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

1,1

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.

Links

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

Formula

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

Example

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.

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 *)

Prog

(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

Crossrefs

Cf. A001223, A274263, A276309, A022885.

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

Sequence in context: A020591 A197188 A265809 * A182320 A119993 A045346

Adjacent sequences: A303109 A303110 A303111 * A303113 A303114 A303115

Keyword

nonn

Author

Andres Cicuttin, Apr 18 2018

Status

approved