Author | Message | Time |
---|---|---|
Yoni | I'm looking for a polynomial P(x) with the following properties: 1. P(N) is partial to N. Meaning, if n is a natural number, P(n) is natural as well. 2. If n is divisible by 3 (n mod 3 = 0), then P(n) is even (P(n) mod 2 = 0). 3. If n isn't divisible by 3 (n mod 3 = 1, or n mod 3 = 2), then P(n) is odd (P(n) mod 2 = 1). So far I haven't been able to find one or prove the inexistence of one. Any ideas? | November 26, 2003, 10:29 PM |
Yoni | Not a polynomial, but I found a magnificent solution to my problem! A(n) = |sin(pi*n/3)|/(sqrt(3)/2) Gives the following sequence: 1, 1, 0, 1, 1, 0, 1, 1, 0, (repeats infinitely) | December 22, 2003, 12:26 AM |
Spht | Try A(n) = |sin(pi*n/3)|/(sqrt(3)/2). Edit - Nevermind, I see you've already found one. | December 22, 2003, 12:30 AM |
Yoni | [quote author=Spht link=board=36;threadid=3882;start=0#msg36664 date=1072053025] Try A(n) = |sin(pi*n/3)|/(sqrt(3)/2). Edit - Nevermind, I see you've already found one. [/quote] Nice try. | December 22, 2003, 1:03 AM |