
Re: 18 consecutive zeros in a power of two
Posted:
Oct 15, 2012 8:15 PM


On Monday, October 15, 2012 11:56:01 AM UTC5, Clive Tooth wrote: > On Oct 15, 2:43 am, Mensanator <mensana...@aol.com> wrote: > > > On Saturday, October 13, 2012 5:17:03 AM UTC5, Clive Tooth wrote: > > > > Some stuff on facebook about all this... > > > > > > >http://www.facebook.com/media/set/?set=a.4178263132511.2153798.116236... > > > > > > Very cool, but the big question is: using that number as a seed, how > > > of a Collatz sequence does it generate? > > > > I may not understand your question correctly, but... > > > > If you start with 4,627,233,991 you reach 1 after 371 steps. > > > > If you start with 2^4,627,233,991 you reach 1 after 4,627,233,991 > > steps. > > > >  > > Clive Tooth
Oh, yeah. A power of 2. Duh.
I should have said that power of 2 minus 1.

