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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.
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%I #8 Feb 17 2018 12:00:41

%S 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,

%T 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,

%U 3,3,2,6,5,3,4,3,4,4,3,3,4,4,3,3,4,4,3

%N 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.

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

%C 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.

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

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

%H Lars Blomberg, <a href="/A299351/b299351.txt">Table of n, a(n) for n = 2..10000</a>

%e For n=2: 2 -> (2+1)=3 -> (3+1)=4=2^2 -> (2+1)=3; 3 is repeated so a(2)=3.

%e 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.

%Y Cf. A299352.

%K nonn

%O 2,1

%A _Lars Blomberg_, Feb 07 2018