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

1,1

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

Links

Table of n, a(n) for n=1..96.

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(A001223(n+1)/A001223(n)).

Example

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.

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 *)

Prog

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

Crossrefs

Cf. A001223.

Sequence in context: A275768 A117167 A117169 * A046920 A107599 A156836

Adjacent sequences: A274260 A274261 A274262 * A274264 A274265 A274266

Keyword

nonn

Author

Andres Cicuttin, Jun 17 2016

Status

approved