[seqfan] Are we going to get anything other than 2’s and 3’s from this algorithm?

[Ali Sada]

Hi Everyone,

Please consider the following algorithm: Divide n by its largest prime factor and add the result to n-1. Continue with the same algorithm until we go back to 1. 

For example, to find a(6):
6/3 = 2
5+2 = 7, 7/7 = 1
4+1 = 5, 5/5 = 1
3+1 = 4, 4/2 = 2
2+2 = 4, 4/2 = 2
1+2 = 3 --> a(6) = 3

The routes to reach the terms produce interesting numbers, but the sequence itself is not that diverse. 

After a(1) = 1, the sequence's terms are either 2’s or 3’s. (The 2 means that the number in the semifinal step is a prime.) Between n = 3 and n = 950, the combination 2, 2, 3, 3 was constant. But it changed after that. The semifinal numbers were 2 if a(n) = 3, and 3 if a(n)= 2. 

I couldn’t go beyond n = 1000 because of problems with the software I am using. I would really appreciate it if you could help me find more terms. 

It seemed logical to try to find what it would need to make a(n) = 4, for example, using a counter-algorithm. But in this case, I had to guess what the prime factors were. Even if the prime factor were 2 in every step, the counter-algorithm goes to high numbers very quickly. 

When I used the same algorithm, replacing “the largest prime factor” with “the least prime factor”, the results were also interesting. After a(1) = 1, The first terms are either 2’s or 3’s. Then, at n = 39, 2 disappears and we get a string of 3’s. The first 4 appears at n = 273. Then, 3 becomes less frequent. Between n = 675 and 1000, there are only 4’s. It seems like we are going to see 5, 6, 7, etc. in this version, but I am not sure. 


Best,

Ali