## RE Primorial and Natural Logarithn

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• ... From: Anton Is there a simple and easy explanation (proof) as to why Ln[p#(i)] nearly equal p(i) p#(i) being the primorial of the
Message 1 of 1 , Sep 3, 2005
----- Original Message -----
From: "Anton" <al_at_i@...>

Is there a simple and easy explanation (proof) as to why

Ln[p#(i)] nearly equal p(i)

p#(i) being the primorial of the i-th prime p(i)

Ln[] being the natural logarithm

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

1) By the Prime Number Theorem, p(i) ~ i*Ln[i]

2) p#(i) = 2*3*5*...*p(i) ~ 1*2*Ln(2)*3*Ln(3)*...*i*Ln(i) = i!* PRODUCT Ln[j]

3) And Ln[i!*PRODUCT Ln[j]] should be approximately equal to i*Ln[i]

Ln[i!] + Ln[PRODUCT Ln[j]] ~ i*Ln[i] ?

4) By Stirling's Formula, Ln[i!] ~ i*Ln[i] - i

5) i*Ln[i] - i + Ln[PRODUCT Ln[j]] ~ i*Ln[i] ? --> Ln[PRODUCT Ln[j]] ~ i ?

Unfortunately, I don't know at this moment how to prove that limits equality. Maybe
another people in the list can solve it. If it doesn't hold, it wouldn't mean that the
original approximation is false, only that we need to solve the exact problem (note that I
changed p#(i) by the PNT inside the logarithm and therefore I changed the problem, but I
think it's a sufficient condition).

Did you come to this approximation by yourself or did you read it anywhere? Maybe it is
proved (or disproved!) elsewhere. Another consecuence of it is:

1) Ln[p#(i)] = Ln[2] + Ln[3] + ... Ln[p(i)]

2) By the PNT, p(i) ~ i*Ln[i]. Conversely, Ln[i] ~ p(i)/i.

3) Ln[p#(i)] ~ 2/1 + 3/2 + 5/3 + ... + p(i) / i ~ p(i)

4) i*SUM p(j)/j ~ (i-1)*p(i) --> SUM p(j)/j ~ p(i) with the sum up to i-1

Anyone has heard about these other result before? I'm thinking of making a chain of
aproximations to p(1), p(2), p(3), ... , p(i) and see if SUM p(j)/j ~ p(i) still holds; if
it does, then I think this gives us a quicker way to estimate the approximate position of
the primes, if we want to find several consecutive primes, instead of i*Ln[i], that
involves a logarithm for every prime we want to find.

Regards. Jose Brox
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