

A276309


Integer part of the ratio of alternate consecutive prime gaps.


7



2, 2, 1, 1, 1, 1, 3, 0, 1, 2, 0, 1, 3, 1, 0, 1, 2, 0, 1, 2, 1, 2, 0, 0, 1, 1, 1, 7, 1, 0, 0, 1, 1, 0, 3, 0, 1, 1, 0, 1, 1, 0, 1, 3, 6, 0, 0, 1, 3, 0, 1, 3, 0, 1, 0, 1, 2, 0, 2, 7, 0, 0, 1, 7, 1, 0, 0, 0, 3, 2, 1, 0, 0, 1, 2, 0, 1, 2, 0, 1, 1, 0, 2, 1, 2, 0, 0, 1, 6, 2, 0, 1, 1, 0, 3, 0
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OFFSET

1,1


COMMENTS

Conjectures: The most frequent ratio among alternate prime gaps is 1, while the most frequent ratios among consecutive prime gaps seems to be 2 and 1/2, both with nearly the same frequency (see links). It also appears that next four most frequent ratios are 2, 1/2, 3 and 1/3, all four with nearly the same frequency (see links).


LINKS

Table of n, a(n) for n=1..96.
Andres Cicuttin, Counts on first 200000 ratios of alternate prime gaps
Andres Cicuttin, Several histograms of the logarithm of the ratio of consecutive prime gaps (prime(n+2)prime(n+1))/(prime(n+1)prime(n)) obtained for the first 2*10^5 primes with different bin sizes


FORMULA

a(n) = floor((prime(n + 3)  prime(n + 2))/(prime(n + 1)  prime(n))) .


EXAMPLE

For n=2, the second prime is 3, and the next three primes are 5, 7, and 11. So the ratio of prime gaps is (117)/(53) = 4/2 = 2, and the integer part of this is a(2) = 2.  Michael B. Porter, Aug 11 2016


MATHEMATICA

Table[Floor[(Prime[j + 3]  Prime[j + 2])/(Prime[j + 1]  Prime[j])], {j, 1, 200}]


PROG

(MAGMA) [Floor((NthPrime(n+3)NthPrime(n+2))/(NthPrime(n+ 1) NthPrime(n))): n in [1..100]]; // Vincenzo Librandi, Aug 30 2016


CROSSREFS

Cf. A001223, A274263, A272863, A274225.
Sequence in context: A156072 A215788 A060990 * A165031 A286634 A099245
Adjacent sequences: A276306 A276307 A276308 * A276310 A276311 A276312


KEYWORD

nonn


AUTHOR

Andres Cicuttin, Aug 06 2016


STATUS

approved



