A299351 For x=n, iterate the map x -> Product_{k is a prime dividing x} (k + 1), a(n) is the number of steps to see a repeated term for the first time.

3, 2, 2, 3, 2, 4, 3, 3, 3, 2, 1, 4, 3, 3, 3, 3, 2, 4, 3, 4, 3, 3, 2, 3, 4, 3, 3, 4, 3, 4, 3, 3, 3, 3, 2, 5, 4, 4, 3, 4, 3, 4, 3, 3, 3, 3, 2, 4, 3, 3, 4, 3, 2, 3, 3, 4, 4, 4, 3, 4, 3, 4, 3, 4, 3, 4, 3, 3, 3, 3, 2, 6, 5, 3, 4, 3, 4, 4, 3, 3, 4, 4, 3, 3, 4, 4, 3

Offset

2,1

Comments

It appears that all n end in the orbit (3,4) or the fixed point 12, verified to n=10^8.

Let p,q,r,... be primes that increased by 1 become a power of 2 (the Mersenne primes, A000668). Then for n = p^a*q^b*r^c*..., a,b,c,...>=1 -> (p+1)*(q+1)*(r+1)... = 2^e, e>=2 -> (2+1)=3.

The case 3^k, k>=2 first yields 4 and then 3: -> (3+1)=4=2^2 -> (2+1)=3.

It appears that these are the only ones entering the orbit (3,4), all other n end in the fixed point 12.

Links

Lars Blomberg, Table of n, a(n) for n = 2..10000

Example

For n=2: 2 -> (2+1)=3 -> (3+1)=4=2^2 -> (2+1)=3; 3 is repeated so a(2)=3.
For n=19: 19 -> (19+1)=20=2^2*5 -> (2+1)*(5+1)=18=2*3^2 -> (2+1)*(3+1)=12=2^2*3 -> (2+1)*(3+1)=12; 12 is repeated so a(19)=4.

Crossrefs

Cf. A299352.

Sequence in context: A305534 A248138 A049234 * A358935 A294299 A125504

Adjacent sequences: A299348 A299349 A299350 * A299352 A299353 A299354

Keyword

nonn

Author

Lars Blomberg, Feb 07 2018

Status

approved