cp's OEIS Frontend

A274263 Integer part of the ratio of consecutive prime gaps.

%I A274263 #34 Nov 18 2024 22:31:07
%S A274263 2,1,2,0,2,0,2,1,0,3,0,0,2,1,1,0,3,0,0,3,0,1,1,0,0,2,0,2,3,0,1,0,5,0,
%T A274263 3,1,0,1,1,0,5,0,2,0,6,1,0,0,2,1,0,5,0,1,1,0,3,0,0,5,1,0,0,2,3,0,1,0,
%U A274263 2,1,1,0,1,0,1,1,0,2,1,0,5,0,3,0,1,1,0,0,2,3,0,0,2,0,1,2
%N A274263 Integer part of the ratio of consecutive prime gaps.
%C A274263 It seems that the distribution of the ratios of consecutive prime gaps exhibits a quite symmetric pattern, in the sense that the relative frequency of each ratio is similar to that of the inverse of that ratio (at least for the first 2*10^5 primes). This is more clearly seen by mean of a histogram of the logarithm of the ratios which is nearly symmetric and nearly centered around zero (see link).
%C A274263 Integer part of 2, 1, 2, 1/2, 2, 1/2, 2, 3/2, 1/3, 3, 2/3, 1/2,.... - _R. J. Mathar_, Jun 26 2016
%H A274263 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 A274263 a(n) = floor(A001223(n+1)/A001223(n)).
%e A274263 For n = 1 we have (prime(3)-prime(2))/(prime(2)-prime(1)) = (5-3)/(3-2) = 2 and its integer part is 2: a(1) = 2.
%e A274263 For n = 4 we have (prime(6)-prime(5))/(prime(5)-prime(4)) = (13-11)/(11-7) = 1/2 and its integer part is 0: a(4) = 0.
%p A274263 A274264 := proc(n)
%p A274263     A001223(n+1)/A001223(n) ;
%p A274263     floor(%) ;
%p A274263 end proc: # _R. J. Mathar_, Jun 26 2016
%t A274263 Table[Floor[(Prime[j+2]-Prime[j+1])/(Prime[j+1]-Prime[j])],{j,1,200}];
%t A274263 IntegerPart[#[[2]]/#[[1]]]&/@Partition[Differences[Prime[Range[200]]],2,1] (* _Harvey P. Dale_, Mar 07 2018 *)
%o A274263 (PARI) a(n) = (prime(n+2)-prime(n+1))\(prime(n+1)-prime(n)); \\ _Michel Marcus_, Jun 18 2016
%Y A274263 Cf. A001223.
%K A274263 nonn
%O A274263 1,1
%A A274263 _Andres Cicuttin_, Jun 17 2016