login
A274263
Integer part of the ratio of consecutive prime gaps.
11
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
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 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
KEYWORD
nonn
AUTHOR
Andres Cicuttin, Jun 17 2016
STATUS
approved