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). 
What is that [B]gp>[/B] console you guys using?

[QUOTE=ramshanker;417237]What is that [B]gp>[/B] console you guys using?[/QUOTE]
[url]http://pari.math.ubordeaux.fr/[/url] 
(20:15) gp > 304795270400000*3*4*5*6*3*2*4*5*7*8
%37 = 737360718151680000000 (20:16) gp > isprime(%1) %38 = 1 (20:17) gp > thank you for the program PARI. Now i can test , this is number 8 but ..... 
(20:40) gp > 304795270400000*3*4*5*6*3*2*4*5*7*8*3*5*7*8*3*7*8*9
%67 = 936507100910085734400000000 (20:41) gp > isprime(%1) %68 = 1 (20:42) gp > and this is the number 9 , but is different , is more complex , thinking 
[QUOTE=Ale;417263](20:40) gp > 304795270400000*3*4*5*6*3*2*4*5*7*8*3*5*7*8*3*7*8*9
%67 = 936507100910085734400000000 (20:41) gp > isprime(%1) %68 = 1 (20:42) gp > and this is the number 9 , but is different , is more complex , thinking[/QUOTE] still wondering what the "mechanism" is . 
chain of numbers, but I think it is impossible now

[QUOTE=Ale;417266]chain of numbers, but I think it is impossible now[/QUOTE]
okay how is this chain produced ? and how can you guarantee it's prime ? edit: could the strong law of small numbers be playing a role ? 
the number * 3 *, is always after a series chain of numbers, example 2 * 4 * 5, but at 8 is not yet chain of numbers, see 2 * 4 * 5 * 7 * 8, there isn't 6.
I thought, example 2 * 4 * 5 repeated three times, 2 * 4 repeated two times , an so up . 
[QUOTE=Ale;417272]the number * 3 *, is always after a series chain of numbers, example 2 * 4 * 5, but at 8 is not yet chain of numbers, see 2 * 4 * 5 * 7 * 8, there isn't 6.
I thought, example 2 * 4 * 5 repeated three times, 2 * 4 repeated two times , an so up .[/QUOTE] okay, so what are the specifics of the algorithm, and what is your level of knowledge of: Va bene, quindi quali sono le specifiche dell'algoritmo, e qual è il vostro livello di conoscenza di: 1) Sieve techniques/tecniche di Sieve 2) remainder math/resto matematica 3) Prime numbers/numeri primi 4) Factorials/fattoriali 5) Programming/Programmazione 
only hobby

[QUOTE=Ale;417277]only hobby[/QUOTE]
Okay well without specifics, it's pretty difficult to give a program equivalent to what needs to be done. do you know the fact that all primes greater than 3 must have a remainder of 1 or 5 when divided by 6 ? are you aware of how/why the sieve of Eratosthenes works ? are you aware of other ways to find prime numbers ? Va bene anche senza specifiche, è piuttosto difficile dare un programma equivalente a quello che deve essere fatto. sapete il fatto che tutti i numeri primi superiore a 3 deve avere un resto di 1 o 5 quando diviso per 6? siete consapevoli di come / perché il crivello di Eratostene funziona? siete a conoscenza di altri modi per trovare i numeri primi? 
this is correct
number 8 : (11:44) gp > 30474925270400000*3*2*4*5*3*6*7*8 %148 = 3686246960707584000000 (11:44) gp > isprime(%1) %149 = 1 (11:44) gp > before at number 8 and 9 i have write ,30479 but was wrong 
this is correct
this is number 9: (11:53) gp > 30474925270400000*3*2*4*5*3*6*7*8*3*2*4*5*6*3*7*8*9 %156 = 4012995891304704245760000000 (11:54) gp > isprime(%1) %157 = 1 (11:54) gp > right now. 
numer 10 not
direct number 11 (12:16) gp > 30474925270400000*3*2*4*5*3*6*7*8*3*2*4*5*6*3*7*8*9*3*5*6*7*3*8*9*11 %187 = 6006973289776185691393228800000000 (12:16) gp > isprime(%1) %188 = 1 (12:17) gp > 
number 12:
*3*7*8*9*3*8*9*11*12 %219 = 258961483526405092926701654192947200000000 (12:38) gp > isprime(%1) %220 = 1 (12:38) gp > 
but number 13 is wrong

[QUOTE=Ale;417307]but number 13 is wrong[/QUOTE]
okay so all I see is random number to me so there's really nothing to program from. 
Move to misc math...

But if now *3* get *5* , the number 13 is right
<*7*8*9*3*8*9*11*12*[B][U]5[/U][/B]*2*4*5*6*7*[B][U]5[/U][/B]*11*12*13 %373 = 18663872040715067857413241620994090598400000000000 (14:05) gp > isprime(%1) %374 = 1 (14:05) gp > 
[QUOTE=Ale;417311]But if now *3* get *5* , the number 13 is right
<*7*8*9*3*8*9*11*12*[B][U]5[/U][/B]*2*4*5*6*7*[B][U]5[/U][/B]*11*12*13 %373 = 18663872040715067857413241620994090598400000000000 (14:05) gp > isprime(%1) %374 = 1 (14:05) gp >[/QUOTE] you know you can use the code tags they use the # symbol in the advanced reply setup. 
okay so all I see is random number to me so there's really nothing to program from.
no probem , is impossible , i know. 
i know the symbol

Riemann:
this is possible: 143 / 0.5 / 0.5 = 572 143  1 / 0.5 / 0.5 = 139 
[QUOTE=Ale;417459]Riemann:
this is possible: 143 / 0.5 / 0.5 = 572 143  1 / 0.5 / 0.5 = 139[/QUOTE] Idiot 
This is idiot??
0,4545 / 0,5 = 0,909 >0000000000........continue 4545 / 0,5 = 9090 > no zero 
or this is idiot?
0,3 >3>0,30 0,5>5>0,50 0,7>7>0,70 00,10,10110 
[QUOTE=Ale;417677]This is idiot??
0,4545 / 0,5 = 0,909 >0000000000........continue 4545 / 0,5 = 9090 > no zero[/QUOTE] What you write (and continue to do so) is idiotic. One nonsensical post does not make one an idiot. Doing so repeatedly does. 
The crank score is rapidly approaching infinity here.

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