Mark Wildon has authored 6 sequences.
A261318
Number of set partitions T'_t(n) of {1,2,...,t} into exactly n parts, with an even number of elements in each part distinguished by marks, and such that no part contains both 1 and t with 1 unmarked or both i and i+1 with i+1 unmarked for some i with 1 <= i < t; triangle T'_t(n), t>=0, 0<=n<=t, read by rows.
Original entry on oeis.org
1, 0, 0, 0, 1, 1, 0, 0, 3, 1, 0, 1, 10, 8, 1, 0, 0, 30, 50, 15, 1, 0, 1, 91, 280, 155, 24, 1, 0, 0, 273, 1491, 1365, 371, 35, 1, 0, 1, 820, 7728, 11046, 4704, 756, 48, 1, 0, 0, 2460, 39460, 85050, 53382, 13020, 1380, 63, 1
Offset: 0
Triangle starts:
1;
0, 0;
0, 1, 1;
0, 0, 3, 1;
0, 1, 10, 8, 1;
0, 0, 30, 50, 15, 1;
0, 1, 91, 280, 155, 24, 1;
0, 0, 273, 1491, 1365, 371, 35, 1;
0, 1, 820, 7728, 11046, 4704, 756, 48, 1;
-
TGF[1, x_] := x^2/(1 - x^2); TGF[n_, x_] := x^n/(1 + x)*Product[1/(1 - (2*j - 1)*x), {j, 1, n}];
T[0, 0] := 1; T[, 0] := 0; T[0,]:=0; T[t_, n_] := Coefficient[Series[TGF[n, x], {x, 0, t}], x^t]
-
T(t, n) = {if ((t==0) && (n==0), return(1)); if (n==0, return(0)); if (n==1, return(1 - t%2)); 1/(2^n*n!)*(sum(k=0,n-1,binomial(n,k)*(-1)^k*(2*(n-k)-1)^t)+(-1)^(n+t));}
tabl(nn) = {for (t=0, nn, for (n=0, t, print1(T(t, n), ", ");); print(););} \\ Michel Marcus, Aug 17 2015
Corrected description in name to agree with section 4.1 in linked paper
Mark Wildon, Mar 11 2019
A261319
Number of set partitions C'_t(n) of {1,2,...,t} into at most n parts, with an even number of elements in each part distinguished by marks and such that no part contains both 1 and t (each unmarked) or both i and i+1 (each unmarked) for some i with 1 <= i < t; triangle C'_t(n), t>=0, 0<=n<=t, read by rows.
Original entry on oeis.org
1, 0, 0, 0, 1, 2, 0, 0, 3, 4, 0, 1, 11, 19, 20, 0, 0, 30, 80, 95, 96, 0, 1, 92, 372, 527, 551, 552, 0, 0, 273, 1764, 3129, 3500, 3535, 3536, 0, 1, 821, 8549, 19595, 24299, 25055, 25103, 25104
Offset: 0
Triangle starts:
1;
0, 0;
0, 1, 2;
0, 0, 3, 4;
0, 1, 11, 19, 20;
0, 0, 30, 80, 95, 96;
0, 1, 92, 372, 527, 551, 552;
0, 0, 273, 1764, 3129, 3500, 3535, 3536;
0, 1, 821, 8549, 19595, 24299, 25055, 25103, 25104;
-
TGF[1, x_] := x^2/(1 - x^2); TGF[n_, x_] := x^n/(1 + x)*Product[1/(1 - (2*j - 1)*x), {j, 1, n}];
T[0, 0] := 1; T[, 0] := 0; T[0, ] := 0; T[t_, n_] := Coefficient[Series[TGF[n, x], {x, 0, t}], x^t];
CC[t_, n_] := Sum[T[t, m], {m, 0, n}]
A261275
Number of set partitions C_t(n) of {1,2,...,t} into at most n parts, with an even number of elements in each part distinguished by marks; triangle C_t(n), t>=0, 0<=n<=t, read by rows.
Original entry on oeis.org
1, 0, 1, 0, 2, 3, 0, 4, 10, 11, 0, 8, 36, 48, 49, 0, 16, 136, 236, 256, 257, 0, 32, 528, 1248, 1508, 1538, 1539, 0, 64, 2080, 6896, 9696, 10256, 10298, 10299, 0, 128, 8256, 39168, 66384, 74784, 75848, 75904, 75905, 0, 256, 32896, 226496, 475136, 586352, 607520, 609368, 609440, 609441
Offset: 0
Triangle starts:
1;
0, 1;
0, 2, 3;
0, 4, 10, 11;
0, 8, 36, 48, 49;
0, 16, 136, 236, 256, 257;
0, 32, 528, 1248, 1508, 1538, 1539;
0, 64, 2080, 6896, 9696, 10256, 10298, 10299;
...
