20210716, 13:03  #1 
May 2017
ITALY
2^{3}·3^{2}·7 Posts 
Lepore factorization nr. 105 (Bruteforce)

20210717, 07:20  #2 
May 2017
ITALY
504_{10} Posts 
ERRATA CORRIGE
[2*(x1)*((x1)+1)] < [2*x*(x+1)y*(y1)/2] <= [2*(x)*(x+1)] 1 <= y < (sqrt(32*x+1)+1)/2 
20210717, 19:29  #3 
May 2017
ITALY
2^{3}×3^{2}×7 Posts 
Code:
/* This algorithm is generic and does not exploit that q / p < 2 Plus it uses a single A and not many A */ A=9+16*a;//choose A with many small factors if(M % 4 ==1){ M=3*M; } if((M % 8 == 3){ N=M; }else{ N=5*M; } while(1){ if([1/4*(sqrt(N+1)+2)] != (int)[1/4*(sqrt(N+1)+2)]){ x=(int)[1/4*(sqrt(N+1)+2)]; }else{ x=[1/4*(sqrt(N+1)+2)]1; } P=4*x+2sqrt[4*(2*x+1)^2N]; Q=N/P; if (P is integer && (P % M) !=0 && (Q % M) !=0){ breack; } N=N*A } p=GCD(P,M); Last fiddled with by Alberico Lepore on 20210717 at 19:37 Reason: } 
20210720, 12:15  #4 
May 2017
ITALY
2^{3}×3^{2}×7 Posts 

20210722, 11:25  #5 
May 2017
ITALY
111111000_{2} Posts 
the new theory applied to Lepore Factorization nr. 105
If the theory is correct to factor RSA with q / p <2 it would seem O ((log_2 (n)) ^ 2) but I am studying how to implement it in O (2 * (log_2 (n))),after lunch I will study this third hypothesis and the implementation tonight 
20210724, 21:10  #6 
May 2017
ITALY
2^{3}·3^{2}·7 Posts 
Hello
Unfortunately O((log_2(n))^2) does not work O(2*(log_2(n))) does not work It would seem that O(K*[sqrt(8*sqrt(n)31)1]/16) work K depends on the number n I still don't know the order of size of K for example for n=390644893234047643 > K=4 
20210725, 10:14  #7 
May 2017
ITALY
2^{3}·3^{2}·7 Posts 
with a parallel distribution on many computers, finding p and q takes as long as possible
Example on 10 computers for n = 390644893234047643 would take 16150 cycles or 1615 for each computer Last fiddled with by Uncwilly on 20210730 at 21:36 Reason: Removed unneeded self quote of immediately preceding post. 
20210728, 20:02  #8 
May 2017
ITALY
111111000_{2} Posts 
use m not 2*m here https://www.academia.edu/50318218/Cr...zation_example
15 digit 3194383 12 digit 63245 9 digit 3195 p= 9 digit [2*(h1)*((h1)+1)] < [25000000625000001/2 (h  x) (1 + h  x + 2 y)+2 (h  x) (1 + h + x)] <= [2*(h)*(h+1)] , 2*(x)*(x+1)y*(y1)/2=2500000062500000 , h=62500001 37500000<=x<37503195 > 3195 p=12 digit [2*(h1)*((h1)+1)] < [25000000000625000000001/2 (h  x) (1 + h  x + 2 y)+2 (h  x) (1 + h + x)] <= [2*(h)*(h+1)] , 2*(x)*(x+1)y*(y1)/2=2500000000062500000000 , h=12500000000 37500000000<=<<37500063245 > 63245 P=15 digit [2*(h1)*((h1)+1)] < [25000000000000625000000000001/2 (h  x) (1 + h  x + 2 y)+2 (h  x) (1 + h + x)] <= [2*(h)*(h+1)] , 2*(x)*(x+1)y*(y1)/2=2500000000000062500000000000 , h=62500000000001 37500000000000<=x<37500003194383 > 3194383 
20210729, 19:25  #9 
May 2017
ITALY
2^{3}×3^{2}×7 Posts 
@CRGreathouse I know you don't talk to me anymore but I have achieved an extraordinary result with your number
N=390644893234047643 sqrt(390644893234047643/(15/10))=a , (15/10*a+a4)/8=x , 2*x*(x+1)y*(y1)/2=(3906448932340476433)/8 > y=63790420,........ [2*(h)*(h1)] < [(3906448932340476433)/8+k*(k1)/2] <= [2*(h)*(h+1)] , 2*(x)*(x+1)y*(y1)/2=(3906448932340476433)/8 , x(sqrt(32*x+1)+1)/2 <h<x+(sqrt(32*x+1)+1)/2 , k=63790420+j*100000 for j=25 see please range h here https://www.wolframalpha.com/input/?...00000%2Cj%3D25 total cost for factorize N=390644893234047643 is 25*10=250 step 
20210730, 09:10  #10  
May 2017
ITALY
2^{3}×3^{2}×7 Posts 
Quote:
yes, I'll try to explain myself better knowing that the ratio q / p <2 then I will test: q / p > 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 and I will execute them at the same time so it will be the actual time for 10 we consider a 30digit number with p and q also not prime numbers 188723059539473758658629052963=323456789054341*583456789054343 q/p=1,8038167965499768547404880957269 N=188723059539473758658629052963 , sqrt(N/(18/10))=a , (18/10*a+a4)/8=x , 2*x*(x+1)y*(y1)/2=(N3)/8 > y=64759908643727,........ N=188723059539473758658629052963 , 2*(h)*(h1)<(N3)/8+k*(k1)/2<=2*(h)*(h+1) , 2*(x)*(x+1)y*(y1)/2=(N3)/8 , x(sqrt(32*x+1)+1)/2<h<x+(sqrt(32*x+1)+1)/2 , k=64759908643727+2399*100000000 I still can't establish the exact size order of the number in red, it would appear from the first tests to be 10 ^ [((digit p) +1) / 2] , but I'm still studying this number. In theory, if the above were confirmed, since k <= y <p with y being the order size of p1 and the first digit of y is given by the 10 ratios we will have our solution in 10 * {[[ digit p] 1] 1  [((digit p) +1) / 2]} I repeat still do not know well the number in red. 

20210730, 11:01  #11 
May 2017
ITALY
2^{3}×3^{2}×7 Posts 
maybe i quantified the r number in red
red value = r N=188723059539473758658629052963 , sqrt(N/(18/10))=a , (18/10*a+a4)/8=x , 2*x*(x+1)y*(y1)/2=(N3)/8 , (sqrt(32*x+1)+1)/2=b , b*(b1)/2(sqrt(32*(xb)+1)+1)/2*[(sqrt(32*(xb)+1)+1)/21]/2=r r=120441770,...... N=188723059539473758658629052963 , 2*(h)*(h1)<(N3)/8+k*(k1)/2<=2*(h)*(h+1) , 2*(x)*(x+1)y*(y1)/2=(N3)/8 , x(sqrt(32*x+1)+1)/2<h<x+(sqrt(32*x+1)+1)/2 , k=64759908643727+1992*120441770 Last fiddled with by Uncwilly on 20210730 at 21:37 Reason: Removed unneeded self quote of immediately preceding post. 
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