A111166 Let p < q be consecutive primes; p is in the sequence if p/(q-p) is a record.

2, 5, 11, 17, 29, 41, 59, 71, 101, 107, 137, 149, 179, 191, 197, 227, 239, 269, 281, 311, 347, 419, 431, 461, 521, 569, 599, 617, 641, 659, 809, 821, 827, 857, 881, 1019, 1031, 1049, 1061, 1091, 1151, 1229, 1277, 1289, 1301, 1319, 1427, 1451, 1481, 1487, 1607

Offset: 1

Author

Bernardo Boncompagni, Oct 21 2005

Keywords

• easy
• nonn

Comments

Conjecture: Except for first term, the sequence coincides with A001359. This is true for all primes < 2*10^7.

Conjecture: Except for first term, the sequence coincides with A001359. This is true for all primes < 7*10^16. Let n >= 2 be an integer, N +- 1 and M +- 1 two consecutive twin pairs where M>n*N. Finding a counterexample is the same as finding two consecutive primes P1 and P2 with n*N

The smallest prime(n) such that prime(n+1)/prime(n) is decreasing. [Thomas Ordowski, May 13 2012]

This sequence corresponds with A001359 for all terms less than 10^100. - Charles R Greathouse IV, May 14 2012

Examples

			a(0)=2 and the record is 2/(3-2)=2; a(1)<>3 because 3/(5-3)=1.5; a(1)=5 because 5/(7-5)=2.5
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A001359.

Programs

Mathematica

rmax = 0; p = 2; seq = {}; Do[q = NextPrime[p]; r = p/(q-p); If[r > rmax, rmax = r; AppendTo[seq, p]]; p = q, {100}]; seq (* Amiram Eldar, Dec 24 2019 *)
DeleteDuplicates[{#[[1]],#[[2]],#[[1]]/(#[[2]]-#[[1]])}&/@Partition[Prime[Range[300]],2,1],GreaterEqual[#1[[3]],#2[[3]]]&][[;;,1]] (* Harvey P. Dale, Jun 16 2025 *)

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