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

%I A111166 #15 Jun 16 2025 15:22:07
%S A111166 2,5,11,17,29,41,59,71,101,107,137,149,179,191,197,227,239,269,281,
%T A111166 311,347,419,431,461,521,569,599,617,641,659,809,821,827,857,881,1019,
%U A111166 1031,1049,1061,1091,1151,1229,1277,1289,1301,1319,1427,1451,1481,1487,1607
%N A111166 Let p < q be consecutive primes; p is in the sequence if p/(q-p) is a record.
%C A111166 Conjecture: Except for first term, the sequence coincides with A001359. This is true for all primes < 2*10^7.
%C A111166 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 A111166 The smallest prime(n) such that prime(n+1)/prime(n) is decreasing. [_Thomas Ordowski_, May 13 2012]
%C A111166 This sequence corresponds with A001359 for all terms less than 10^100. - _Charles R Greathouse IV_, May 14 2012
%H A111166 Amiram Eldar, <a href="/A111166/b111166.txt">Table of n, a(n) for n = 1..10000</a>
%e A111166 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 A111166 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 *)
%t A111166 DeleteDuplicates[{#[[1]],#[[2]],#[[1]]/(#[[2]]-#[[1]])}&/@Partition[Prime[Range[300]],2,1],GreaterEqual[#1[[3]],#2[[3]]]&][[;;,1]] (* _Harvey P. Dale_, Jun 16 2025 *)
%Y A111166 Cf. A001359.
%K A111166 easy,nonn
%O A111166 1,1
%A A111166 _Bernardo Boncompagni_, Oct 21 2005