A274263 Integer part of the ratio of consecutive prime gaps.

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, 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, 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

Offset: 1

Author

Andres Cicuttin, Jun 17 2016

Keywords

• nonn

Comments

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).

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

Examples

			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.
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.
		

Links

• 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

Crossrefs

Cf. A001223.

Programs

Maple

A274264 := proc(n)
    A001223(n+1)/A001223(n) ;
    floor(%) ;
end proc: # R. J. Mathar, Jun 26 2016

Mathematica

Table[Floor[(Prime[j+2]-Prime[j+1])/(Prime[j+1]-Prime[j])],{j,1,200}];
IntegerPart[#[[2]]/#[[1]]]&/@Partition[Differences[Prime[Range[200]]],2,1] (* Harvey P. Dale, Mar 07 2018 *)

PARI

a(n) = (prime(n+2)-prime(n+1))\(prime(n+1)-prime(n)); \\ Michel Marcus, Jun 18 2016

Formula

a(n) = floor(A001223(n+1)/A001223(n)).