[seqfan] Re: question from Alexander P-sky

franktaw at netscape.net franktaw at netscape.net
Thu May 13 22:08:44 CEST 2010

Given two sequences a and b, where one is always at least as large as 
the other, one generally considers b a bound on a if b is easier to 
compute, and a is considered a bound on b if it is simpler. In this 
case, A002024 is considerably simpler than A063655, so I would consider 
it to be the bound.

(From the formulas section, A002024 is floor( 1/2 + sqrt(2n) ) = 
round(sqrt(2n)). By contrast, computing A063655 requires factoring n.)

Franklin T. Adams-Watters

-----Original Message-----
From: Alexander P-sky <apovolot at gmail.com>

>If it's just because A002024 is a lower bound for A063655
Sort of - but I rather thought about it in terms of A063655 being a
high bound of A002024

On 5/12/10, franktaw at netscape.net <franktaw at netscape.net> wrote:
> No, it isn't. I'm not sure why it should be.
> If it's just because A002024 is a lower bound for A063655, then 
> is a better one along the same lines.
> Franklin T. Adams-Watters
> -----Original Message-----
> From: N. J. A. Sloane <njas at research.att.com>
> Dear Sequence Fans,  Alexander asked me:
> Hi Neil,
> Is A063655 - A002024 in OEIS ?
> If not - is it worth putting it there ?
> Thanks,
> Best Regards,
> Alex
> I don't have time to look at this - could someone reply to him?


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