You know how the factorial function has the recurrence
(n+1)! = (n+1) (n!)
and the 'subfactorial' function ---- which counts the derangements of length n --- has the recurrence
!(n+1) = (n+1) (!n) + (-1)^(n+1)
right?
Well, what happens if I put some other root of unity in for that -1? Like, let me say
f(n+1) = (n+1) (f(n)) + (i)^(n+1)
and for the hell of it set f(1) = 1. Then what do I get?
n f(n) 1 1 2 1 3 3-i 4 13-4i 5 65-19i 6 389-114i 7 2723-799i 8 21785-6392i 9 196065-57527i
Aaaand the imaginary part of f(n) is apparently A186359, the "number of permutations of {1,2,...,n} having no up-down cycles". Can't find anything interestingly combinatorial for its real part, but it's (up to a sign flip) the egf of cos(x)/(1+x). (A009102).