20151124, 14:15  #1 
2^{4}×401 Posts 
This simple algorithm incomplete can only calculate prime numbers?
Ciao,
3 x 3 x 2 x 4 = 72 1 = 71 (is a prime number) now 5: 3 x 3 x 2 x 4 x 3 x 2 x 3 x 4 x 5 = 25920 1 = 25919 (is a prime number) . now 6 : 3 x 3 x 2 x 4 x 3 x 2 x 3 x 4 x 5 x 3 x 2 x 4 x 3 x 4 x 5 x 6 = 223948800 1 = 223948799 ( is a prime number) ho isolato il 2 x 4 now 7 : 3 x 3 x 2 x 4 x 3 x 2 x 3 x 4 x 5 x 3 x 2 x 4 x 3 x 4 x 5 x 6 x 3 x 4 x 5 x 3 x 4 x 5 x 6 x 3 x 5 x 6 x 7 = 3047495270400000 1 = 3047495270399999 ( is a prime number) Do you understand the mechanism? then work to the 8 910 ........ should be all primes larger. 
20151124, 15:57  #2  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:


20151124, 16:28  #3 
"Curtis"
Feb 2005
Riverside, CA
1001111001111_{2} Posts 

20151124, 16:39  #4 
2^{3}×331 Posts 
I'm italian and not speak very well english, but because I can not test them are too large after 7, they have 23 decimals, for example 8

20151124, 17:18  #5  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:
translated with google: finora, ho il secondo è le prime volte quadrato 5, il terzo è il secondo quadrato superiore a 3, il quarto sembra essere la terza tempi quadrato 35/576. così che cosa mi manca? 

20151124, 17:58  #6 
13×347 Posts 
this is 7
3 x 3 x 2 x 4 x 3 x 2 x 3 x 4 x 5 x 3 x 2 x 4 x 3 x 4 x 5 x 6 x 3 x 4 x 5 x 3 x 4 x 5 x 6 x 3 x 5 x 6 x 7 = 3047495270400000 1 = 3047495270399999 now 8 is: 3 x 3 x 2 x 4 x 3 x 2 x 3 x 4 x 5 x 3 x 2 x 4 x 3 x 4 x 5 x 6 x 3 x 4 x 5 x 3 x 4 x 5 x 6 x 3 x 5 x 6 x 7 x3 x 2 x 4 x 5 x 6 x 3 x 4 x 5 x 6 x 3 x ? 6 x 7 x 8 = n  1 = prime number or x ? 5 x 6 x 7 x 8 = n  1 = prime number I can not test the true mechanism , but i think ( you see 4 x 5 x 6 , 3 times , and the number moltiplication 3 always after a series of numbers ) , i think that ,in this mechanism there is peralps hide one algorithm particular for to find a prime numbers. Last fiddled with by Ale on 20151124 at 18:03 
20151124, 18:08  #7  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:
Code:
(13:37) gp > %32*2 * 4 * 5 * 6 * 3 * 4 * 5 * 6 * 3 * 6 * 7 * 8 %39 = 265410020093460480000000 (14:03) gp > for(x=1,8,print(isprime(x*%391))) 0 0 0 0 0 0 0 0 

20151124, 18:14  #8 
15212_{8} Posts 
or this
3047495270400000 x 3 x 4 x 5 x 6 x 3 x 5 x 6 x 7 x 8 = n  1 = prime number 
20151124, 18:18  #9  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:
Code:
(14:04) gp > 3047495270400000*3*4*5*6*3*5*6*7*8 %41 = 5529375418613760000000 (14:16) gp > isprime(%1) %42 = 0 (14:16) gp > factor(%411) %43 = [ 580824613 1] [9519871050323 1] Last fiddled with by science_man_88 on 20151124 at 18:19 

20151124, 23:48  #10  
∂^{2}ω=0
Sep 2002
República de California
2^{2}×3×7×139 Posts 
Quote:
? factor(3*3*2*4*3*2*3*4*5*3*2*4*3*4*5*6*3*4*5*3*4*5*6*3*5*6*7*3*2*4*5*6*3*4*5*6*3*6*7*81) %2 = [71 1] [11214507891272978028169 1] And you still have not given an actual *algorithm* for how you generate these smallnumber product sequences. Please do so  it should only require very rudimentary English. Do it in Italian and then post it here, if you prefer  I'm sure one of our Italianspeaking regular readers could translate it. If by '?' you mean 'I am not sure here', then in fact you have no algorithm, just a vague supposition  in that case, download PARI, learn its basic operations (*,+,, isprime and factor are the main operators and functions you need), and see if you can work out an actual *algorithm* which generates more than 3 or 4 primes in succession. Last fiddled with by ewmayer on 20151124 at 23:49 

20151125, 05:12  #11 
Romulan Interpreter
"name field"
Jun 2011
Thailand
2^{4}×613 Posts 
Rearranging those small numbers, you only have a product of small primorials, n# multiplied by m# etc, and subtract 1.
For small n, m, etc, this has higher chance to generate a prime, because n#1 does not have prime factors under n. As n grows, your chances to find a prime by this method are as much as picking a random odd number and test it if it is prime or not. Therefore is a fallacy. As suggested above, if you have and algorithm, post it in Italian and I can handle the translation (I am Romanian). Last fiddled with by LaurV on 20151125 at 05:14 
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