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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. 2
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 (list; graph; refs; listen; history; text; internal format)
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 * A294299 A125504 A243929

Adjacent sequences: A299348 A299349 A299350 * A299352 A299353 A299354

KEYWORD

nonn

AUTHOR

Lars Blomberg, Feb 07 2018

STATUS

approved

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Last modified November 29 18:13 EST 2022. Contains 358431 sequences. (Running on oeis4.)