[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.
>
>
>
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