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A303112 Primes p such that (r-q)/(q-p) = 2 or 1/2, and p < q < r are three consecutive primes.

%I A303112 #31 Mar 14 2022 19:31:19
%S A303112 2,5,7,11,13,17,37,41,67,89,97,101,103,107,191,193,223,227,277,307,
%T A303112 311,347,389,397,449,457,461,479,487,491,503,613,641,739,757,761,821,
%U A303112 823,853,857,877,881,907,929,991,1087,1091,1231,1277,1297,1301,1423,1427,1439,1447,1453
%N A303112 Primes p such that (r-q)/(q-p) = 2 or 1/2, and p < q < r are three consecutive primes.
%C A303112 Conjecture: The two most frequent ratios between consecutive prime gaps are 2 and 1/2, and both ratios occur with about the same frequency.
%H A303112 Harvey P. Dale, <a href="/A303112/b303112.txt">Table of n, a(n) for n = 1..1000</a>
%F A303112 Conjecture: lim_{n->inf} n/primepi(a(n)) > k > 0 for some k.
%e A303112 The first three consecutive primes are 2, 3 and 5, and (5-3)/(3-2)=2, so the first term is a(1)=2, that is, the first prime of (2,3,5).
%e A303112 The next three consecutive primes are 3, 5 and 7, and (7-5)/(5-3)=1, so the first prime of (3,5,7) is not in the list.
%e A303112 The next three consecutive primes are 5, 7 and 11, and (11-7)/(7-5)=2, so the second term is a(2)=5, that is, the first prime of (5,7,11).
%e A303112 The prime 13 is also in the list because (19-17)/(17-13)=1/2.
%t A303112 b={};
%t A303112 Do[If[Abs[Log[2,(Prime[j+2]-Prime[j+1])/(Prime[j+1]-Prime[j])]]==1,AppendTo[b,Prime[j]]],{j,1,200}];
%t A303112 Print@b
%t A303112 Select[Partition[Prime[Range[250]],3,1],(#[[3]]-#[[2]])/(#[[2]]-#[[1]]) == 2||(#[[3]]-#[[2]])/(#[[2]]-#[[1]])==1/2&][[All,1]] (* _Harvey P. Dale_, Mar 14 2022 *)
%o A303112 (PARI) isok(p) = my(q = nextprime(p+1), r = nextprime(q+1), f = (r-q)/(q-p)); (f == 2) || (f == 1/2);
%o A303112 forprime(p=2, 1000, if (isok(p), print1(p, ", "))); \\ _Michel Marcus_, Apr 23 2018
%Y A303112 Cf. A001223, A274263, A276309, A022885.
%Y A303112 Cf. A257762 (indices of primes with above ratio = 2).
%K A303112 nonn
%O A303112 1,1
%A A303112 _Andres Cicuttin_, Apr 18 2018