Gunter Lassmann
Hallo Joseph,
the question looks similar to the Collatz Conjecture, but it is much easier to prove.
My proof is in the annex. But I guess my proof is not helpful for the real Collatz Conjecture.
With greetings from Germany
Gunter Lassmann
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Lucy Kozeratska
Thank you for this beautiful question
I've got almost the same proof.
I still think that formula
a_k=2(((3/2)^k)(n+1)-1) (*)
under the assumption
Odd(a_k)=a_k/2 for all k>1 and all odd n>1
could simplify Gunter's proof:
since a_k is integer and even, then
((3/2)^k)(n+1) is integer and
n+1 is divided by 2^k for all k>1 and all odd n>1.
It contradicts the fact that
for any odd n>1 there exists k>0 such that
(n+1)/ (2^k)n
such that Odd(a_k)
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