## Re: [PrimeNumbers] An equation for primes

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• ... And it is p + g ... [h] = [p + g + O(1/p)] = p + g = q Best regards, Ignacio Larrosa Cañestro A Coruña (España)
Message 1 of 3 , Aug 29 2:18 AM
Ignacio Larrosa Cañestro wrote:
> Sebastian Martin Ruiz wrote:
>> Hello all:
>>
>> I have obtained the following equation for primes:
>>
>> Let p a prime number p>=127. Let g=q-p where q=the next prime to p.
>>
>> We have:
>>
>> Floor((p^(g/2+1))/((p-2)^(g/2)))=g+p=q
>>
>> ( floor(x) = ⌊x⌋ is the largest integer not greater than x. Example:
>> Floor (3.459)=3)
>>
>
> Dear Sebastian,
>
>
> Let
>
> h = (p^(g/2+1))/((p-2)^(g/2)) = p(p/(p-2))^(g/2)
>
> = p(1/(1 - 2/p)^(g/2) ~= p(1 + 2/p + 4/p^2 + ...)^g/2
>
> ~= p(1 + g/p + O(1/p^2) = p + g + O(1/p)

And it is > p + g

>
> Then, from certain p, h = p + g = q

[h] = [p + g + O(1/p)] = p + g = q

Best regards,

Ignacio Larrosa Cañestro
A Coruña (España)
> ilarrosa@...
>
>
>
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