2 ms·
Seems to work fine for -1: f(-1, -1) = sum_{n=0}^b 2^n (floor(-1/2^n) mod 2) (floor(-1/2^n) mod 2) = sum_{n=0}^b 2^n * 1 * 1 = 1 + 10 + 100 + 1
by Strilanc 10y ago
Seems to work fine for -1:
f(-1, -1)
= sum_{n=0}^b 2^n (floor(-1/2^n) mod 2) (floor(-1/2^n) mod 2)
= sum_{n=0}^b 2^n * 1 * 1
= 1 + 10 + 100 + 1000 + ... [in binary]
= ...11111 [in the 2-adics]
= -1