[seqfan] Re: A000960

Chris Thompson cet1 at cam.ac.uk
Tue Dec 1 16:23:11 CET 2015

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 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

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