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

[Ami Eldar]

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.
>
>
>
> Best,
> Ali
>
>
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