A266928 Number of steps needed to reach 1 or 2 where a step is x -> 3x+1 if x is odd, or x -> x/2p if x is even, 2p being the smallest semiprime dividing x.

0, 0, 2, 1, 3, 1, 2, 1, 4, 1, 2, 3, 3, 1, 2, 2, 5, 3, 2, 4, 4, 1, 4, 2, 4, 1, 2, 3, 3, 4, 2, 2, 6, 1, 2, 5, 5, 1, 2, 2, 4, 3, 5, 3, 3, 1, 2, 4, 7, 4, 4, 4, 4, 5, 2, 2, 7, 1, 2, 3, 3, 1, 4, 3, 9, 3, 2, 6, 6, 3, 2, 4, 4, 1, 2, 3, 3, 4, 7, 5, 5, 1, 6, 5, 5, 1, 2

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

Cf. A006577.

Sequence in context: A265331 A347705 A244569 * A285324 A308551 A194550

Adjacent sequences: A266925 A266926 A266927 * A266929 A266930 A266931

Keyword

nonn

Author

Michel Lagneau, Jan 06 2016

Status

approved