In a Collatz-inspired tree comprising solely odd integers, an 'ant' simulates random walks utilizing three operations and their inverses:
- $ D $: $ i' = (i-1)/4 $, requiring $ i' $ to be odd
- $ V $: $ i' = (3i+1)/2 $, requiring $ i' $ to be odd
- $ W $: $ i' = (3i+1)/4 $, requiring $ i' $ to be odd
The ant embarks from 'home' at number 1, employing random choices among valid operations to ascend the tree.
For instance, when at level 7 with the value step at 739, the ant's valid options are D, W, and the inverse of V. A random selection yields 1109, propelling it to level 8.
The ant's ascent is unrestricted until it reaches a pre-set level, exemplified here as level 25, where it ostensibly finds a 'grain of rice' and opts to return to its origin.
The above graph delineates the ant's stochastic trajectory halted upon attaining level 25. The blue curve denotes the levels traversed, whereas the red curve (on a logarithmic scale) indicates the odd integer values encountered per step. Black markers pinpoint the steps along the optimum descent path within the random walk to level 25.
During this particular excursion, the ant traversed 210 steps to reach level 25, with merely 57 unique integers, signifying frequent detours. Notably, the optimum 25-step return path is encapsulated within these 57 integers.
On reaching level 25, the ant retraces to the start using only inverse operations, thus delineating an optimal path back to 1. This optimal return route, intriguingly, is a subset of the ant's broader random journey. It suggests that although the random walk overshoots with extraneous meandering, the most direct path to return is a condensed version of the walk.
The central query for the Math.StackExchange community is whether every odd integer can be accessed in this tree, given that the ant can navigate to any level via a random walk and return through an optimal pathway inherent in that journey. If so, this would imply the ant's potential to randomly approach any odd integer and methodically retreat to 1.
NB 1: if you're using a spreadsheet, the following formula gives the successor of i in the direction of the return to 1. Replace "i" with the cell containing the odd number whose successor you want : =IF(AND(ISODD((3i+1)/4), INT((3i+1)/4)=(3i+1)/4), (3i+1)/4, IF(AND(ISODD((i-1)/4), INT((i-1)/4)=(i-1)/4), (i-1)/4, IF(AND(ISODD((3i+1)/2), INT((3i+1)/2)=(3i+1)/2), (3i+1)/2, "")))
NB 2 : As the ant has reached the odd i =19583636365 at level 25, the optimum path to 1 is : 4895909091, 7343863637, 1835965909, 458991477, 114747869, 28686967, 43030451, 64545677, 16136419, 24204629, 6051157, 1512789, 378197, 94549, 23637, 5909, 1477, 369, 277, 69, 17, 13, 3, 5, 1