[seqfan] Re: Doomed by odd integers

Torleiv.Klove at ii.uib.no Torleiv.Klove at ii.uib.no
Thu Nov 25 12:38:41 CET 2010


Clearly, the exact power of 2 dividing numbers in the sequence decrease 
by one in each step (you multiply by 3/2). Hence, starting with N, the 
number of steps equals m where 2^m is the exact power of 2 dividing N.
The smallest number N ending after n steps is therefore 2^n.

Best,
Torleiv


Eric Angelini wrote:
> Hello SeqFans,
> Say I start with an even integer and successively add
> halves like this:
> 
> 16+8=24
>      24+12=36
>            36+18=54
>                  54+27=81 END
> 
> ... am I doomed to always END (on an odd integer)?
> 
> Is there a sequence of terms a(n) where a(n) is the
> smallest integer ENDING in n steps?
> 
> Best,
> É.
> 
> 
> 
> _______________________________________________
> 
> Seqfan Mailing list - http://list.seqfan.eu/





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