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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).

 ramshanker 2015-11-25 15:07

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

 science_man_88 2015-11-25 15:12

[QUOTE=ramshanker;417237]What is that [B]gp>[/B] console you guys using?[/QUOTE]

[url]http://pari.math.u-bordeaux.fr/[/url]

 Ale 2015-11-25 19:22

(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 .....

 Ale 2015-11-25 19:46

(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

 science_man_88 2015-11-25 19:56

[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 .

 Ale 2015-11-25 20:01

chain of numbers, but I think it is impossible now

 science_man_88 2015-11-25 20:04

[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 ?

 Ale 2015-11-25 20:23

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 .

 science_man_88 2015-11-25 20:42

[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

 Ale 2015-11-25 20:57

only hobby

 science_man_88 2015-11-25 21:15

[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?

 Ale 2015-11-26 10:48

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

 Ale 2015-11-26 10:56

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.

 Ale 2015-11-26 11:20

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 >

 Ale 2015-11-26 11:41

number 12:

*3*7*8*9*3*8*9*11*12
%219 = 258961483526405092926701654192947200000000
(12:38) gp > isprime(%-1)
%220 = 1
(12:38) gp >

 Ale 2015-11-26 12:33

but number 13 is wrong

 science_man_88 2015-11-26 12:37

[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.

 LaurV 2015-11-26 12:43

Move to misc math...

 Ale 2015-11-26 13:10

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 >

 science_man_88 2015-11-26 13:14

[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.

 Ale 2015-11-26 13:37

okay so all I see is random number to me so there's really nothing to program from.

no probem , is impossible , i know.

 Ale 2015-11-26 13:38

i know the symbol

 Ale 2015-11-28 07:46

Riemann:

this is possible:

143 / 0.5 / 0.5 = 572

143 - 1 / 0.5 / 0.5 = 139

 R.D. Silverman 2015-11-28 15:42

[QUOTE=Ale;417459]Riemann:

this is possible:

143 / 0.5 / 0.5 = 572

143 - 1 / 0.5 / 0.5 = 139[/QUOTE]

Idiot

 Ale 2015-11-29 21:52

This is idiot??

0,4545 / 0,5 = 0,909 -------->0000000000........continue

4545 / 0,5 = 9090 ---------> no zero

 Ale 2015-11-29 22:59

or this is idiot?

0,3 --------->3--------->0,30
0,5---------->5--------->0,50
0,7---------->7--------->0,70

0---------0,1--------0,10---------1---------10

 R.D. Silverman 2015-11-29 23:14

[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.

 Uncwilly 2015-11-29 23:27

The crank score is rapidly approaching infinity here.

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