

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.


1



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
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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: A001511 A265331 A244569 * A285324 A194550 A242923
Adjacent sequences: A266925 A266926 A266927 * A266929 A266930 A266931


KEYWORD

nonn


AUTHOR

Michel Lagneau, Jan 06 2016


STATUS

approved



