

A274263


Integer part of the ratio of consecutive prime gaps.


9



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
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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))) = (53)/(32) = 2 and its integer part is 2: a(1) = 2.
For n = 4 we have (prime(6)prime(5))/(prime(5)prime(4))) = (1311)/(117) = 1/2 an 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



