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

2, 5, 7, 11, 13, 17, 37, 41, 67, 89, 97, 101, 103, 107, 191, 193, 223, 227, 277, 307, 311, 347, 389, 397, 449, 457, 461, 479, 487, 491, 503, 613, 641, 739, 757, 761, 821, 823, 853, 857, 877, 881, 907, 929, 991, 1087, 1091, 1231, 1277, 1297, 1301, 1423, 1427, 1439, 1447, 1453

Offset: 1

Author

Andres Cicuttin, Apr 18 2018

Keywords

• nonn

Comments

Conjecture: The two most frequent ratios between consecutive prime gaps are 2 and 1/2, and both ratios occur with about the same frequency.

Examples

			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).
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.
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).
The prime 13 is also in the list because (19-17)/(17-13)=1/2.
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A001223, A274263, A276309, A022885.

Cf. A257762 (indices of primes with above ratio = 2).

Programs

Mathematica

b={};
Do[If[Abs[Log[2,(Prime[j+2]-Prime[j+1])/(Prime[j+1]-Prime[j])]]==1,AppendTo[b,Prime[j]]],{j,1,200}];
Print@b
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 *)

PARI

isok(p) = my(q = nextprime(p+1), r = nextprime(q+1), f = (r-q)/(q-p)); (f == 2) || (f == 1/2);
forprime(p=2, 1000, if (isok(p), print1(p, ", "))); \\ Michel Marcus, Apr 23 2018

Formula

Conjecture: lim_{n->inf} n/primepi(a(n)) > k > 0 for some k.