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

Author

J. M. Bergot, Jun 22 2007

Keywords

• nonn

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.

Examples

			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).
		

Programs

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:

Formula

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

Extensions

Edited by R. J. Mathar, Jul 13 2007