A270701
Total sum T(n,k) of the sizes of all blocks with maximal element k in all set partitions of {1,2,...,n}; triangle T(n,k), n>=1, 1<=k<=n, read by rows.
Original entry on oeis.org
1, 1, 3, 2, 4, 9, 5, 9, 16, 30, 15, 25, 41, 67, 112, 52, 82, 127, 195, 299, 463, 203, 307, 456, 670, 979, 1429, 2095, 877, 1283, 1845, 2623, 3702, 5204, 7307, 10279, 4140, 5894, 8257, 11437, 15717, 21485, 29278, 39848, 54267, 21147, 29427, 40338, 54692, 73561, 98367, 131007, 174029, 230884, 306298
Offset: 1
Row n=3 is [2, 4, 9] = [0+0+0+1+1, 0+2+1+0+1, 3+1+2+2+1] because the set partitions of {1,2,3} are: 123, 12|3, 13|2, 1|23, 1|2|3.
Triangle T(n,k) begins:
: 1;
: 1, 3;
: 2, 4, 9;
: 5, 9, 16, 30;
: 15, 25, 41, 67, 112;
: 52, 82, 127, 195, 299, 463;
: 203, 307, 456, 670, 979, 1429, 2095;
: 877, 1283, 1845, 2623, 3702, 5204, 7307, 10279;
: 4140, 5894, 8257, 11437, 15717, 21485, 29278, 39848, 54267;
-
b:= proc(n, m, t) option remember; `if`(n=0, [1, 0], add(
`if`(t=1 and j<>m+1, 0, (p->p+`if`(j=-t or t=1 and j=m+1,
[0, p[1]], 0))(b(n-1, max(m, j), `if`(t=1 and j=m+1, -j,
`if`(t<0, t, `if`(t>0, t-1, 0)))))), j=1..m+1))
end:
T:= (n, k)-> b(n, 0, max(0, 1+n-k))[2]:
seq(seq(T(n, k), k=1..n), n=1..12);
-
b[n_, m_, t_] := b[n, m, t] = If[n == 0, {1, 0}, Sum[If[t == 1 && j != m+1, 0, Function[p, p + If[j == -t || t == 1 && j == m+1, {0, p[[1]]}, 0]][b[ n-1, Max[m, j], If[t == 1 && j == m+1, -j, If[t < 0, t, If[t > 0, t-1, 0] ]]]]], {j, 1, m+1}]];
T[n_, k_] := b[n, 0, Max[0, 1+n-k]][[2]];
Table[Table[T[n, k], {k, 1, n}], {n, 1, 12}] // Flatten (* Jean-François Alcover, Apr 24 2016, translated from Maple *)
A270702
Total sum T(n,k) of the sizes of all blocks with minimal element k in all set partitions of {1,2,...,n}; triangle T(n,k), n>=1, 1<=k<=n, read by rows.
Original entry on oeis.org
1, 3, 1, 9, 4, 2, 30, 16, 9, 5, 112, 67, 41, 25, 15, 463, 299, 195, 127, 82, 52, 2095, 1429, 979, 670, 456, 307, 203, 10279, 7307, 5204, 3702, 2623, 1845, 1283, 877, 54267, 39848, 29278, 21485, 15717, 11437, 8257, 5894, 4140, 306298, 230884, 174029, 131007, 98367, 73561, 54692, 40338, 29427, 21147
Offset: 1
Row n=3 is [9, 4, 2] = [3+2+2+1+1, 0+0+1+2+1, 0+1+0+0+1] because the set partitions of {1,2,3} are: 123, 12|3, 13|2, 1|23, 1|2|3.
