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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 (list; graph; refs; listen; history; text; internal format)
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/(n-1))*Sum_{i=1..n-1} (prime(i+1) - prime(i)) or simply by (prime(n) - 2)/(n-1).

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 3-2 = 1 and 5-3 = 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 3-2 = 1, 5-3 = 2 and 7-5 = 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)/(n-1)), 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

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Last modified April 20 02:06 EDT 2019. Contains 322291 sequences. (Running on oeis4.)