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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).
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%I #30 Feb 17 2022 13:20:40

%S 1,1,2,5,1,17,7,74,46,394,311,15,2484,2241,315,18108,17627,4585,

%T 149904,152839,57897,2240,1389456,1460944,705600,72800,14257440,

%U 15326180,8673060,1660120,160460640,175421214,110271546,31600800,1247400,1965444480,2177730270,1469308698,559402272,55135080

%N 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).

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

%H Alois P. Heinz, <a href="/A350015/b350015.txt">Rows n = 0..140, flattened</a>

%H Steven Finch, <a href="http://arxiv.org/abs/2202.07621">Second best, Third worst, Fourth in line</a>, arxiv:2202.07621 [math.CO], 2022.

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

%e Triangle begins:

%e [0] 1;

%e [1] 1;

%e [2] 2;

%e [3] 5, 1;

%e [4] 17, 7;

%e [5] 74, 46;

%e [6] 394, 311, 15;

%e [7] 2484, 2241, 315;

%e [8] 18108, 17627, 4585;

%e [9] 149904, 152839, 57897, 2240;

%e ...

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

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

%p end:

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

%p seq(lprint(T(n)), n=0..14); # _Alois P. Heinz_, Dec 11 2021

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

%t Table[T[n], {n, 0, 14}] // Flatten (* _Jean-François Alcover_, Dec 28 2021, after _Alois P. Heinz_ *)

%Y Column 0 gives 1 together with A000774.

%Y Row sums give A000142.

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

%K nonn,tabf

%O 0,3

%A _Steven Finch_, Dec 08 2021