[ponzi]

Are those really equivalent? What if ix == 15?

It would work with a 2,8-fizzbuzz, though.

[krallja]

Whoops, I wrote that far too late last night. 15&15 = 15, so of course it wouldn't work :)

A 2,8-fizzbuzz would only need 8 items in the circular array since the least common multiple of 2 and 8 is 8:

    ix = ix & 8;