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 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. 1
 3, 5, 7, 13, 23, 29, 31, 37, 43, 53, 59, 71, 73, 89, 97, 103, 109, 113, 137, 139, 149, 157, 163, 179, 181, 197, 211, 223, 239, 263, 269, 293, 307, 313, 317, 337, 373, 389, 409, 419, 421, 431, 433, 449, 457, 463, 467, 479, 491, 521, 523, 547, 563, 577, 593, 599 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS For any three consecutive primes p, q and r, is it reasonable to say that a countless number of pairs (p1,p2) forming the 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. LINKS FORMULA Look for an r/s so that p/q < r/s < q/r. EXAMPLE 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). MAPLE 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: CROSSREFS Sequence in context: A173145 A060274 A005235 * A321671 A085013 A164939 Adjacent sequences:  A107661 A107662 A107663 * A107665 A107666 A107667 KEYWORD nonn AUTHOR J. M. Bergot, Jun 22 2007 EXTENSIONS Edited by R. J. Mathar, Jul 13 2007 STATUS approved

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Last modified May 19 22:56 EDT 2019. Contains 323411 sequences. (Running on oeis4.)