Prizes for the 3n+1 Problem

Hello,

I am offering a 1,000-Euro Prize for solving the
Collatz Problem before the end of 2037. And a
few additional smaller prizes for variants.

Details here:
https://www.althofer.de/collatz-prizes.html

Cheers, ingo.

Probably a base 2 solution
Might take a bit of shifting

To get longer runs on average start with odd numbers 2n-1
2n-1 with an up and down process produces 3n-1 which is half odd and even
6n-1 been the odd list produces 9n-1 with 18n-1 been the odd list
2*3^p*n-1 will lead to 2*3^(p+1)*n-1
So starting with an odder number will always lead to needing to start with a more odder number which would never get you to infinity

2(3^p)n-1