%I A107664
%S A107664 3,5,7,13,23,29,31,37,43,53,59,71,73,89,97,103,109,113,137,139,149,157,
%T A107664 163,179,181,197,211,223,239,263,269,293,307,313,317,337,373,389,409,
%U A107664 419,421,431,433,449,457,463,467,479,491,521,523,547,563,577,593,599
%N A107664 The first member p of a triple (p,q,r) of consecutive primes such that
a solution to p/q<r/s<q/r or p/q>r/s>q/r with s prime exists.
%C A107664 For any three consecutive primes p,q and r, is it reasonable to say that
a countless number of pairs (p1,p2) forming fraction p1/p2 will fit
inside the interval p/q to q/r?
%C A107664 Equivalent definition: smallest p in a triple (p,q,r) of consecutive
primes such that there is at least one prime in the interval spanned
by the minimum and maximum of r^2/q and rq/p.
%F A107664 Look for an r/s so that p/q < r/s < q/r.
%e A107664 For p = 103, we have primes 103,107 and 109 to form fractions
%e A107664 103/107 = 0.9439 and 107/109 = 0.9817. Will a prime greater than
%e A107664 109 form a fraction that fits? Try 109/113 = 0.9646 and it fits
%e A107664 inside the interval.
%e A107664 p=103 is in the sequence because p=103, q=107, r=109 solve p/q<r/s<q/
r choosing s=113 (a prime).
%p A107664 isA107664 := proc(p) local q,r,s ; if isprime(p) then q := nextprime(p)
; r := nextprime(q) ; if p*r < q^2 then for s from ceil(r^2/q) to
floor(r*q/p) do if isprime(s) then RETURN(true) ; fi ; od ; elif
p*r > q^2 then for s from ceil(r*q/p) to floor(r^2/q) do if isprime(s)
then RETURN(true) ; fi ; od ; fi ; RETURN(false) ; else RETURN(false)
; fi ; end: for i from 1 to 300 do p := ithprime(i) ; if isA107664(p)
then printf("%d,",p) ; fi ; od:
%Y A107664 Sequence in context: A051507 A060274 A005235 this_sequence A085013 A164939
A125272
%Y A107664 Adjacent sequences: A107661 A107662 A107663 this_sequence A107665 A107666
A107667
%K A107664 nonn
%O A107664 1,1
%A A107664 J. M. Bergot (thekingfishb(AT)yahoo.ca), Jun 22 2007
%E A107664 Edited by R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jul 13 2007
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