-
with(combinat):
b:= proc(n, i) option remember; expand(`if`(n=0, 1,
`if`(i<1, 0, add(x^j*multinomial(n, n-i*j, i$j)/j!*add(
binomial(i, 2*k), k=0..i/2)^j*b(n-i*j, i-1), j=0..n/i))))
end:
T:= n-> (p-> seq(add(coeff(p, x, j), j=0..i), i=0..n))(b(n$2)):
seq(T(n), n=0..12); # Alois P. Heinz, Aug 13 2015
-
CC[t_, n_] := Sum[2^(t - m)*StirlingS2[t, m], {m, 0, n}];
Table[CC[t, n], {t, 0, 12}, {n, 0, t}] // Flatten
(* Second program: *)
multinomial[n_, k_List] := n!/Times @@ (k!);
b[n_, i_] := b[n, i] = If[n == 0, 1, If[i < 1, 0, Sum[x^j*multinomial[n, Join[{n - i*j}, Table[i, j]]]/j!*Sum[Binomial[i, 2*k], {k, 0, i/2}]^j*b[n - i*j, i - 1], {j, 0, n/i}]]];
T[n_] := Function[p, Table[Sum[Coefficient[p, x, j], {j, 0, i}], {i, 0, n} ] ][b[n, n]];
Table[T[n], {n, 0, 12}] // Flatten (* Jean-François Alcover, Nov 07 2017, after Alois P. Heinz *)
A261137
Number of set partitions B'_t(n) of {1,2,...,t} into at most n parts, so that no part contains both 1 and t, or both i and i+1 with 1 <= i < t; triangle B'_t(n), t>=0, 0<=n<=t, read by rows.
Original entry on oeis.org
1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 3, 4, 0, 0, 0, 5, 10, 11, 0, 0, 1, 11, 31, 40, 41, 0, 0, 0, 21, 91, 147, 161, 162, 0, 0, 1, 43, 274, 568, 694, 714, 715, 0, 0, 0, 85, 820, 2227, 3151, 3397, 3424, 3425, 0, 0, 1, 171, 2461, 8824, 14851, 17251, 17686, 17721, 17722
Offset: 0
Triangle starts:
1;
0, 0;
0, 0, 1;
0, 0, 0, 1;
0, 0, 1, 3, 4;
0, 0, 0, 5, 10, 11;
0, 0, 1, 11, 31, 40, 41;
0, 0, 0, 21, 91, 147, 161, 162;
0, 0, 1, 43, 274, 568, 694, 714, 715;
0, 0, 0, 85, 820, 2227, 3151, 3397, 3424, 3425;
...
- Alois P. Heinz, Rows n = 0..140, flattened
- John R. Britnell and Mark Wildon, Bell numbers, partition moves and the eigenvalues of the random-to-top shuffle in Dynkin Types A, B and D, arXiv:1507.04803 [math.CO], 2015.
- D. E. Knuth and O. P. Lossers, Partitions of a circular set, Problem 11151 in Amer. Math. Monthly 114 (3), (2007), p 265, E_4.
-
g:= proc(t, l, h) option remember; `if`(t=0, `if`(l=1, 0, x^h),
add(`if`(j=l, 0, g(t-1, j, max(h,j))), j=1..h+1))
end:
B:= t-> (p-> seq(add(coeff(p, x, j), j=0..i), i=0..t))(g(t, 0$2)):
seq(B(t), t=0..12); # Alois P. Heinz, Aug 10 2015
-
StirPrimedGF[0, x_] := 1; StirPrimedGF[1, x_] := 0;
StirPrimedGF[n_, x_] := x^n/(1 + x)*Product[1/(1 - j*x), {j, 1, n - 1}];
StirPrimed[0, 0] := 1; StirPrimed[0, _] := 0;
StirPrimed[t_, n_] := Coefficient[Series[StirPrimedGF[n, x], {x, 0, t}], x^t];
BPrimed[t_, n_] := Sum[StirPrimed[t, m], {m, 0, n}]
(* Second program: *)
g[t_, l_, h_] := g[t, l, h] = If[t == 0, If[l == 1, 0, x^h], Sum[If[j == l, 0, g[t - 1, j, Max[h, j]]], {j, 1, h + 1}]];
B[t_] := Function[p, Table[Sum[Coefficient[p, x, j], {j, 0, i}], {i, 0, t}] ][g[t, 0, 0]];
Table[B[t], {t, 0, 12}] // Flatten (* Jean-François Alcover, May 20 2016, after Alois P. Heinz *)
A261139
S'_t(n) is the number of set partitions of {1,2,...,t} into exactly n parts such that no part contains both 1 and t or both i and i+1 for some i with 1 <= i < t; triangle S'_t(n), t >= 0, 0 <= n <= t, read by rows.
