[seqfan] Re: A000960

[Chris Thompson]

On Nov 26 2015, David Wilson wrote:
>Does A000960(n) ~ (pi/4)n^2?

On Nov 26 2015, israel at math.ubc.ca wrote:
>It certainly appears to, and the first Formula says it does. Do you doubt 
>it?

It looks as thought the error term proved in the Andersson paper ought
to be capable of improvement, though. He shows 2*sqrt(n/pi)+O(n^(1/6))
terms less than n, which is equivalent to A000960(n)=(pi/4)*n^2+O(n^(4/3)).
Looking at the 10000 terms in the b-file, it seems as though the error
is closer to O(n) than O(n^(4/3)) - maybe n*(something logarithmic)?

Specifically, the extreme values for error/n in this range are -2.190
and +2.208, while 9189 of the 10000 entries have |error/n| < 1.

-- 
Chris Thompson
Email: cet1 at cam.ac.uk