I had a little insight, and I'm curious whether or not it's true.(adsbygoogle = window.adsbygoogle || []).push({});

What I conjecture:

Say we have a set of functions S = {f(a, n) = b_n * a^n + b_n-1 * a^n-1 + ... + b_0} where a and n must be both natural numbers, and b must belong to {-1, 0, 1}. Then for fixed a = a_0 and n = n_0, we have that for every integer m from 0 to s = 1 + a_0 + a_0 ^2 + .... + a_0 ^n_0, there are values of b satisfying f(a_0, n_0) = m. For example, choosing a = 2 and n = 2, we have the integers 0, 1, 2, 3, 4, 5, 6, 7 which can be written as

0 = 0*1 + 0*2 + 0*4

1 = 1*1 + 0*2 + 0*4

2 = 0*1 + 1*2 + 0*4

3 = 1*1 + 1*2 + 0*4

4 = 0 *1 + 0*2 + 1*4

5 = 1*1 + 0*2 + 1*4

6 = 0*1 + 1*2 + 1*4

7 = 1*1 + 1*2 + 1*4

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Interesting finding

**Physics Forums | Science Articles, Homework Help, Discussion**