

A286888


Floor of the average gap between consecutive primes among the first n primes, for n > 1.


1



1, 1, 1, 2, 2, 2, 2, 2, 3, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5
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OFFSET

2,4


COMMENTS

The minimum n to evaluate a gap is 2, then n must be larger than 1. The average gap between consecutive primes computed over the first n primes is given by (1/(n1))*Sum_{i=1..n1} (prime(i+1)  prime(i)) or simply by (prime(n)  2)/(n1).


LINKS

Robert Israel, Table of n, a(n) for n = 2..10000


FORMULA

a(n)= floor((prime(n)  2)/(n  1)).
floor(log(n) + log(log(n))  1) <= a(n) <= floor(log(n) + log(log(n)) + 1).  Robert Israel, Aug 04 2017


EXAMPLE

a(3)=1 because the two gaps between consecutive primes among the first three primes are 32 = 1 and 53 = 2, the average gap is (1+2)/2 = 3/2, and the floor of 3/2 is 1.
a(4)=1 because the three gaps between consecutive primes among the first four primes are 32 = 1, 53 = 2 and 75 = 2, the average gap is (1+2+2)/3 = 5/3, and the floor of 5/3 is 1.


MAPLE

seq(floor((ithprime(n)2)/(n1)), n=2..200); # Robert Israel, Aug 04 2017


MATHEMATICA

nmax=132;
Table[Floor[(Prime[n]  2)/(n  1)], {n, 2, nmax}]


CROSSREFS

Cf. A001223.
Sequence in context: A125973 A227196 A189172 * A257212 A001031 A035250
Adjacent sequences: A286885 A286886 A286887 * A286889 A286890 A286891


KEYWORD

nonn


AUTHOR

Andres Cicuttin, Jul 22 2017


STATUS

approved



