OFFSET
1,1
COMMENTS
The sequence is column c in the link funct.pdf with obvious adjustments.
"Adjusted" means that, since for every four terms the first two are 1's they are omitted and then the sequence is relabeled. The 3x+1 function is defined:
For x a positive integer. f(x) := 3x + 1 with all positive powers of 2 remove.Note 1 is a fixed point of f.
The range of the 3x+1 function is two disjoint sets 6N+1 and 6N+5 for N nonnegative integers.
See the link to paper ammprob-f4-1-2, for a proof for range of the 3x+1 function.
Observations, Conjectures:
The famous 3x+1 problem would be solved if and only if ALL stopping time values are finite.
a(n)=2 iff Mod_{8}(n) is in {3, 6} a(n)=3 iff Mod_{16}(n) is in {7,9, 2,4} a(n)=4 iff Mod_{64}(n) is in {1,31,45,10,24,44}
a(n)=5 iff Mod_{128}(n) is in {13,29,33,49,63,79,101,16,56,72,76,79,92,106,122}
a(n)=6 iff Mod_{512}(n) is in {61,97,241,255,293,333,337,389,399,437,477,495,48,58,96,136,154,232,268,412,426,464,504,508}
Pattern seems to be a(n)=c iff there exist k and sets A,B such that
Mod_{2^k}(n) is in A union B, where |A|=|B| and A are odd and B are even numbers, where A is associated with 6N+1 and B with 6N+5.
Conjecture: Ultimately every positive integer appears in the stopping time sequence. (verified up to 100, examples: a(6802394)=160, a(31229269)=161) And each positive integer is in the sequence an infinite number of times.
REFERENCES
Ultimate Challenge: the 3x+1 problem, J.C. Lagarias - editor, AMS 2010.
LINKS
Sam E. Speed, Table of n, a(n) for n = 1..8194
Sam E. Speed, ammprob-f4-1-2.tex, proof of range of 3x+1 function
Sam E. Speed, Maple 12 program cutoff.mw used to make b-file
Wikipedia, Collatz conjecture
FORMULA
For n a positive integer,
a(n) = Min_{e=1,2,...} f^e(x(n)) < x(n), where f is the 3x+1 function defined above and
x(n) = 6n+1 if n=1,3,5,.. (odd) and x(n) = 6n-1 if n=2,4,6,... (even).
See original stopping time file, funct, before adjustments.
MAPLE
See links cutoff.mw.
CROSSREFS
KEYWORD
nonn
AUTHOR
Sam E. Speed, Jan 18 2021
STATUS
approved