Puzzles for Feb. 16. by Terrence Bisson | Feb 17, 2022 | Uncategorized | 1 comment We chose the Collatz-like Problem 338. Based on experiments, we conjecture that starting with any number leads into a repeating loop… 1 Comment Terrence Bisson on February 23, 2022 at 7:29 pm If $n\equiv 1\pmod 3$ then $n+20\equiv 0\pmod 3$ and $n\to (n+20)/3$, which is less than $n$ when $n > 10$.
Terrence Bisson on February 23, 2022 at 7:29 pm If $n\equiv 1\pmod 3$ then $n+20\equiv 0\pmod 3$ and $n\to (n+20)/3$, which is less than $n$ when $n > 10$.
If $n\equiv 1\pmod 3$ then $n+20\equiv 0\pmod 3$ and $n\to (n+20)/3$, which is less than $n$ when $n > 10$.