A332852 -id:A332852

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A322384

Number T(n,k) of entries in the k-th cycles of all permutations of [n] when cycles are ordered by decreasing lengths (and increasing smallest elements); triangle T(n,k), n>=1, 1<=k<=n, read by rows.

+10
15

1, 3, 1, 13, 4, 1, 67, 21, 7, 1, 411, 131, 46, 11, 1, 2911, 950, 341, 101, 16, 1, 23563, 7694, 2871, 932, 197, 22, 1, 213543, 70343, 26797, 9185, 2311, 351, 29, 1, 2149927, 709015, 275353, 98317, 27568, 5119, 583, 37, 1, 23759791, 7867174, 3090544, 1141614, 343909, 73639, 10366, 916, 46, 1

(list; table; graph; refs; listen; history; text; internal format)

OFFSET

1,2

LINKS

Alois P. Heinz, Rows n = 1..141, flattened

Andrew V. Sills, Integer Partitions Probability Distributions, arXiv:1912.05306 [math.CO], 2019.

Wikipedia, Permutation

EXAMPLE

The 6 permutations of {1,2,3} are:

(1) (2) (3)

(1,2) (3)

(1,3) (2)

(2,3) (1)

(1,2,3)

(1,3,2)

so there are 13 elements in the first cycles, 4 in the second cycles and only 1 in the third cycles.

Triangle T(n,k) begins:

3, 1;

13, 4, 1;

67, 21, 7, 1;

411, 131, 46, 11, 1;

2911, 950, 341, 101, 16, 1;

23563, 7694, 2871, 932, 197, 22, 1;

213543, 70343, 26797, 9185, 2311, 351, 29, 1;

...

MAPLE

b:= proc(n, l) option remember; `if`(n=0, add(l[-i]*

x^i, i=1..nops(l)), add(binomial(n-1, j-1)*

b(n-j, sort([l[], j]))*(j-1)!, j=1..n))

end:

T:= n-> (p-> (seq(coeff(p, x, i), i=1..n)))(b(n, [])):

seq(T(n), n=1..12);

MATHEMATICA

b[n_, l_] := b[n, l] = If[n == 0, Sum[l[[-i]]*x^i, {i, 1, Length[l]}], Sum[Binomial[n-1, j-1]*b[n-j, Sort[Append[l, j]]]*(j-1)!, {j, 1, n}]];

T[n_] := CoefficientList[b[n, {}], x] // Rest;

Array[T, 12] // Flatten (* Jean-François Alcover, Feb 26 2020, after Alois P. Heinz *)

CROSSREFS

Columns k=1-10 give: A028418, A332851, A332852, A332853, A332854, A332855, A332856, A332857, A332858, A332859.

Row sums give A001563.

T(2n,n) gives A332928.

Cf. A185105, A322383.

KEYWORD

nonn,tabl

AUTHOR

Alois P. Heinz, Dec 05 2018

STATUS

approved

A350015

Irregular triangle read by rows: T(n,k) is the number of n-permutations whose third-longest cycle has length exactly k; n >= 0, 0 <= k <= floor(n/3).

+10
7

1, 1, 2, 5, 1, 17, 7, 74, 46, 394, 311, 15, 2484, 2241, 315, 18108, 17627, 4585, 149904, 152839, 57897, 2240, 1389456, 1460944, 705600, 72800, 14257440, 15326180, 8673060, 1660120, 160460640, 175421214, 110271546, 31600800, 1247400, 1965444480, 2177730270, 1469308698, 559402272, 55135080

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,3

COMMENTS

If the permutation has no third cycle, then its third-longest cycle is defined to have length 0.

LINKS

Alois P. Heinz, Rows n = 0..140, flattened

Steven Finch, Second best, Third worst, Fourth in line, arxiv:2202.07621 [math.CO], 2022.

FORMULA

Sum_{k=0..floor(n/3)} k * T(n,k) = A332852(n) for n >= 3. - Alois P. Heinz, Dec 12 2021

EXAMPLE

Triangle begins:

[0] 1;

[1] 1;

[2] 2;

[3] 5, 1;

[4] 17, 7;

[5] 74, 46;

[6] 394, 311, 15;

[7] 2484, 2241, 315;

[8] 18108, 17627, 4585;

[9] 149904, 152839, 57897, 2240;

...

MAPLE

b:= proc(n, l) option remember; `if`(n=0, x^l[1], add((j-1)!*

b(n-j, sort([l[], j])[2..4])*binomial(n-1, j-1), j=1..n))

end:

T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n, [0$3])):

seq(lprint(T(n)), n=0..14); # Alois P. Heinz, Dec 11 2021

MATHEMATICA

b[n_, l_] := b[n, l] = If[n == 0, x^l[[1]], Sum[(j - 1)!*b[n - j, Sort[Append[l, j]][[2 ;; 4]]]*Binomial[n - 1, j - 1], {j, 1, n}]];

T[n_] := With[{p = b[n, {0, 0, 0}]}, Table[Coefficient[p, x, i], {i, 0, Exponent[p, x]}]];

Table[T[n], {n, 0, 14}] // Flatten (* Jean-François Alcover, Dec 28 2021, after Alois P. Heinz *)

CROSSREFS

Column 0 gives 1 together with A000774.

Row sums give A000142.

Cf. A126074, A145877, A332852, A349979, A349980, A350016, A350273, A350274.

KEYWORD

nonn,tabf

AUTHOR

Steven Finch, Dec 08 2021

STATUS

approved

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