English

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

Complex Variables 2015-10-27 v1 Number Theory

Abstract

The 3x+k3x+k function Tk(n)T_{k}(n) sends nn to (3n+k)/2(3n+k)/2 resp. n/2,n/2, according as nn is odd, resp. even, where k±1 (mod6)k \equiv \pm 1~(\bmod \, 6). The map Tk()T_k(\cdot) sends integers to integers, and for m1m \ge 1 let nmn \rightarrow m mean that mm is in the forward orbit of nn under iteration of Tk().T_k(\cdot). We consider the generating functions fk,m(z)=n>0,nmzn,f_{k,m}(z) = \sum_{n>0, n \rightarrow m} z^{n}, which are holomorphic in the unit disk. We give sufficient conditions on (k,m)(k,m) for the functions fk,m(z)f_{k, m}(z) have the unit circle {z=1}\{|z|=1\} as a natural boundary to analytic continuation. For the 3x+13x+1 function these conditions hold for all m1m \ge 1 to show that f1,m(z)f_{1,m}(z) has the unit circle as a natural boundary except possibly for m=1,2,4m= 1, 2, 4 and 88. The 3x+13x+1 Conjecture is equivalent to the assertion that f1,m(z)f_{1, m}(z) is a rational function of zz for the remaining values m=1,2,4,8m=1,2, 4, 8.

Cite

@article{arxiv.1408.6884,
  title  = {$3x+1$ inverse orbit generating functions almost always have natural boundaries},
  author = {Jason P. Bell and Jeffrey C. Lagarias},
  journal= {arXiv preprint arXiv:1408.6884},
  year   = {2015}
}

Comments

15 pages

R2 v1 2026-06-22T05:43:30.708Z