• ## Re: [PrimeNumbers] Prime Numbers Equation

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• ... That s, in its shorter form, is the oldest one in the book! You ve attempted to obfuscate it by separating it into two sums, but we can see through that.
Message 1 of 7 , Nov 8, 2008
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--- On Sat, 11/8/08, Sebastian Martin <sebi_sebi@...> wrote:
> Let x a integer number>1.
>
> x is a prime number if and only if it is a solution of this
> equation:
>
> Sum[Floor[x/i],{i,1,Sqrt[x]}]=1+Sum[Floor[(x-1)/i],{i,1,Sqrt[x]}]

That's, in its shorter form, is the oldest one in the book!

You've attempted to obfuscate it by separating it into two sums, but we can see through that.

Sum[Floor[x/i]-Floor[(x-1)/i],{i,1,Sqrt[x]}] = 1

The difference of floors is either 0 or 1, and only 1 if i divides x.
So the left hand side is just a count of the divisors between 1 and sqrt(x) inclusive. Only for primes can that be 1.

I'm sure you've posted this to the list in the past, but am too lazy to check.

Phil
• Hello all: n positive integer n=/=1 and 4   n is prime  if and only if 8*Product[(k*n-1)/(Floor[n/k]+1), {k,1,n}]=Floor[8*Product[(k*n-1)/(Floor[n/k]+1),
Message 2 of 7 , Mar 30 7:14 AM
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Hello all:
n positive integer n=/=1 and 4
n is prime
if and only if
8*Product[(k*n-1)/(Floor[n/k]+1), {k,1,n}]=Floor[8*Product[(k*n-1)/(Floor[n/k]+1), {k,1,n}]]

Sincerely

Sebastian Martin Ruiz

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• Hello all:   I send you a equation whit primes numbers solutions.   Let x an integer x =5 and pi=i-th prime number.   If x is solution of the equation:
Message 3 of 7 , May 12, 2009
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Hello all:

I send you a equation whit primes numbers solutions.

Let x an integer x>=5 and pi=i-th prime number.

If x is solution of the equation:

x = Product[Numerator[(x+pi)/(x-pi)]+Denominator[(x+pi)/(x-pi)],
{i=2 to Pi[x]-1}]^(1/(Pi[x]-2))
Pi[x] is the prime counting function.
Then x IS PRIME; x=p(Pi[x])

Note that i in the primes of the product is to Pi[x]-1
and x is the next prime with i=Pi[x], then we can obtain a recurrent algorithm
to obtain the prime numbers.

Mathematica Software:

F[x_]:=Product[
Numerator[(x+Prime[i])/(x-Prime[i])]+
Denominator[(x+Prime[i])/(x-Prime[i])],
{i,2,PrimePi[x]-1}]^(1/(PrimePi[x]-2))

Do[If[IntegerQ[F[x]],Print[F[x]," ",x]],{x,5,40}]
5   5
4   6
7   7
16   8
11   11
13   13
32   16
17   17
19   19
23   23
29   29
31   31
64   32
37   37

Sincerely

Sebastián Martín Ruiz

PD:  2x=F[x] has solutions powers of 2.

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• Hello:   If Goldbach Conjecture is True then:   For all p prime number p =7 exists q and r also primes such that:   p**2+4(q+r)**2=4p(q+r)+1   (It is easy
Message 4 of 7 , Jun 12, 2009
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Hello:

If Goldbach Conjecture is True then:

For all p prime number p>=7 exists q and r also primes such that:

p**2+4(q+r)**2=4p(q+r)+1

(It is easy to prove)

Sincerely

Sebastian Martin Ruiz

[Non-text portions of this message have been removed]
• Dear Sebastain, Did you mean by  p**2+4(q+r)* *2=4p(q+r) +1 this                    p^2 + 4(q+r)^2 = 4p(q+r) + 1    ??? if so, maybe
Message 5 of 7 , Jun 12, 2009
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Dear Sebastain,

Did you mean by  p**2+4(q+r)* *2=4p(q+r) +1
this                    p^2 + 4(q+r)^2 = 4p(q+r) + 1    ???

if so, maybe you need to clarify that:

r or q can equal p
or
1 is prime

see below:

11^2 + 4(q+r)^2 - 44(q+r) + 1 = 0

let w = q+r

121 + 4w^2 - 44w - 1 = 0

120 + 4w^2 - 44w = 0

divide both sides by 4

30 + w^2 - 11w = 0

w^2 - 11w + 30 = 0
a = 1
b = -11
c = 30

w = [-b +- sqrt(b^2 - 4ac)] / 2a
w = [11 +- sqrt(-11^2 - 4*1*30)] / 2*1
w = [11 +- sqrt(121 - 120)] / 2
w = [11 +- sqrt(1)] / 2
w = [11 +- 1] / 2

w1 = 12 / 2 = 5
w2 = 10 / 2 = 10

so

either:
w1 = q1 + r1
12 = 1 + 11    or    5 + 7
but but q != p and r != p
and 1 is not 1 since 1899
so this option is out

or:
w2 = q2 + r2
10 = 3 + 7   or   5 + 5
but q != p and r != p
and r != q
so this fails too unless you calrify that they can :)

Sorry if misunderstood ** and * * oparators.

Ali
<prime numbers are God's signature>

________________________________
From: Sebastian Martin Ruiz <s_m_ruiz@...>
To: Claudi Alsina <claudio.alsina@...>; Azmy Ariff <azmyarif@...>; Chris Caldwell <caldwell@...>; Pierre Deligne <deligne@...>; Gaussianos <gaussianos@...>; Andrew Granville <andrew@...>; Lista NMBRTHRY <nmbrthry@...>; Antonio Pérez Sanz <aperez4@...>; primenumbers@yahoogroups.com; Carlos Rivera <crivera@...>; CarlosB Rivera <cbrfgm@...>
Sent: Saturday, June 13, 2009 12:05:30 AM

Hello:

If Goldbach Conjecture is True then:

For all p prime number p>=7 exists q and r also primes such that:

p**2+4(q+r)* *2=4p(q+r) +1

(It is easy to prove)

Sincerely

Sebastian Martin Ruiz

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[Non-text portions of this message have been removed]
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