Original entry on oeis.org
1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 2, 1, 0, 0, 0, 5, 5, 1, 0, 0, 1, 10, 20, 9, 1, 0, 0, 0, 21, 70, 56, 14, 1, 0, 0, 1, 42, 231, 294, 126, 20, 1, 0, 0, 0, 85, 735, 1407, 924, 246, 27, 1, 0, 0, 1, 170, 2290, 6363, 6027, 2400, 435, 35, 1
Offset: 0
Triangle starts:
1;
0, 0;
0, 0, 1;
0, 0, 0, 1;
0, 0, 1, 2, 1;
0, 0, 0, 5, 5, 1;
0, 0, 1, 10, 20, 9, 1;
0, 0, 0, 21, 70, 56, 14, 1;
0, 0, 1, 42, 231, 294, 126, 20, 1;
0, 0, 0, 85, 735, 1407, 924, 246, 27, 1;
...
- Alois P. Heinz, Rows n = 0..140, flattened
- John R. Britnell and Mark Wildon, Bell numbers, partition moves and the eigenvalues of the random-to-top shuffle in Dynkin Types A, B and D, arXiv:1507.04803 [math.CO], 2015.
- D. E. Knuth and O. P. Lossers, Partitions of a circular set, Problem 11151 in Amer. Math. Monthly 114 (3), (2007), p 265, E_4.
- Sophie Morier-Genoud, Counting Coxeter's friezes over a finite field via moduli spaces, arXiv:1907.12790 [math.CO], 2019.
- Augustine O. Munagi, Two Applications of the Bijection on Fibonacci Set Partitions, Fibonacci Quart. 55 (2017), no. 5, 144-148. See c(n,k) p. 145 giving shifted triangle.
The same as
A105794, except for the first two columns.
-
g:= proc(t, l, h) option remember; `if`(t=0, `if`(l=1, 0, x^h),
add(`if`(j=l, 0, g(t-1, j, max(h,j))), j=1..h+1))
end:
S:= t-> (p-> seq(coeff(p, x, i), i=0..t))(g(t, 0$2)):
seq(S(t), t=0..12); # Alois P. Heinz, Aug 10 2015
-
StirPrimedGF[n_, x_] := x^n/(1 + x)*Product[1/(1 - j*x), {j, 1, n - 1}]; T[0, 0] = 1; T[, 0] = T[, 1] = 0; T[n_, k_] := SeriesCoefficient[ StirPrimedGF[k, x], {x, 0, n}]; Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* script completed by Jean-François Alcover, Jan 31 2016 *)
-
a(n,k)=if(k==0, n==0, sum(j=0, k, binomial(k, j) * (-1)^(k-j) * ((j-1)^n + (-1)^n * (j-1))) / k!);
for(n=0, 10, for(k=0, n, print1( a(n, k), ", "); ); print(); ); \\ Andrew Howroyd, Apr 08 2017
A192983
a(n) is the number of pairs (g, h) of elements of the symmetric group S_n such that g and h have conjugates that commute.
Original entry on oeis.org
1, 4, 24, 264, 5640, 151200, 5722920, 282868992, 18371308032, 1504791561600, 148978034686800, 18007146260231040, 2528615024682544512, 426310052282058252672, 81830910530970671616000, 18305445786667543107072000, 4570435510076312321728158720
Offset: 1
For n = 3 the probability that two elements of S_3 have conjugates that commute is a(3)/3!^2 = 2/3. Proof: only the transpositions and three cycles fail to have conjugates that commute; the probability of choosing one permutation from each of these classes is 2*1/2*1/3 = 1/3.
- Simon R. Blackburn, John R. Britnell, and Mark Wildon, The probability that a pair of elements of a finite group are conjugate, arXiv:1108.1784 [math.GR], 2011-2012.
- J. R. Britnell and M. Wildon, Commuting elements in conjugacy classes: an application of Hall's Marriage Theorem to group theory, J. Group Theory, 12 (2009), 795-802.
- Mark Wildon, Haskell source code for computing values of the sequence.
Cf.
A087132 (the sum of squares of the sizes of the conjugacy classes of S_n).
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