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 2016-07-12, 08:55 #34 Godzilla     May 2016 7·23 Posts 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
2016-07-12, 11:11   #35
axn

Jun 2003

23·19·31 Posts

Quote:
 Originally Posted by Godzilla 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.

2016-07-12, 14:24   #36
Godzilla

May 2016

7·23 Posts

Quote:
 Originally Posted by axn 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

 2016-07-12, 14:34 #37 retina Undefined     "The unspeakable one" Jun 2006 My evil lair 5,807 Posts Godzilla: Good job. You managed to completely ignore a proof from axn that your method doesn't work. Currently this is my favourite thread.
2016-07-12, 14:40   #38
Godzilla

May 2016

A116 Posts

Quote:
 Originally Posted by retina 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

2016-07-12, 20:10   #39
Godzilla

May 2016

7×23 Posts

Quote:
 Originally Posted by firejuggler 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.

2016-07-13, 05:37   #40
LaurV
Romulan Interpreter

Jun 2011
Thailand

22×3×739 Posts

Quote:
 Originally Posted by Godzilla 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.

Last fiddled with by LaurV on 2016-07-13 at 05:40

2016-07-13, 06:30   #41
Godzilla

May 2016

7×23 Posts

Quote:
 Originally Posted by LaurV 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

 2016-07-13, 11:36 #42 firejuggler     Apr 2010 Over the rainbow 2·1,217 Posts So the point is not to factor huge number? what is the purpose of your algorithm?
2016-07-13, 11:40   #43
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

100000101100012 Posts

Quote:
 Originally Posted by firejuggler 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.

2016-07-13, 11:51   #44
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

5,807 Posts

Quote:
 Originally Posted by science_man_88 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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