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

Chris Thompson cet1 at cam.ac.uk
Thu Oct 20 18:09:03 CEST 2016

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 )) );

[In passing, I can't help feeling A062300 would be more transparently
defined as floor(cosec(Pi/n)) with the offset set to 2.]

>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 ...

and on Oct 19 2016, Robert Israel replied:

>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.

Fine, as long as Pi is "random" ... :-)

The economical way to find the exceptions is to use continued fractions.
As Robert says, we have

  cosec(Pi/n) = n/Pi + (Pi/6)n^{-1} + (7*Pi/360)n^{-3} + ...

so if there is to be an integer m between n/Pi and cosec(Pi/n) we need

  0 < (m/n - 1/Pi) < (Pi/6)n^{-2} + {terms we can forget for large n}

If Pi/6 were less than 1/2, then this would ensure that m/n was equal
to a covergent of 1/Pi [NB - not necessarily in its lowest terms].
Well it's pretty close, so lets look at the convergents anyway:

  Partial        m        n      A  
        3        1        3     7.396  **
        7        7       22   -16.133
       15      106      333     1.069
        1      113      355  -293.573
      292    33102   103993     1.579
        1    33215   104348    -2.733
        1    66317   208341     1.858
        1    99532   312689    -3.464
        2   265381   833719     1.629
        1   364913  1146408    -4.662
        3  1360120  4272943     1.338
        1  1725033  5419351   -15.175
       14 25510582 80143857     2.653  **
        2 ...

See A001203 for the first column. Here A is defined by

  m/n - 1/Pi = 1/(A*n^2)

and can be computed from the surrounding partial quotients - in particular
it is in the range a < |A| < a+2 where a is the next partial quotient.

What we need is for A to be positive and greater than 6/Pi = 1.90985...
The table above shows that we are rather unlucky early on - all the
really good (or even fairly good) approximations to 1/Pi are on the
wrong side of it. Hence Kevin's big gap between 6 (I did warn you that
m/n didn't need to be in lowest terms! 2/6 just squeaks in as well as 1/3)
and 80143857.

Maybe someone would like to compute the next few exceptions? As Pi/6
is actually a bit more than 1/2, it should be possible to find some
which aren't actually convergents (but can be found by fiddling the
partial quotients slightly).

And are there infinitely many? There certainly would be if the partial
quotients for 1/Pi behaved like those for "almost all" reals. But
although one of every consecutive pair of convergents m/n must be
as close as 1/(2*n^2), it is possible for it always to be the wrong one.
If A001203 eventually turns into 1,big,1,big,1,big,... indefinitely,
where the big's only have to by >=2, then the alternate approximations
will only be as good as 1/(sqrt(3)*n^2) > Pi/(6*n^2) and there will be
no more exceptions. Of course this is very "unlikely" ....

Chris Thompson
Email: cet1 at cam.ac.uk

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