20160712, 08:55  #34 
May 2016
7·23 Posts 
or test this 21780
101 * 3 = 303 sum = 104 sqrt(21780*303) / 22 = 116 sqrt(21780*(30399)) / 22 = 95 Last fiddled with by Godzilla on 20160712 at 08:59 
20160712, 11:11  #35  
Jun 2003
2^{3}·19·31 Posts 
Quote:
Code:
10007*3 = 30021 10007+3 = 10010 sqrt(21780*30021) / 22 = 1162 sqrt(21780*(300219999)) / 22 = 949 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. 

20160712, 14:24  #36  
May 2016
7·23 Posts 
Quote:
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 take 2199978 three 9s not four ok each time a 9 in less now formula sqrt (2199978 *30021) / 22 = 11681 sqrt (2199978 * 300219999)) / 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 * 30399)) / 22 = 96 Last fiddled with by Godzilla on 20160712 at 14:26 

20160712, 14:34  #37 
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. 
20160712, 14:40  #38  
May 2016
A1_{16} Posts 
Quote:
correction : don't work . Last fiddled with by Godzilla on 20160712 at 15:25 

20160712, 20:10  #39 
May 2016
7×23 Posts 

20160713, 05:37  #40  
Romulan Interpreter
Jun 2011
Thailand
2^{2}×3×739 Posts 
Quote:
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 20160713 at 05:40 

20160713, 06:30  #41 
May 2016
7×23 Posts 
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 > 159899999999999881999999999965199999999999999999999999999999999 %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 20160713 at 06:32 
20160713, 11:36  #42 
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?

20160713, 11:40  #43 
"Forget I exist"
Jul 2009
Dumbassville
10000010110001_{2} Posts 

20160713, 11:51  #44 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
5,807 Posts 
Not really. It is purely a guesswork and trialanderror 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 20160713 at 11:52 
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