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 2018-05-03, 00:14 #1 marcusvdl   "Marcus Vinicius" May 2018 Brazil 22 Posts Beginner prime search Friends, first apologize my english, I'm better listening or reading, not so good in speach and write. And I'm not a mathematician, so I very probably will say something stupid/dumb. I recently have interested in the Prime Searchs because the challenge and the beauty from mathematics. I don't believe the prime numbers can be aleatory, the pattern maybe be found in future. In my reasearch, I came to any "functions" from type "may be prime". I want to keep the search, but I want and need opinion so I'll know if I'm in the right way. What is the good "may be prime" functions hit percentage? Because of course many functions probably will found some amount of primes, so what is the percentage that we see the primes find was not aleatory? I don't wana waste anybody's time, but someone can tell if think that is a good search? There is a computer program which allow me to do calculations using a function and return to me the results like of the numbers generated how many is prime? My functions: 1) x^2+3x+1 Sequence: 5, 11, 19, 29, 41, 55, 71, 89, 109, 131... Obs.: the multiples of five, except 5, obviously can be excluded Obs. 2: example of a probably prime number with 100 million digits: x=10^50million, the result will be 1(49millionZeros)3(49millionZeros)1 2) x^2-1x+1 Sequence: 1, 3, 7, 13, 21, 31, 43, 57, 73, 91... Obs.: not so good 3) 2x^2-2x+1 Sequence: 1, 5, 13, 25, 41, 61, 85, 113, 145, 181... Obs.: the multiples of five, except 5, obviously can be excluded 4) 0,5x^2-0,5x+1 Sequence: 1, 2, 4, 7, 11, 16, 22, 29, 37, 46... Obs.: generated many even numbers, so have to exclude the multiples of two, except 2, but the results in odd numbers is good And now one not prime function: 1) 0,5x^2+0,5x Sequence: 1, 3, 6, 10, 15, 21, 28, 36, 45, 55... Obs.: until where I look (x<50) just 3 was prime, and i recognize that a function to found composite numbers isn't so interesting I appreciate any help and sorry again if I'm posting something very basics and without a good mathematics base. Thanks, Marcus
 2018-05-03, 04:18 #2 CRGreathouse     Aug 2006 2×11×271 Posts A great deal is known about the number of primes in quadratic polynomials, although proofs have been hard to come by. I suggest looking up G. H. Hardy and J. E. Littlewood, "Some Problems of 'Partitio Numerorum.' III. On the Expression of a Number as a Sum of Primes." Acta Math. 44, 1-70, 1923. and you can find dozens or hundreds of others citing it.
2018-05-03, 15:11   #3
Dr Sardonicus

Feb 2017
Nowhere

22×1,049 Posts

A lot can be told from the quadratic discriminant, because this pretty much determines which primes can divide the function values. Discriminants which exclude a number of small primes as divisors result in quadratic polynomials that may appear "richer" in prime values.

Quote:
 Originally Posted by marcusvdl [snip] My functions: 1) x^2+3x+1 [snip]
D = 3^2 - 4*1*1 = 5. (D/p) = 0 for p = 5, divisible by 5 for x == 1 (mod 5); (D/p) = + 1 for p == 1 or 9 (mod 10). Values never divisible by the small primes 2, 3, or 7. Divisible by 5 for x == 1 (mod 5).

Quote:
 2) x^2-1x+1 [snip
D = (-1)^2 - 4*1*1 = -3. (D/p) = 0 for p = 3; divisible by 3 for x == 2 (mod 3); (D/p) = + 1 for p == 1 (mod 6). Values never divisible by the small primes 2, 5, or 11.
Quote:
 3) 2x^2-2x+1 [snip]
D = (-2)^2 - 4*2*1 = -4; (D/p) = 0 for p = 2, values never even. (D/p) = +1 for p == 1 (mod 4). values never divisible by small primes 2, 3, or 7.

Quote:
 4) 0,5x^2-0,5x+1 [snip]
D = (-1/2)^2 - 4*(1/2)*1 = -7/4; (D/p) undefined for p = 2; even for x == 2, 3 (mod 4). (D/p) = 0 for p = 7; divisible by 7 for x == 4 (mod 7). (D/p) = + 1 for p == 1, 2, 4 (mod 7). Never divisible by the small primes p = 3, 5, or 13.

Quote:
 And now one not prime function: 1) 0,5x^2+0,5x [snip]
D = (1/2)^2 - 4*(1/2)*0 = (1/2)^2, a square. Algebraic factorization.

If x is even, (x/2) * (x + 1); both factors > 1 for even x > 2.

If x is odd, x * (x + 1)/2; both factors > 1 for odd x > 1.

