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Old 2019-05-23, 17:53   #1
Alberico Lepore
 
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Default 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?
Attached Files
File Type: c 2_lepore_sieve.c (3.2 KB, 190 views)
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Old 2019-05-23, 21:39   #2
VBCurtis
 
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I think anyone who names something mathematical after themselves should not be taken seriously.
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Old 2019-05-23, 23:34   #3
Alberico Lepore
 
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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 2019-05-23 at 23:38
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Old 2019-05-24, 03:20   #4
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How would you say it differs from a standard sieve of Eratosthenes on arithmetic progressions?
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Old 2019-05-24, 03:35   #5
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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
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Old 2019-05-24, 09:17   #6
Alberico Lepore
 
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Quote:
Originally Posted by CRGreathouse View Post
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
p are primes

**************************************************************************************************

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 2019-05-24 at 09:21 Reason: EDIT
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Old 2019-05-24, 09:23   #7
jnml
 
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Quote:
Originally Posted by Alberico Lepore View Post
if you find the method to factorize these special numbers into O (1) you can find very large prime numbers
Any program that reads some input number n is at least O(log n).
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Old 2019-05-24, 09:37   #8
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Quote:
Originally Posted by Alberico Lepore View Post
p are primes
Quote:
Originally Posted by Alberico Lepore
p=99
99 = 11 * 3 * 3
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Old 2019-05-24, 09:53   #9
Alberico Lepore
 
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Quote:
Originally Posted by lukerichards View Post
99 = 11 * 3 * 3
sorry I posted the wrong file
this is correct
Attached Files
File Type: c 2_lepore_sieve.c (3.2 KB, 186 views)
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Old 2019-05-24, 15:41   #10
Alberico Lepore
 
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which means particular solutions?


https://www.wolframalpha.com/input/?...x+%3D786+mod+x


https://www.wolframalpha.com/input/?...x+%3D306+mod+x
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Old 2019-05-25, 16:48   #11
Alberico Lepore
 
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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

(3465/15-h) 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 2019-05-25 at 17:35
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