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).

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

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.

Sequence in context: A352010 A350016 A162975 * A187244 A120294 A186766

Adjacent sequences: A350012 A350013 A350014 * A350016 A350017 A350018

Keyword

nonn,tabf

Author

Steven Finch, Dec 08 2021

Status

approved