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Old 2016-07-12, 08:55   #34
Godzilla
 
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or test this 21780

101 * 3 = 303
sum = 104

sqrt(21780*303) / 22 = 116
sqrt(21780*(303-99)) / 22 = 95

Last fiddled with by Godzilla on 2016-07-12 at 08:59
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Old 2016-07-12, 11:11   #35
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Quote:
Originally Posted by Godzilla View Post
or test this 21780

101 * 3 = 303
sum = 104

sqrt(21780*303) / 22 = 116
sqrt(21780*(303-99)) / 22 = 95
Code:
10007*3 = 30021
10007+3 = 10010

sqrt(21780*30021) / 22 = 1162
sqrt(21780*(30021-9999)) / 22 = 949
Now, consider the worst-case scenario:

N=2*p
S=p+2 = N/2+2. To find c such that sqrt(c*N)/22 = N/2+2
we get c = 121*N+968. Thus, there is no fixed c that will work for all N.
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Old 2016-07-12, 14:24   #36
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Quote:
Originally Posted by axn View Post
Code:
10007*3 = 30021
10007+3 = 10010

sqrt(21780*30021) / 22 = 1162
sqrt(21780*(30021-9999)) / 22 = 949
Now, consider the worst-case scenario:

N=2*p
S=p+2 = N/2+2. To find c such that sqrt(c*N)/22 = N/2+2
we get c = 121*N+968. Thus, there is no fixed c that will work for all N.
I think i have found c , now listen , my program in c++ in this point "test it"

Code:
if (ii == 0 )

{
// num = (6534 * (inversione(num ) - num ) ) / 22; // produce serie di numeri che si annullano o ritornano , con qualsiasi numero not it
//num = ((15657 * (inversione(num ) - num ) ))/22; // not it
 num = (2178 * (inversione(num ) - num ) ) /22; // produce 2178 o 21....9....78 e l'opposto 6534 o 65...9...34 con qualsiasi numero , test it

//num = (2178 * ( num ) )/22; not it
ii++;
now 30021 * 2178 = 65385738 put it in my program and the result is 21999978 and 65999934

now take 2199978 three 9s not four ok each time a 9 in less

now formula

sqrt (2199978 *30021) / 22 = 11681
sqrt (2199978 * 30021-9999)) / 22 = 9539

take 101 * 3 = 303
303* 2178 = 659934 put it in my program and the result is 219978 and 659934

now take 21978

now formula

sqrt (21978 *303) / 22 = 117
sqrt (21978 * 303-99)) / 22 = 96

Last fiddled with by Godzilla on 2016-07-12 at 14:26
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Old 2016-07-12, 14:34   #37
retina
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Godzilla: Good job. You managed to completely ignore a proof from axn that your method doesn't work.

Currently this is my favourite thread.

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Old 2016-07-12, 14:40   #38
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Quote:
Originally Posted by retina View Post
Godzilla: Good job. You managed to completely ignore a proof from axn that your method doesn't work.

Currently this is my favourite thread.

I understood what he said axn, c is variable but c = ??? , now c = 21…9…..78 three , four five 9s , and with my program i think to know c.
correction : don't work .

Last fiddled with by Godzilla on 2016-07-12 at 15:25
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Old 2016-07-12, 20:10   #39
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Quote:
Originally Posted by firejuggler View Post
One factor HAVE to be close to sqrt(product) otherwise it won't work. wich we could assume to work with RSA-100
But we can say that the confidence interval (with the formula), it only works if the numbers are similar in size, just that.
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Old 2016-07-13, 05:37   #40
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Quote:
Originally Posted by Godzilla View Post
But we can say that the confidence interval (with the formula), it only works if the numbers are similar in size, just that.
How similar? You didn't answer that question yet. Can you demonstrate for any of the two numbers given by CRG in posts #2 and #6?

edit: Or, say for the next number, where the two prime factors are quite close indeed: 15999992799999439999954599999031.
How about 15989999999999988199999999996519 ?

Last fiddled with by LaurV on 2016-07-13 at 05:40
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Old 2016-07-13, 06:30   #41
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Quote:
Originally Posted by LaurV View Post
How similar?
How about 15989999999999988199999999996519 ?
factors are similar size and the sum is an interval of confidence , but more is the size of the numbers and more is the interval of confidence (max min)

%4 =
[3899999999999941 1]
[4100000000000059 1]
(08:08) gp > sqrt(2178*3899999999999941*4100000000000059)/22
%5 = 8482629309359212.291194025494
(08:10) gp > 4100000000000059+3899999999999941
%6 = 8000000000000000
(08:11) gp > 15989999999999988199999999996519-9999999999999999999999999999999
%7 = 5989999999999988199999999996520
(08:22) gp > sqrt(2178*5989999999999988199999999996520)/22
%8 = 5191820489963029.037623624881
(08:23) gp >

min = 5191820489963029
max = 8482629309359212
Sum = 8000000000000000

Last fiddled with by Godzilla on 2016-07-13 at 06:32
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Old 2016-07-13, 11:36   #42
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So the point is not to factor huge number? what is the purpose of your algorithm?
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Old 2016-07-13, 11:40   #43
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Quote:
Originally Posted by firejuggler View Post
So the point is not to factor huge number? what is the purpose of your algorithm?
I see how it might be used to estimate the sum of divisors for aliquot sequences full of semiprimes but that's about all.
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Old 2016-07-13, 11:51   #44
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Quote:
Originally Posted by science_man_88 View Post
I see how it might be used to estimate the sum of divisors for aliquot sequences full of semiprimes but that's about all.
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.

Last fiddled with by retina on 2016-07-13 at 11:52
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