OFFSET
0,3
COMMENTS
The cycle descent number of a permutation is computed by writing each cycle with its smallest element first, and then counting up the number of pairs (x,y) where x is the element before y in its cycle and x > y.
Conjecture: T(n,n-3) = 2^(n-1) for n >= 4.
Conjecture: T(n,0) = A000110(n).
LINKS
Alois P. Heinz, Rows n = 0..150, flattened
FORMULA
Conjecture: T(n,k) = [z^k] Sum_{i=0..n} Stirling2(n,i)*(1 - z)^(n - i) Product_{j=0..i-1} (j*z + 1). - Mikhail Kurkov, Jun 13 2023
From Alois P. Heinz, Jun 13 2023: (Start)
Sum_{k=0..max(0,n-2)} k * T(n,k) = A321853(n-1) for n>=2.
Sum_{k=0..max(0,n-2)} (-1)^k * T(n,k) = A011782(n). (End)
EXAMPLE
Table begins:
n\k | 0 1 2 3 4 5 6 7
----+---------------------------------------------------
0 | 1;
1 | 1;
2 | 2;
3 | 5, 1;
4 | 15, 8, 1;
5 | 52, 51, 16, 1;
6 | 203, 312, 172, 32, 1;
7 | 877, 1926, 1611, 561, 64, 1;
8 | 4140, 12224, 14289, 7744, 1794, 128, 1;
9 | 21147, 80401, 124410, 95255, 35755, 5655, 256, 1;
...
For example, the permutation (1)(2735)(4986) has a cycle descent number of 3 because 7 > 3, 9 > 8, and 8 > 6.
The T(9,7) = 1 permutation in S_9 with cycle descent number 7 is (198765432).
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Peter Kagey, Dec 30 2021
EXTENSIONS
T(0,0)=1 prepended by Alois P. Heinz, Jun 13 2023
STATUS
approved