%I #10 Sep 08 2022 08:46:17
%S 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,
%T 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,
%U 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
%N Integer part of the ratio of alternate consecutive prime gaps.
%C 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).
%H Andres Cicuttin, <a href="/A276309/a276309.pdf">Counts on first 200000 ratios of alternate prime gaps</a>
%H Andres Cicuttin, <a href="/A274263/a274263.pdf">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</a>
%F a(n) = floor((prime(n + 3) - prime(n + 2))/(prime(n + 1) - prime(n))) .
%e 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 (11-7)/(5-3) = 4/2 = 2, and the integer part of this is a(2) = 2. - _Michael B. Porter_, Aug 11 2016
%t Table[Floor[(Prime[j + 3] - Prime[j + 2])/(Prime[j + 1] - Prime[j])], {j, 1, 200}]
%o (Magma) [Floor((NthPrime(n+3)-NthPrime(n+2))/(NthPrime(n+ 1)- NthPrime(n))): n in [1..100]]; // _Vincenzo Librandi_, Aug 30 2016
%Y Cf. A001223, A274263, A272863, A274225.
%K nonn
%O 1,1
%A _Andres Cicuttin_, Aug 06 2016
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