11:27 PM
@Steffan for ur next power of b answer, log_b(n) can be written as log_bn
also log_bn+1 -> log_bbn for another -1 byte
so, f(n,b)=b^{floor(log_bbn)}-n (27 bytes)
(log_bbn works because log_b(n) + 1 = log_b(n) + log_b(b) = log_b(bn))
also ive noticed that only two ppl contributed to lotm so far, i feel like no one else is interested :(
