cp's OEIS Frontend

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.

%I A299351 #8 Feb 17 2018 12:00:41
%S A299351 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 A299351 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 A299351 3,3,2,6,5,3,4,3,4,4,3,3,4,4,3,3,4,4,3
%N 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.
%C A299351 It appears that all n end in the orbit (3,4) or the fixed point 12, verified to n=10^8.
%C A299351 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 A299351 The case 3^k, k>=2 first yields 4 and then 3: -> (3+1)=4=2^2 -> (2+1)=3.
%C A299351 It appears that these are the only ones entering the orbit (3,4), all other n end in the fixed point 12.
%H A299351 Lars Blomberg, <a href="/A299351/b299351.txt">Table of n, a(n) for n = 2..10000</a>
%e A299351 For n=2: 2 -> (2+1)=3 -> (3+1)=4=2^2 -> (2+1)=3; 3 is repeated so a(2)=3.
%e A299351 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 A299351 Cf. A299352.
%K A299351 nonn
%O A299351 2,1
%A A299351 _Lars Blomberg_, Feb 07 2018