# [seqfan] Re: nearly equal floor(n/Pi) A032615 and A062300

israel at math.ubc.ca israel at math.ubc.ca
Wed Oct 19 17:56:58 CEST 2016

```since 1/sin(1/n/Pi)) ~ n/Pi + Pi/(6*n) as n -> infinity, and frac(n/Pi) is
equidistributed in [0,1), the probability that their floors differ is
approximately Pi/(6*n). Since the sum of that diverges, albeit slowly, we
should expect infinitely many differences.

Cheers,
Robert

On Oct 19 2016, Kevin Ryde via SeqFan wrote:

>On the topic of the "nearly equal", searching recently for something
>else I found
>
>    A032615(n) = floor(             n  /Pi    );
>    A062300(n) = floor( 1/sin(1/( (n+1)/Pi )) );
>
>which are nearly the same (up to offset +1) after the first few values,
>since sin(x)~=x when x small.
>
>I make the next difference (after n=6) at n=80143857.  Maybe a fan of
>the nearly-equal would be interested.
>
>I imagine n/Pi can fall arbitrarily close to an integer, and with the
>sin bigger that may push up across the integer boundary to different
>floor().  But the extra the sin adds decreases with n ...
>
>--
>Seqfan Mailing list - http://list.seqfan.eu/
>
>

```