I meant the second of Marc's interpretations, not the first. (I should've put scare-quotes around the phrase in the subject line. Then I wouldn't have wasted Simon's time.) My (non-mainstream) take on base-b representation is that it's unduly restrictive to limit oneself to digits less than b and to limit oneself to the greedy algorithm for constructing those digits. I permit any nonnegative integer to be a "digit". But I neglected to mention this, so I can see how my question caused confusion. Note that 4/5 = (2/3)^2 + (2/3)^4 + (2/3)^6 + ..., which one could write as 0.010101... in "permissive base three-halves". My question is whether there is a similar representation of 1/3. But I am also interested in the general questions (a) what rational numbers can be written in the form \sum_{-\infty < k \leq n} a_k (3/2)^k (with nonnegative integer coefficients) using only finitely-many nonzero digits (1/3 can't be), and (b) what rational numbers can be written in this form using eventually periodic digits. Jim Propp On Sunday, September 25, 2016, Marc LeBrun <mlb@well.com <javascript:_e(%7B%7D,'cvml','mlb@well.com');>> wrote:
Jim, your Subject line asks about “base 3/2” but the body asks for a function F(x) = P(x)/Q(x), with P & Q polynomials having non-negative integer coefficients, such that F(2/3) = 1/3. Which?
On Sep 24, 2016, at 8:39 PM, James Propp <jamespropp@gmail.com> wrote:
Is it possible to write 1/3 as an infinite sum of the form a_1 (2/3)^1 + a_2 (2/3)^2 + ... where the sequence of nonnegative integer coefficients a_1, a_2, ... is eventually periodic?
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