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cousin_it comments on A Proof of Occam's Razor - Less Wrong
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Such a g(m) cannot exist. Here's a proof:
Assume that g(m) exists. Then there exists an m0 such that g(m0)=f(1)=1. Take any m1>m0. Then g(m1)>1 (because g is strictly increasing). But there also exists an n1 such that g(m1)=f(n1). This cannot be, because f(n) is always less than or equal to 1 when n is a positive integer. Contradiction, QED.
If you have a proof that g(m) exists (computable or uncomputable, no matter), we have a difficult situation on our hands. Could you exhibit your proof now?