[seqfan] Re: Colorless integers
Jack Brennen
jfb at brennen.net
Wed Feb 23 01:02:22 CET 2011
There is no other solution of F(F(n))=n, other than the trivial 0.
The sequence can only go up by an even delta, and only down by
an odd delta. So no two-step cycles, since one would have to be
an up delta, and the other a down delta.
A three-step cycle would have to be UP, DOWN, DOWN. For instance,
99->117->108->99.
On 2/22/2011 3:52 PM, Robert Gerbicz wrote:
> 2011/2/23 Robert Gerbicz<robert.gerbicz at gmail.com>
>
>>
>>
>> 2011/2/22 N. J. A. Sloane<njas at research.att.com>
>>
>> Start at n, let a(n) be the smallest number in the cycle if it
>>> goes into a cycle, or -1 if it diverges. Is this sequence
>>> in the OEIS? and if not, you know what to do!
>>> Neil
>>>
>>> _______________________________________________
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>>> Seqfan Mailing list - http://list.seqfan.eu/
>>>
>>
>> The idea behind this that by O(1/log(n)) probability F(F(n))=n is true.
>>
>> Need another idea, up to n=10^7 there is no solution of F(F(n))=n, other
> than the trivial n=0.
> Does it have a loop of length two?
>
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