[seqfan] Re: Just a quick (but hard?) funny sequence idea

Chris Thompson cet1 at cam.ac.uk
Fri Sep 1 16:33:26 CEST 2017


On Aug 31 2017, Joseph Myers wrote:

>On Wed, 30 Aug 2017, Chris Thompson wrote:
>
>>   a(13) = 11598859508648
>
>My program gives 6759747953135, which does not come from a convergent.
>
>>   a(16) = 21996765548122104
>
>My program gives 19027704941439533, not a convergent.
>
>>   a(27) = 1837855483715284868285348842
>
>My program gives 1367172464457274205192687057, not a convergent.

I can confirm these, and that there are no other corrections necessary
up to a(29).

>q*log10(pi) - p must (it is easy to see) be closer to 0 than for any 
>smaller q *that gives a value of that expression of the same sign*.  By 
>allowing approximations to log10(pi) that are only best (in that sense) 
>for a given sign of the error, you get approximations that are not 
>convergents (but it's still not hard to find all of them, I think - I 
>believe that if q1 and q2 give two successive best approximations (which 
>must be of opposite sign) then you just need also to consider the 
>approximations given by q1+q2, q1+2*q2, q1+3*q2, ... until you get one 
>that is better than q2 and so comes from the next convergent, and that's 
>probably exactly the semiconvergents though I haven't checked that in 
>detail).

I agree with all of this. It follows from the fact that all rational
numbers in the interval (a/b,c/d) where bc-ad=1 [e.g. two successive
convergents] are obtained by recursively adding mediants (a+c)/(b+d).

>If other people agree with this reasoning and those values I'll upload a 
>b-file of a(1) through a(1000).

I would encourage you to go ahead with that!

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



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