1,1

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?

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.

Look for an r/s so that p/q < r/s < q/r.

For p = 103, we have primes 103,107, and 109 to form fractions

103/107 = 0.9439 and 107/109 = 0.9817. Will a prime greater than

109 form a fraction that fits? Try 109/113 = 0.9646 and it fits

inside the interval.

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).

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:

Adjacent sequences: A107661 A107662 A107663 this_sequence A107665 A107666 A107667

Sequence in context: A051507 A060274 A005235 this_sequence A085013 A125272 A127443

nonn

J. M. Bergot (thekingfishb(AT)yahoo.ca), Jun 22 2007

Edited by R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jul 13 2007