[seqfan] Re: need help for a calculation
David Applegate
david at research.att.com
Sat Jun 28 21:03:04 CEST 2014
I'm pretty sure your sequence is A098294; I haven't checked carefully, so
there may be a fence-post, but since (3/2)^n=2^j can only happen for n=j=0,
I believe they are equal.
David Applegate
> From seqfan-bounces at list.seqfan.eu Sat Jun 28 14:58:36 2014
> To: seqfan at list.seqfan.eu
> From: Frank Adams-Watters <franktaw at netscape.net>
> Date: Sat, 28 Jun 2014 14:58:16 -0400 (EDT)
> Subject: [seqfan] Re: need help for a calculation
> You computed right up to where they diverge. floor(3/2)^17) = 985 has
> 10 digits, but A005378(17) = 11. Sorry.
> Franklin T. Adams-Watters
> -----Original Message-----
> From: David Newman <davidsnewman at gmail.com>
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Sent: Sat, Jun 28, 2014 1:42 pm
> Subject: [seqfan] need help for a calculation
> I'm away from home and don't have access to my copy of Mathematica,
> otherwise I'd do this myself.
> How many binary digits are there in Floor [ (3/2) ^n ] ?
> The first 16 values that I've computed by hand seem to agree with
> A005378.
> ex. for n=0 Floor[ (3/2)^0]=1 which is 1 in binary, so a(0)=1
> for n=1 Floor[3/2]=1, which is 1 in binary, so a(1)=1
> for n=2 Floor[9/4]=2, which is 10 in binary, so a(2)=2
> I'd be very pleasantly surprised if this continues since the quantity
> being
> calculated is related to several unsolved problems.
> If anyone out there has a few minutes to continue the calculations I'd
> appreciate it.
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