Prime numbers and primality testing
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Creating primes

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cashogor     Message 1 of 9  May 12, 2004
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Hi all!

(Please forgive my english)

I´ve noticed that given p1, p2,...,pn correlative prime numbers:

p1*p2*p3*....*pi + p(i+1)*p(i+2)*....*pn

usually gives a prime number.

Example:

2*3*5*7*11 + 13*17*19*23*29 = 2803043

I expected to have a result divisible by a 'near' prime such as 31 or 
37, but I obtained a prime number.

Any observations?

N.
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Payam Samidoost     Message 2 of 9  May 12, 2004
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Hi N.

> 2*3*5*7*11 + 13*17*19*23*29 = 2803043 
= 311 * 9013

P.
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cashogor     Message 3 of 9  May 12, 2004
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Ups!!!

2803043 is not prime!!! 

:P

I was looking for the smallest factor of the result to see how 'near' 
it is from the others used in the calculus.

2 + 3 = 5 ---------> 5
2*3 + 5*7 = 41 -----------> 41
2*3*5 + 7*11*13 = 1031 -----------> 1031
2*3*5*7 + 11*13*17*19 = 46399 -----------> 46399
2*3*5*7*11 + 13*17*19*23*29 = 2803043 ---------> 311*9013

N.
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Payam Samidoost     Message 4 of 9  May 12, 2004
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for each n<=17
I have written n followed by the factorization of
(p[1]*...*p[n])+(p[n+1]*...*p[2n])

(1) 5
(2) 41
(3) 1031
(4) 46399
(5) 311 * 9013
(6) 127 * 1945991
(7) 137 * 187060999
(8) 179 * 37897 * 495289
(9) 181 * 2904630622811
(10) 98299 * 358373 * 2448049
(11) 501923753 * 31963547227
(12) 10499 * 305078184752140339
(13) 619 * 827 * 1495106677900766629
(14) 18016051 * 10888330227638071729
(15) 571252434769 * 89991171579989521
(16) 8353 * 84982897 * 181369373 * 125339844677
(17) 631 * 1317703 * 6264170627636194115699149

It is prime for n=1,2,3,4,24,25,45,59
and no mor for n<100

further observations:

(a) the numbers have no small factor.(small means less than p[2n]).
(b) they are pairwise coprime
(c) they are squarefree

Proof is needed.

Payam
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David Cleaver     Message 5 of 9  May 12, 2004
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How about when you just use + and * operations on the first k primes, using
one plus operator and the rest as multiplications?  Like this:

2 + 3 = 5
2*3 + 5 = 11
2 + 3*5 = 17
2*3*5 + 7 = 37
2*3 + 5*7 = 41
2 + 3*5*7 = 107
2*3*5 + 7*11 = 107  (the rest with 5 primes are all composite)
2*3*5*7 + 11*13 = 353
2*3*5 + 7*11*13 = 1031
2*3 + 5*7*11*13 = 5011
2 + 3*5*7*11*13 = 15017
2*3*5*7*11*13 + 17 = 30047
2*3*5*7*11 + 13*17 = 2531
2*3*5 + 7*11*13*17 = 17047
2*3 + 5*7*11*13*17 = 85091
...

I guess we could check with all combinations of addition and
multiplication.  [obviously we coulnt' have ALL multiplications cuz that
would negate what this group is all about ;) ]

2 + 3 = 5
2 + 3 + 5 = 10 (no good)
2 * 3 + 5 = 11
2 + 3 * 5 = 17
2 + 3 + 5 + 7 = 17
2 + 3 + 5 * 7 = 40 (no good)
2 + 3 * 5 + 7 = 24 (no good)
2 + 3 * 5 * 7 = 107
2 * 3 + 5 + 7 = 18 (no good)
2 * 3 + 5 * 7 = 41
2 * 3 * 5 + 7 = 37
...

So, to name these, we could call them N_k [ N for the creator, k for the
number of primes] so that the above list could be written as:
N_2(+) = 5
N_3(+,+) = 10 (no good)
N_3(*,+) = 11
N_3(+,*) = 17
N_4(+,+,+) = 17
... (or maybe use 1 for addition and 0 for multiplication) ...
N_2(1) = 5
N_3(11) = 10 (no good)
N_3(01) = 11
N_3(10) = 17
N_4(111) = 17
... (or maybe now, convert the 1's and 0's from binary to decimal #'s) ...
N_2(1) = 5
N_3(3) = 10 (no good)
N_3(1) = 11
N_3(2) = 17
N_4(7) = 17
...

This way, if anyone was interested in searching these types of numbers, we
could all have an easy system to reference these numbers.  Just an idea I
had.  Its late and I'm about to fall asleep, so if this is all just
jibba-jabba, then feel free to ignore me.

-David C.

cashogor wrote:
 > 
> Ups!!!
> 
> 2803043 is not prime!!!
> 
> :P
> 
> I was looking for the smallest factor of the result to see how 'near'
> it is from the others used in the calculus.
> 
> 2 + 3 = 5 ---------> 5
> 2*3 + 5*7 = 41 -----------> 41
> 2*3*5 + 7*11*13 = 1031 -----------> 1031
> 2*3*5*7 + 11*13*17*19 = 46399 -----------> 46399
> 2*3*5*7*11 + 13*17*19*23*29 = 2803043 ---------> 311*9013
> 
> N. 
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Payam Samidoost     Message 6 of 9  May 12, 2004
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> (a) the numbers have no small factor.(small means less than p[2n]).

> Proof is needed.

It is trivial!

If p<p[2n] divides p[1]*...*p[n]+p[n+1]*...*p[2n], since it divides one side
of the +, it must divide the other side which is impossible. QED

Payam
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Payam Samidoost     Message 7 of 9  May 12, 2004
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The same proof works as well for the numbers proposed by David Cleaver.

Payam
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Payam Samidoost     Message 8 of 9  May 12, 2004
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> (p[1]*...*p[n])+(p[n+1]*...*p[2n])

> (b) they are pairwise coprime

> Proof is needed.

It is NOT true:

Let f(n)=(p[1]*...*p[n])+(p[n+1]*...*p[2n])
Then
1283 divides both f(57) and f(80)
3221 divides both f(81) and f(109)
14939 divides both f(241) and f(242)
5569 divides both f(164) and f(262)

Payam
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Jens Kruse Andersen     Message 9 of 9  May 12, 2004
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Payam Samidoost wrote:
> (p[1]*...*p[n])+(p[n+1]*...*p[2n])
> It is prime for n=1,2,3,4,24,25,45,59
> and no mor for n<100

Primeform/GW can search much deeper with a simple input file:

ABC2 p($a)#+p(2*$a)#/p($a)#
a: from 1 to 1600

pfgw agrees: It is 3-prp for n=1,2,3,4,24,25,45,59
... and for n=1238. That number is 10091# + 22091#/10091# (5193 digits).
Not much of a prime creator.
There is nothing unusual about n=1,2,3,4 when they are small numbers without the
smallest prime factors.

--
Jens Kruse Andersen
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