OFFSET
1,6
COMMENTS
In the case of n = 1, the single bead is considered to be cyclically adjacent to itself giving T(1,1) = 0. If compatibility with A208535 is wanted then T(1,1) should be 1.
LINKS
Andrew Howroyd, Table of n, a(n) for n = 1..1275 (first 50 rows)
FORMULA
T(n,k) = Sum_{j=1..k} (-1)^(k-j)*binomial(k,j)*A208535(n,j) for n > 1.
T(n,n) = (n-1)! for n > 1.
EXAMPLE
Triangle begins:
0;
0, 1;
0, 0, 2;
0, 1, 3, 6;
0, 0, 6, 24, 24;
0, 1, 11, 80, 180, 120;
0, 0, 18, 240, 960, 1440, 720;
0, 1, 33, 696, 4410, 11340, 12600, 5040;
0, 0, 58, 1960, 18760, 73920, 137760, 120960, 40320;
...
PROG
(PARI) \\ here U(n, k) is A208535(n, k) for n > 1.
U(n, k)={sumdiv(n, d, eulerphi(n/d)*(k-1)^d)/n - if(n%2, k-1)}
T(n, k)={sum(j=1, k, (-1)^(k-j)*binomial(k, j)*U(n, j))}
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Andrew Howroyd, Dec 20 2019
STATUS
approved