 336, 147, 0, 11, 223, 1222, 2692, 2580, 899, 0, 13, 508, 4467, 15512, 25330, 19566, 5777, 0, 19, 1193, 15540, 78819, 194075, 248976, 160377, 41024, 0, 27, 2822, 52981, 375440, 1303250, 2463534, 2593339, 1430288, 322488
Offset: 0
T(3,1) = 3: 3aaa, 2aa1a, 1a2aa.
T(3,2) = 10: 3aab, 3abb, 2aa1b, 2ab1a, 2ab1b, 2bb1a, 1a2ab, 1a2bb, 1b2aa, 1b2ab.
T(3,3) = 8: 3abc, 2ab1c, 2ac1b, 2bc1a, 1a2bc, 1b2ac, 1c2ab, 1a1b1c.
Triangle T(n,k) begins:
1;
0, 1;
0, 1, 2;
0, 3, 10, 8;
0, 3, 27, 54, 31;
0, 5, 70, 255, 336, 147;
0, 11, 223, 1222, 2692, 2580, 899;
0, 13, 508, 4467, 15512, 25330, 19566, 5777;
0, 19, 1193, 15540, 78819, 194075, 248976, 160377, 41024;
...
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C:= binomial:
b:= proc(n, i, k, p) option remember; `if`(n=0, p!, `if`(i<1, 0, add(
b(n-i*j, min(n-i*j, i-1), k, p+j)/j!*C(C(k+i-1,i),j), j=0..n/i)))
end:
T:= (n, k)-> add(b(n$2, i, 0)*(-1)^(k-i)*C(k, i), i=0..k):
seq(seq(T(n, k), k=0..n), n=0..10);
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c = Binomial;
b[n_, i_, k_, p_] := b[n, i, k, p] = If[n == 0, p!, If[i < 1, 0, Sum[b[n - i*j, Min[n - i*j, i-1], k, p+j]/j!*c[c[k + i - 1, i], j], {j, 0, n/i}]]];
T[n_, k_] := Sum[b[n, n, i, 0]*(-1)^(k-i)*c[k, i], {i, 0, k}];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Apr 28 2020, after Alois P. Heinz *)
A196301
The number of ways to linearly order the cycles in each permutation of {1,2,...,n} where two cycles are considered identical if they have the same length.
Original entry on oeis.org
1, 1, 2, 9, 44, 270, 2139, 18837, 186808, 2070828, 25861140, 350000640, 5145279611, 81492295079, 1381583542234, 25097285838765, 484602684624080, 9894705390149400, 213418984780492164, 4842425874827849868, 115231446547162291200, 2874808892527026177240
Offset: 0
a(4) = 44 because in the conjugacy classes of S(4): (4), (3)(1), (2)(2), (2)(1)(1), (1)(1)(1)(1) there are (respectively) 6 permutations times 1 arrangement, 8 permutations times 2 arrangements, 3 permutations times 1 arrangement, 6 permutations times 3 arrangements, and 1 permutation times 1 arrangement. So 6*1+8*2+3*1+6*3+1*1 = 44.
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b:= proc(n, i, p) option remember; `if`(n=0 or i=1,
(p+n)!/n!, add(b(n-i*j, i-1, p+j)*(i-1)!^j*combinat
[multinomial](n, n-i*j, i$j)/j!^2, j=0..n/i))
end:
a:= n-> b(n$2, 0):
seq(a(n), n=0..25); # Alois P. Heinz, Apr 27 2017
-
Needs["Combinatorica`"]; f[{x_, y_}]:= x^y y!; Table[Total[Table[n!, {PartitionsP[n]}]/Apply[Times, Map[f, Map[Tally, Partitions[n]], {2}], 2] * Apply[Multinomial, Map[Last, Map[Tally, Partitions[n]], {2}], 2]], {n, 0, 20}]
A179233
Irregular triangle T(n,k) = A049019(n,k)/A096162(n,k) read along rows, 1<=k <= A000041(n).
Original entry on oeis.org
1, 1, 1, 1, 6, 1, 1, 8, 3, 18, 1, 1, 10, 20, 30, 45, 40, 1, 1, 12, 30, 10, 45, 360, 15, 80, 270, 75, 1, 1, 14, 42, 70, 63, 630, 210, 315, 140, 2520, 420, 175, 1050, 126, 1, 1, 16, 56, 112, 35, 84, 1008, 1680, 630, 840, 224, 5040, 1680
Offset: 1
A049019(.,.) begins 1; 1; 2, 1; 6, 6, 1; 8, 6, 36, 24, ...
A096162(.,.) begins 1; 1; 2, 1; 1, 6, 1; 1, 2, 2, 24 ...
so
T(.,.) begins ..... 1; 1; 1, 1; 6, 1, 1; 8, 3, 18, 1 ...
Extended, and bivariate indices restored -
R. J. Mathar, Jul 13 2010
Showing 1-5 of 5 results.
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