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Let p < q be consecutive primes; p is in the sequence if p/(q-p) is a record.
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%I #13 Dec 24 2019 08:27:20

%S 2,5,11,17,29,41,59,71,101,107,137,149,179,191,197,227,239,269,281,

%T 311,347,419,431,461,521,569,599,617,641,659,809,821,827,857,881,1019,

%U 1031,1049,1061,1091,1151,1229,1277,1289,1301,1319,1427,1451,1481,1487,1607

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

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

%C 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<P1<M and P2-P1 <= n. However, such gaps are unknown even for n=2.

%C The smallest prime(n) such that prime(n+1)/prime(n) is decreasing. [_Thomas Ordowski_, May 13 2012]

%C This sequence corresponds with A001359 for all terms less than 10^100. - _Charles R Greathouse IV_, May 14 2012

%H Amiram Eldar, <a href="/A111166/b111166.txt">Table of n, a(n) for n = 1..10000</a>

%e 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

%t 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 *)

%Y Cf. A001359.

%K easy,nonn

%O 1,1

%A _Bernardo Boncompagni_, Oct 21 2005