20190523, 17:53  #1 
May 2017
ITALY
2^{2}×127 Posts 
2° Lepore sieve
this sieve riddles in A + B * log_2 (B)
where A is the number of special numbers generated and B the number of discards what do you think? 
20190523, 21:39  #2 
"Curtis"
Feb 2005
Riverside, CA
5·1,013 Posts 
I think anyone who names something mathematical after themselves should not be taken seriously.

20190523, 23:34  #3 
May 2017
ITALY
2^{2}·127 Posts 
I found something really exceptional.
Tomorrow I will find the largest prime number ever. Now in Italy it is late, but I leave you one of the formulas valid for 3 + 20 * h 939=(3+20*h)^2+(30+40*k)*(3+20*h) , (236(26+280*h))/(10*(3+10*h))=k I'm very happy thank you Last fiddled with by Alberico Lepore on 20190523 at 23:38 
20190524, 03:20  #4 
Aug 2006
3×1,993 Posts 
How would you say it differs from a standard sieve of Eratosthenes on arithmetic progressions?

20190524, 03:35  #5 
Aug 2006
13533_{8} Posts 
Is this claiming that 99, 219, 259, 299, 339, 459, 539, 579, 699, 779, 819, 899, 939, and 979 are prime? Or am I misunderstanding?
Code:
Fino a che numero? (max 10000): 1000 0 99 99 219 259 299 339 459 459 459 539 539 579 699 779 819 819 819 819 819 899 939 979 p=19 p=59 p=99 p=139 p=179 p=219 p=259 p=299 p=339 p=379 p=419 p=459 p=499 p=539 p=579 p=619 p=659 p=699 p=739 p=779 p=819 p=859 p=899 p=939 p=979 
20190524, 09:17  #6  
May 2017
ITALY
2^{2}·127 Posts 
Quote:
************************************************************************************************** building a list (already ordered) I can sift the prime numbers in A by eliminating part B * log_2 (B). The list is sorted like this C 6 16 26 36 46 .... Finding the B (waste) C = (B + 5) / 4 Example (99 + 5) / 4 = 26 the rest are primes therefore the computational cost is A = p + B moreover, there is the problem of duplication that I am still studying P.S. if you find the method to factorize these special numbers into O (1) you can find very large prime numbers ********************************************************************************************************************* EDIT: I have solved the problem of duplication Last fiddled with by Alberico Lepore on 20190524 at 09:21 Reason: EDIT 

20190524, 09:23  #7 
Feb 2012
Prague, Czech Republ
181 Posts 

20190524, 09:37  #8 
"Luke Richards"
Jan 2018
Birmingham, UK
2^{5}·3^{2} Posts 

20190524, 09:53  #9 
May 2017
ITALY
2^{2}×127 Posts 
sorry I posted the wrong file
this is correct 
20190524, 15:41  #10 
May 2017
ITALY
111111100_{2} Posts 
which means particular solutions?
https://www.wolframalpha.com/input/?...x+%3D786+mod+x https://www.wolframalpha.com/input/?...x+%3D306+mod+x 
20190525, 16:48  #11 
May 2017
ITALY
2^{2}·127 Posts 
In some cases it is very simple to factor the sieve numbers into polynomial time.
I show you an example N=13899 (N+5)/4=3476 If it is feasible one of the 8 is true so let's say the real one is [(N+5)/4 [((13+20*h)*3+5)/4]] mod (13+20*h) =0 > [3476 [((13+20*h)*3+5)/4]] mod (13+20*h) =0 > (3465 15*h) mod (13+20*h) =0 this is when 3465 divides 15 is the case in which it is factorizable in polynomial time (3465/15h) mod (13+20*h) =0 > (231 h) mod (13+20*h) =0 > 20*(231 h) mod (13+20*h) =0 > (4620 20*h) mod (13+20*h) =0 (4633) mod (13+20*h) =0 GCD(13899,4633)=131 I don't know in which and how many cases this is valid what do you think? Edit: [(N+5)/4 [((3+20*h)*13+5)/4]] mod (3+20*h) =0 [(N+5)/4 [((13+20*h)*3+5)/4]] mod (13+20*h) =0 [(N+5)/4 [((7+20*h)*17+5)/4]] mod (7+20*h) =0 [(N+5)/4 [((17+20*h)*7+5)/4]] mod (17+20*h) =0 [(N+5)/4 [((21+20*h)*39+5)/4]] mod (21+20*h) =0 [(N+5)/4 [((11+20*h)*9+5)/4]] mod (11+20*h) =0 [(N+5)/4 [((9+20*h)*11+5)/4]] mod (9+20*h) =0 [(N+5)/4 [((19+20*h)*21+5)/4]] mod (19+20*h) =0 Last fiddled with by Alberico Lepore on 20190525 at 17:35 
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