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

Joseph Myers jsm at polyomino.org.uk
Thu Aug 31 03:53:36 CEST 2017

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.

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 

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

Joseph S. Myers
jsm at polyomino.org.uk

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