Triangle T(n,k) begins:
: 1;
: 3, 1;
: 9, 4, 2;
: 30, 16, 9, 5;
: 112, 67, 41, 25, 15;
: 463, 299, 195, 127, 82, 52;
: 2095, 1429, 979, 670, 456, 307, 203;
: 10279, 7307, 5204, 3702, 2623, 1845, 1283, 877;
: 54267, 39848, 29278, 21485, 15717, 11437, 8257, 5894, 4140;
-
b:= proc(n, m, t) option remember; `if`(n=0, [1, 0], add(
`if`(t=1 and j<>m+1, 0, (p->p+`if`(j=-t or t=1 and j=m+1,
[0, p[1]], 0))(b(n-1, max(m, j), `if`(t=1 and j=m+1, -j,
`if`(t<0, t, `if`(t>0, t-1, 0)))))), j=1..m+1))
end:
T:= (n, k)-> b(n, 0, k)[2]:
seq(seq(T(n, k), k=1..n), n=1..12);
-
b[n_, m_, t_] := b[n, m, t] = If[n == 0, {1, 0}, Sum[If[t == 1 && j != m + 1, 0, Function[p, p + If[j == -t || t == 1 && j == m + 1, {0, p[[1]]}, 0] ][b[n - 1, Max[m, j], If[t == 1 && j == m + 1, -j, If[t < 0, t, If[t > 0, t - 1, 0]]]]]], {j, 1, m + 1}]];
T[n_, k_] := b[n, 0, k][[2]];
Table[Table[T[n, k], {k, 1, n}], {n, 1, 12}] // Flatten (* Jean-François Alcover, Apr 24 2016, translated from Maple *)
A270236
Triangle T(n,p) read by rows: the number of occurrences of p in the restricted growth functions of length n.
Original entry on oeis.org
1, 3, 1, 9, 5, 1, 30, 21, 8, 1, 112, 88, 47, 12, 1, 463, 387, 253, 97, 17, 1, 2095, 1816, 1345, 675, 184, 23, 1, 10279, 9123, 7304, 4418, 1641, 324, 30, 1, 54267, 48971, 41193, 28396, 13276, 3645, 536, 38, 1, 306298, 279855, 243152, 183615, 102244, 36223, 7473, 842, 47, 1
Offset: 1
A283424
Number T(n,k) of blocks of size >= k in all set partitions of [n], assuming that every set partition contains one block of size zero; triangle T(n,k), n>=0, 0<=k<=n, read by rows.
Original entry on oeis.org
1, 2, 1, 5, 3, 1, 15, 10, 4, 1, 52, 37, 17, 5, 1, 203, 151, 76, 26, 6, 1, 877, 674, 362, 137, 37, 7, 1, 4140, 3263, 1842, 750, 225, 50, 8, 1, 21147, 17007, 9991, 4307, 1395, 345, 65, 9, 1, 115975, 94828, 57568, 25996, 8944, 2392, 502, 82, 10, 1
Offset: 0
T(3,2) = 4 because the number of blocks of size >= 2 in all set partitions of [3] (123, 12|3, 13|2, 1|23, 1|2|3) is 1+1+1+1+0 = 4.
Triangle T(n,k) begins:
1;
2, 1;
5, 3, 1;
15, 10, 4, 1;
52, 37, 17, 5, 1;
203, 151, 76, 26, 6, 1;
877, 674, 362, 137, 37, 7, 1;
4140, 3263, 1842, 750, 225, 50, 8, 1;
21147, 17007, 9991, 4307, 1395, 345, 65, 9, 1;
...
Columns k=0-10 give:
A000110(n+1),
A138378 or
A005493(n-1),
A124325,
A288785,
A288786,
A288787,
A288788,
A288789,
A288790,
A288791,
A288792.
-
T:= proc(n, k) option remember; `if`(k>n, 0,
binomial(n, k)*combinat[bell](n-k)+T(n, k+1))
end:
seq(seq(T(n, k), k=0..n), n=0..14);
-
T[n_, k_] := Sum[Binomial[n, j]*BellB[j], {j, 0, n - k}];
Table[T[n, k], {n, 0, 14}, {k, 0, n}] // Flatten (* Jean-François Alcover, Apr 30 2018 *)
A175716
The total number of elements(ordered pairs) in all equivalence relations on {1,2,...,n}.
Original entry on oeis.org
0, 1, 6, 27, 120, 560, 2778, 14665, 82232, 488403, 3062980, 20221520, 140134404, 1016698813, 7703878042, 60833235795, 499592325152, 4259301450652, 37634032670886, 344092369602461, 3250925202629100
Offset: 0
a(2) = 6 because the equivalence relations on {1,2}: {(1,1), (2,2)}, {(1,1), (2,2), (1,2), (2,1)} contain 6 ordered pairs.
-
f[list_] := Length[list]^2; Table[Total[Map[f, Level[SetParttions[n], {2}]]], {n, 0, 12}] (* or *)
Range[0,20]! CoefficientList[Series[(x + x^2)Exp[x] * Exp[Exp[x] - 1], {x, 0, 20}], x]
Showing 1-5 of 5 results.
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