## RE Prime gaps (it was Prime GAP of 364,188)

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• ... From: Jose Ramón Brox We can get a prime gap of L = N between the numbers (N+1)!-(N+1) and (N+1)!-2. The merit of such a gap is m ~
Message 1 of 1 , Dec 30, 2005
----- Original Message -----
From: "Jose Ramón Brox" <ambroxius@...>

We can get a prime gap of L = N between the numbers (N+1)!-(N+1) and (N+1)!-2. The merit
of such a gap is

m ~ N / log [(N+1)!] ~ N / [(N+1)*(log(N+1)-1)] ~ 1/log N

Indeed, what a poor merit! Does anyone know a way to ensure a minimal merit (say, m = 1)
with a closed formula to get a fixed N gap?

Jose Brox

-------------------------------------------------------

(Well, I didn't say that that was the minimum assured gap, but as N increases, a lot more
of terms will be composite as well. But with the formula we can only be sure of N
composite terms).

After a bit of easy thinking I can answer myself in the affirmative:

Consider N# (the primorial of N): this number has the same properties than N! for the
gaps: if p is a prime, p <= N, then N#-p is a composite. If n < N, then it exists a prime
p <= n < N such that p|n, so N#-n is composite as well. So:

m >= N / log (N#) ~ 1

and we at least get the "expected gap", the "challenge" I was proposing. But actually the
merit is bigger than 1 because of other composites not predicted by the formula. For
example:

(n is the number whose primorial is calculated, N the actual gap, m the merit)

n = 5 --> N = 5 --> m = 1.47
n = 7 --> N = 9 --> m = 1.68
n = 11 --> N = 11 --> m = 1.42
n = 13 --> N = 15 --> m = 1.46
n = 17 --> N = 27 --> m = 2.05
n = 19 --> N = 21 --> m = 1.31
n = 29 --> N = 39 --> m = 1.77
n = 71 --> N = 124 --> m = 2.03
n = 113 --> N = 249 --> m = 2.33
n = 229 --> N = 255 --> m = 1.21
n = 541 --> N = 1011 --> m = 2.00
n = 1223 --> N = 2271 --> m = 1.92

New challenge: Can we extend this further and get better merits with closed formulas?

Jose Brox
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