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.

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

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

Table of n, a(n) for n=1..56.

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: A345130 A005235 A353284 * 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