3x+1 inverse orbit generating functions almost always have natural boundaries

Acta Arithmetica (2015)

• Volume: 170, Issue: 2, page 101-120
• ISSN: 0065-1036

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Abstract

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The 3x+k function ${T}_{k}\left(n\right)$ sends n to (3n+k)/2, resp. n/2, according as n is odd, resp. even, where k ≡ ±1 (mod 6). The map ${T}_{k}\left(·\right)$ sends integers to integers; for m ≥1 let n → m mean that m is in the forward orbit of n under iteration of ${T}_{k}\left(·\right)$. We consider the generating functions ${f}_{k,m}\left(z\right)={\sum }_{n>0,n\to m}{z}^{n}$, which are holomorphic in the unit disk. We give sufficient conditions on (k,m) for the functions ${f}_{k,m}\left(z\right)$ to have the unit circle |z|=1 as a natural boundary to analytic continuation. For the 3x+1 function these conditions hold for all m ≥1 to show that ${f}_{1,m}\left(z\right)$ has the unit circle as a natural boundary except possibly for m= 1, 2, 4 and 8. The 3x+1 Conjecture is equivalent to the assertion that ${f}_{1,m}\left(z\right)$ is a rational function of z for the remaining values m = 1,2,4,8.

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