This simple algorithm incomplete can only calculate prime numbers?
Ciao,
3 x 3 x [B][U]2 x 4 [/B][/U]= 72 1 = 71 (is a prime number) now 5: 3 x 3 x [B][U]2 x 4 [/B][/U]x 3 x [B][U]2[/B][/U] x 3 x 4 x 5 = 25920 1 = 25919 (is a prime number) . now 6 : 3 x 3 x [B][U]2 x 4 [/B][/U]x 3 x [B][U]2[/B][/U] x 3 x 4 x 5 x 3 x [B][U]2 x 4 [/B][/U]x 3 x 4 x 5 x 6 = 223948800 1 = 223948799 ( is a prime number) ho isolato il [B][U]2 x 4 [/B][/U] now 7 : [B][U] [/B][/U]3 x 3 x [B][U]2 x 4 [/B][/U]x 3 x [B][U]2[/B][/U] x 3 x 4 x 5 x 3 x [B][U]2 x 4 [/B][/U]x 3 x 4 x 5 x 6 x 3 x [B][U]4 x 5 [/B][/U]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. 
[QUOTE=Ale;417114]Ciao,
3 x 3 x [B][U]2 x 4 [/B][/U]= 72 1 = 71 (is a prime number) now 5: 3 x 3 x [B][U]2 x 4 [/B][/U]x 3 x [B][U]2[/B][/U] x 3 x 4 x 5 = 25920 1 = 25919 (is a prime number) . now 6 : 3 x 3 x [B][U]2 x 4 [/B][/U]x 3 x [B][U]2[/B][/U] x 3 x 4 x 5 x 3 x [B][U]2 x 4 [/B][/U]x 3 x 4 x 5 x 6 = 223948800 1 = 223948799 ( is a prime number) ho isolato il [B][U]2 x 4 [/B][/U] now 7 : [B][U] [/B][/U]3 x 3 x [B][U]2 x 4 [/B][/U]x 3 x [B][U]2[/B][/U] x 3 x 4 x 5 x 3 x [B][U]2 x 4 [/B][/U]x 3 x 4 x 5 x 6 x 3 x [B][U]4 x 5 [/B][/U]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.[/QUOTE] I like to find patterns and all I see for the most part is that they will always produce numbers that have remainder of 5 when dividing by 6. 
[QUOTE=Ale;417114] then work to the 8 910 ........ should be all primes larger.[/QUOTE]
Why? 
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

[QUOTE=Ale;417133]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[/QUOTE]
so far I have the second is the first squared times 5 the third is the second squared over 3 the fourth appears to be the third squared times 35/576. so what am I missing ? 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? 
this is 7
3 x 3 x [B][U]2 x 4 [/U][/B]x 3 x [B][U]2[/U][/B] x 3 x 4 x 5 x 3 x [B][U]2 x 4 [/U][/B]x 3 x 4 x 5 x 6 x 3 x [B][U]4 x 5 [/U][/B]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 [B][U]2 x 4 [/U][/B]x 3 x [B][U]2[/U][/B] x 3 x 4 x 5 x 3 x [B][U]2 x 4 [/U][/B]x 3 x [U][I][B]4 x 5 x 6[/B][/I][/U] x 3 x [B][U]4 x 5 [/U][/B]x 3 x [U][I][B]4 x 5 x 6[/B][/I][/U] x 3 x 5 x 6 x 7 x3 x [B][U]2 x 4 x 5 x 6[/U][/B] x 3 x [U][B][I]4 x 5 x 6[/I][/B][/U] 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 [U][B]4 x 5 x 6[/B][/U] , 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. 
[QUOTE=Ale;417143]this is 7
3 x 3 x [B][U]2 x 4 [/U][/B]x 3 x [B][U]2[/U][/B] x 3 x 4 x 5 x 3 x [B][U]2 x 4 [/U][/B]x 3 x 4 x 5 x 6 x 3 x [B][U]4 x 5 [/U][/B]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 [B][U]2 x 4 [/U][/B]x 3 x [B][U]2[/U][/B] x 3 x 4 x 5 x 3 x [B][U]2 x 4 [/U][/B]x 3 x [U][I][B]4 x 5 x 6[/B][/I][/U] x 3 x [B][U]4 x 5 [/U][/B]x 3 x [U][I][B]4 x 5 x 6[/B][/I][/U] x 3 x 5 x 6 x 7 x3 x [B][U]2 x 4 x 5 x 6[/U][/B] x 3 x [U][B][I]4 x 5 x 6[/I][/B][/U] 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 [U][B]4 x 5 x 6[/B][/U] , 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.[/QUOTE] I see 4*5*6 4 times. I'm still missing something or I don't see why this should always produce primes in fact I can telll you for ? not more than 8 the first multiplication you suggest never gives a prime: [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[/CODE] 
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 
[QUOTE=Ale;417145]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[/QUOTE] No it isn't. [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] [/CODE] 
[QUOTE=Ale;417143]now 8 is:
3 x 3 x [B][U]2 x 4 [/U][/B]x 3 x [B][U]2[/U][/B] x 3 x 4 x 5 x 3 x [B][U]2 x 4 [/U][/B]x 3 x [U][I][B]4 x 5 x 6[/B][/I][/U] x 3 x [B][U]4 x 5 [/U][/B]x 3 x [U][I][B]4 x 5 x 6[/B][/I][/U] x 3 x 5 x 6 x 7 x3 x [B][U]2 x 4 x 5 x 6[/U][/B] x 3 x [U][B][I]4 x 5 x 6[/I][/B][/U] 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 [U][B]4 x 5 x 6[/B][/U] , 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.[/QUOTE] I assume the '?' in there is a stray character  deleting it and replacing the 'x' with '*', PARI (which is a freely downloadable for you as it was for me, hint, hint) shows this number is composite: ? 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. 
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). 
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