[seqfan] Re: Do we need an escape clause here?

Ali Sada pemd70 at yahoo.com
Wed Feb 19 09:22:21 CET 2020

 Hi Dr. Eldar,
Thank you very much for your email. I really appreciate it. It would be great if you could add this note to A331410. 
Or, with your permission, I will add it happily. 


    On Wednesday, February 19, 2020, 2:17:17 AM EST, Ami Eldar <amiram.eldar at gmail.com> wrote:  
 Let f(n) = A000265(n) be the odd part of n.
Let p be the largest prime factor of k, and say k = p * m.
Suppose that k is not a power of 2, i.e., p > 2, then f(k) = p * f(m).
The iteration is k -> k + k/p = p*m + m = (p+1) * m. So, p * f(m) -> f(p+1) * f(m).
Since for p > 2, f(p+1) < p, the odd part in each each iteration decreases, until it becomes 1, i.e. until we reach a power of 2.

On Wed, Feb 19, 2020 at 8:09 AM Ali Sada via SeqFan <seqfan at list.seqfan.eu> wrote:

Hi Everyone,
Please see the definition below. It' A331410.
"a(n) is the number of iterations needed to reach a power of 2 starting at n and using the map k -> (k+(k/p), where p is the largest prime factor of k)."

It seems so basic that we will reach a power of 2 eventually. I am not sure I can prove it, but I think there is a proof somewhere. We will also reach a power of 2 if we use k-(k/p), or if we divide by any odd prime factor. 


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