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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. 1
 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 (list; graph; refs; listen; history; text; internal format)
 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

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