The number of ways two "digits" can be multiplied to get (0 mod k) in base k. This number can never be less than k. In base 4, 2*2 = (0 mod 4), so there are 5 ways to get 0 in base 4.
1, 2, 3, 5, 5, 8, 7, 11, 12, 14, 11, 21, 13, 20, 23, 26, 17, 33, 19, 37, 33, 32, 23, 51, 35
1 seqfan posts
Mon Sep 6 05:40:16 CEST 2010 [seqfan] Re: Number of solutions of a*b + c*d + ... + y*z = 0 (mod n)