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-   -   This simple algorithm incomplete can only calculate prime numbers? (https://www.mersenneforum.org/showthread.php?t=20695)

Ale 2015-11-24 14:15

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 -9-10 ........ should be all primes larger.

science_man_88 2015-11-24 15:57

[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 -9-10 ........ 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.

VBCurtis 2015-11-24 16:28

[QUOTE=Ale;417114] then work to the 8 -9-10 ........ should be all primes larger.[/QUOTE]

Why?

Ale 2015-11-24 16:39

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

science_man_88 2015-11-24 17:18

[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?

Ale 2015-11-24 17:58

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.

science_man_88 2015-11-24 18:08

[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*%39-1)))
0
0
0
0
0
0
0
0[/CODE]

Ale 2015-11-24 18:14

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

science_man_88 2015-11-24 18:18

[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(%41-1)
%43 =
[ 580824613 1]

[9519871050323 1]
[/CODE]

ewmayer 2015-11-24 23:48

[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*8-1)
%2 =
[71 1]

[11214507891272978028169 1]

And you still have not given an actual *algorithm* for how you generate these small-number 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 Italian-speaking 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.

LaurV 2015-11-25 05:12

Re-arranging 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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