OFFSET
1,3
COMMENTS
Conjecture: all positive integers n not of the form 2^(2m+1) eventually reach 1.
This is a variant of the Collatz problem: start with any number n. If n is even, divide it by 2p where 2p is the smallest semiprime dividing n, otherwise multiply it by 3 and add 1.
Property: a(2^(2m+1)) = m and the last element of the corresponding trajectory is the number 2.
It seems that initially about 17% of the terms satisfy a(i) = a(i+1). For example, up to 1000000, 169961 terms satisfy this condition.
LINKS
Michel Lagneau, Table of n, a(n) for n = 1..10000
EXAMPLE
a(83)=6 because 83 -> 250 -> 25 -> 76 -> 19 -> 58 -> 1 where:
250 = 3*83 + 1, 25 = 250/2*5, 76 = 3*25 + 1, 19 = 76/2*2, 58 = 3*19 + 1 and 1 = 58/2*29.
MATHEMATICA
f[n_]:=Module[{a=n, k=0}, While[a>2, k++; If[EvenQ[a], m=1; While[PrimeOmega[Divisors[a][[m]]]!=2, m++]; a=a/Divisors[a][[m]], a=a*3+1]]; k]; Table[f[n], {n, 1, 100}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Jan 06 2016
STATUS
approved