Last fiddled with by Dr Sardonicus on 2018-05-03 at 15:13 Reason: Fixing typos

 2018-05-03, 22:54 #4 marcusvdl   "Marcus Vinicius" May 2018 Brazil 410 Posts Thanks for the tip CRGreathouse! I found it and I'll read for learn more about prime numbers.. And thanks Dr Sardonicus, now I understand that really the quadratic discriminant had a big responsibility in make the function appear as a great prime searcher... So, as I said, I'll learn more and keep searching for good functions, but now not just with the quadratic discriminant to help improve results, I'll try found something diferent to add in the function and make it a better primes founder Last fiddled with by marcusvdl on 2018-05-03 at 22:55 Reason: complementing
2018-05-04, 00:00   #5
sweety439

Nov 2016

22×691 Posts

Quote:
 Originally Posted by marcusvdl Friends, first apologize my english, I'm better listening or reading, not so good in speach and write. And I'm not a mathematician, so I very probably will say something stupid/dumb. I recently have interested in the Prime Searchs because the challenge and the beauty from mathematics. I don't believe the prime numbers can be aleatory, the pattern maybe be found in future. In my reasearch, I came to any "functions" from type "may be prime". I want to keep the search, but I want and need opinion so I'll know if I'm in the right way. What is the good "may be prime" functions hit percentage? Because of course many functions probably will found some amount of primes, so what is the percentage that we see the primes find was not aleatory? I don't wana waste anybody's time, but someone can tell if think that is a good search? There is a computer program which allow me to do calculations using a function and return to me the results like of the numbers generated how many is prime? My functions: 1) x^2+3x+1 Sequence: 5, 11, 19, 29, 41, 55, 71, 89, 109, 131... Obs.: the multiples of five, except 5, obviously can be excluded Obs. 2: example of a probably prime number with 100 million digits: x=10^50million, the result will be 1(49millionZeros)3(49millionZeros)1 2) x^2-1x+1 Sequence: 1, 3, 7, 13, 21, 31, 43, 57, 73, 91... Obs.: not so good 3) 2x^2-2x+1 Sequence: 1, 5, 13, 25, 41, 61, 85, 113, 145, 181... Obs.: the multiples of five, except 5, obviously can be excluded 4) 0,5x^2-0,5x+1 Sequence: 1, 2, 4, 7, 11, 16, 22, 29, 37, 46... Obs.: generated many even numbers, so have to exclude the multiples of two, except 2, but the results in odd numbers is good And now one not prime function: 1) 0,5x^2+0,5x Sequence: 1, 3, 6, 10, 15, 21, 28, 36, 45, 55... Obs.: until where I look (x<50) just 3 was prime, and i recognize that a function to found composite numbers isn't so interesting I appreciate any help and sorry again if I'm posting something very basics and without a good mathematics base. Thanks, Marcus
x^2+3x+1 have many numbers which are multiples of 5, thus can be excluded, similarly, x^2-1x+1 have many numbers which are multiples of 3, thus can also be excluded.

Besides, if Bunyakovsky's conjecture is true, then every polynomial a0+a1*x+a2*x^2+a3*x^3+... have infinitely many prime values for positive integer inputs, if this polynomial is irreducible over the integers and the values of this polynomial are relatively prime as x runs over the integers.

 2018-05-04, 00:29 #6 science_man_88     "Forget I exist" Jul 2009 Dumbassville 100000110000002 Posts Just my random addition, you can show certain integer polynomials will always be even or always be odd.
 2018-05-04, 00:47 #7 marcusvdl   "Marcus Vinicius" May 2018 Brazil 1002 Posts Yes, and this is not necessarily found something like a "pattern".
2018-05-04, 01:05   #8
sweety439

Nov 2016

1010110011002 Posts

Quote:
 Originally Posted by science_man_88 Just my random addition, you can show certain integer polynomials will always be even or always be odd.
Yes, like cyclotomic polynomials, if Bunyakovsky's conjecture is true, then for every integer n>=1, there are infinitely many integers k>=2 such that Phi_n(k) is prime, see the thread http://mersenneforum.org/showthread.php?t=23313.

If n is not of the form 2^r with integer r>=0, then Phi_n(k) is always odd.

Last fiddled with by sweety439 on 2018-05-04 at 01:05

2018-05-04, 01:08   #9
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26·131 Posts

Quote:
 Originally Posted by marcusvdl Yes, and this is not necessarily found something like a "pattern".
Fine with that : http://mathworld.wolfram.com/Prime-G...olynomial.html

2018-05-04, 01:26   #10
marcusvdl

"Marcus Vinicius"
May 2018
Brazil

22 Posts

Quote:
 Originally Posted by science_man_88 Fine with that : http://mathworld.wolfram.com/Prime-G...olynomial.html
Great! This is amazing!

But as "Goldbach showed that no polynomial with integer coefficients can give a prime for all integer values" so we maybe have to search with polynomials plus something else...

Riemann zeta values also are very interesting.

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