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

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# Interesting finding

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