Similarity of A000799 & A064355

[franktaw at netscape.net]

The converse is really easy enough.  For any number not in one of these forms, the largest (absolute) two terms of the sum total less than 2^n - n.  The next term (if there is one) is also negative, and since the absolute values of the remaining terms sum to less than the absolute value of that term (all terms have absolute value a power of two), the sum is overall less than 2^n - n, so A064355(n) <= 2^n/n - 1.
 
Franklin T. Adams-Watters

-----Original Message-----
From: Max <maxale at gmail.com>
To: franktaw at netscape.net <franktaw at netscape.net>
Cc: pauldhanna at juno.com; seqfan at ext.jussieu.fr
Sent: Tue, 14 Feb 2006 14:14:03 -0800
Subject: Re: Similarity of A000799 & A064355


On 2/14/06, franktaw at netscape.net <franktaw at netscape.net> wrote:
>
> A000799(n) = A064355(n) only for:
> (1) n a power of 2.
> (2) n = p * 2^k, where p is a prime with p > 2^{2^k - k}.  (Including the
> case k=0.)
> (3) n = 9.

Did I get you correctly that you have a proof only in one direction?
Namely,
if n satisfy (1), (2), or (3) then A000799(n) = A064355(n)

What's about the converse?

Max
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