[seqfan] Re: need help for a calculation
Frank Adams-Watters
franktaw at netscape.net
Sat Jun 28 20:58:16 CEST 2014
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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