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Old 2016-07-20, 09:21   #45
Godzilla
 
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Quote:
Originally Posted by retina View Post
Not really. It is purely a guesswork and trial-and-error approach. Any numbers it produces have no mathematical rigour whatsoever. And for numbers of unknown character there is no way to know if there are only two divisors, or more -- so you can't apply the results until you already know all the factors.
retina is right but

I found this, now it works with any two numbers, but the product must be three digits.



n=101⋅3=303

n1=303++303=303303

303303/11=n2

sqrt(n2⋅(n−99))/22=n3=( maximum limit)
the 9 in proportion of the product ex. -99 (two digits) because the product is 303 (three digits)
sqrt(n2⋅(n−99−99))/22=n4=(minimum limit)

just that the margin of error is always high.

just that.

Last fiddled with by Godzilla on 2016-07-20 at 09:23
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Old 2016-08-13, 19:06   #46
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I have this now :

P1 greater than P2

( 2178 * ( P1 * P2 ) ) / 22 = N1

99 / (P1 * P2) = N2

N1 / N2 = N3

N3 / (prime number random but only P1 is correct because is the bigger and only it) = N4

N4 / N1 = P2

Example :

10007 * 3 = 30021

( 2178 * ( 10007 * 3 ) ) / 22 = 2972079

99 / 30021 = 0,00329769161

2972079 / 0,003297 ( I have take only six digits ) = 901449499,545

901449499 / 10007 (considered trying, Random P1 ) = 90081,89

90081 / 30021 = 3,000599 and this is P2 = 3

Now it is always difficult but possible.
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Old 2016-08-13, 19:44   #47
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I can't cancel , there is an error , sorry , the operations are too much

Last fiddled with by Godzilla on 2016-08-13 at 20:00 Reason: error
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Old 2016-08-13, 19:59   #48
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Quote:
Originally Posted by Godzilla View Post
I have this now :

P1 greater than P2

( 2178 * ( P1 * P2 ) ) / 22 = N1

99 / (P1 * P2) = N2

N1 / N2 = N3

N3 / (prime number random but only P1 is correct because is the bigger and only it) = N4

N4 / N1 = P2

Example :

10007 * 3 = 30021

( 2178 * ( 10007 * 3 ) ) / 22 = 2972079

99 / 30021 = 0,00329769161

2972079 / 0,003297 ( I have take only six digits ) = 901449499,545

901449499 / 10007 (considered trying, Random P1 ) = 90081,89

90081 / 30021 = 3,000599 and this is P2 = 3

Now it is always difficult but possible.
so you have ( effectively, before floors, ceilings, and I won't follow my own math operations):
99*(P1*P2)=N1
99/(P1*P2)=N2
N1/N2 = N3 = (99*(P1*P2))*((P1*P2)/99) = (P1 ^2)*(P2 ^2)
N4 = P1*(P2 ^2)
P2 = P2/99 ?

what you have in the example is:

y = P1*P2;
99*y = N1;
99/y = N2;
N1/trunc(N2,x)= N3
floor(N3)/P1 =N4
floor(floor(N4)/y) = P2

Last fiddled with by science_man_88 on 2016-08-13 at 20:11
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Old 2016-08-13, 20:32   #49
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But how many steps do you to find the product of 3 * 5 = 15? Starting at the highest number?

13 * 11
13 * 7
13 * 5
13 * 3
11 * 7
11 * 5
11 * 3
7 * 5
7 * 3
5 * 7
5 * 3 and stop

Instead I :

(N3 / 13) / 15
(N3 / 11) / 15
(N3 / 7) / 15
(N3 / 5) / 15 and stop but I have to check with a multiplication but not all operations

Last fiddled with by Godzilla on 2016-08-13 at 20:48
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Old 2016-08-13, 20:50   #50
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I have wrong to write before, this is correct N4 / (P1*P2) = P2

not this N4 / N1 = P2


P1 greater than P2

( 2178 * ( P1 * P2 ) ) / 22 = N1

99 / (P1 * P2) = N2

N1 / N2 = N3

N3 / (prime number random but only P1 is correct because is the bigger and only it) = N4

N4 / (P1 *P2) = P2

Example :

10007 * 3 = 30021

( 2178 * ( 10007 * 3 ) ) / 22 = 2972079

99 / 30021 = 0,00329769161

2972079 / 0,003297 ( I have take only six digits ) = 901449499,545

901449499 / 10007 (considered trying, Random P1 ) = 90081,89

90081 / 30021 = 3,000599 and this is P2 = 3

Last fiddled with by Godzilla on 2016-08-13 at 21:02
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Old 2016-08-13, 21:02   #51
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Quote:
Originally Posted by Godzilla View Post
I have wrong before to write , this is correct N4 / (P1*P2) = P2

not this N4 /N1 = P2


P1 greater than P2

( 2178 * ( P1 * P2 ) ) / 22 = N1

99 / (P1 * P2) = N2

N1 / N2 = N3

N3 / (prime number random but only P1 is correct because is the bigger and only it) = N4

N4 / (P1 *P2) = P2

Example :

10007 * 3 = 30021

( 2178 * ( 10007 * 3 ) ) / 22 = 2972079

99 / 30021 = 0,00329769161

2972079 / 0,003297 ( I have take only six digits ) = 901449499,545

901449499 / 10007 (considered trying, Random P1 ) = 90081,89

90081 / 30021 = 3,000599 and this is P2 = 3
You do realize, if you do a bit of algebra within your steps, you can avoid floor functions right ? I just showed you why. also your whole procedure is dependant on knowing one of the prime factors.
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Old 2016-08-13, 21:08   #52
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Quote:
Originally Posted by science_man_88 View Post
You do realize, if you do a bit of algebra within your steps, you can avoid floor functions right ? I just showed you why. also your whole procedure is dependant on knowing one of the prime factors.

I know this, but do you think the time is lower or higher?

P.S.
But only one factor and the higher factor and test it with multiplication.

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Old 2016-08-13, 21:51   #53
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Quote:
Originally Posted by Godzilla View Post
I know this, but do you think the time is lower or higher?

P.S.
But only one factor and the higher factor and test it with multiplication.
the problem with that PS is if we know the higher factor of a semiprime we also know the lower factor and therefore can do an addition instead of this convoluted procedure.
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Old 2016-08-27, 18:41   #54
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I conclude and resume with the first algorithm of this thread saying that:


\frac{\sqrt{2178*(p1*p2)}}{22,5} = Limit (max)

and

\frac{\sqrt{2178*((p1*p2)-9...9))}}{22,5} = Limit (min)



The algorithm is a conclusion rather trivial because it is found that the sum of two prime numbers similar is "around" the double of their average : p,q with n for p+q \sim = 2n and p \cdot q \sim = n^{2}.



N_{max} = \frac{\sqrt{2178 \cdot n^{2}}}{22,5} \sim \frac{10}{\sqrt{22,5}} \cdot n \sim = 2,108 n > 2n


N_{min} = \frac{\sqrt{2178 \cdot (n^{2} - 10^{\log(n^{2})-1})}}{22,5} < \frac{\sqrt{2178 \cdot 9 \cdot 10^{(\log(n^{2})-1})}}{22,5} \sim = 2,108 \cdot 3 \cdot \sqrt{\frac{n^{2}}{10}}= \frac{2,108 \cdot 3 \cdot n}{3,162} = 2n is equal to n .


P.S.

I replaced 22 with 22,5

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Old 2016-08-27, 19:06   #55
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Quote:
Originally Posted by Godzilla;440870


[B
The algorithm is a conclusion rather trivial because it is found that the sum of two prime numbers similar is "around" the double of their average : p,q with n for p+q \sim = 2n and p \cdot q \sim = n^{2}.
[/B]
p+q = (2*(p+q)/2) is true it's not about and then by that math you get n^2 = ((p+q)^2)/4 = (p^2+2*p*q+q^2)/4 assuming p=2*k+1 and q=2*j+1 we then can first expand the numerator to:

(2*k+1)^2+2*(2*k+1)*(2*j+1)+(2*j+1)^2 = (4*k^2+4*k+1)+(8*j*k+4*j+4*k+2) + (4*j^2+4*j+1) = 4*(k^2+k+2*j*k+j+k+j^2+j+1) which when divided by four gives (k^2+k+2*j*k+j+k+j^2+j) so your second statement is equivalent to 4*j*k+2*j+2*k+1 ~ = (k^2+k+2*j*k+j+k+j^2+j) lets reduce this a bit we can get 2*j*k+1 ~=(k^2+j^2) which uses less variables overall. let's see what you can get out of that first I want to work on other things right now ( like being lazy).

Last fiddled with by science_man_88 on 2016-08-27 at 19:21
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