- Milton Brown wrote:
>

All of these are algebraic factorizations. Specifically,

> I think that the following is an interesting pattern.

> (It may have been studied before, I would appreciate

> a reference.)

>

> 10^18+1 is divisble by 10^6+1 = 101*9901

> 10^20+1 " " 10^4+1 = 73*137

> 10^24+1 " " 10^8+1 = 17*5882353

> 10^35+1 " " 10^5+1 = 11* 9091

a^(2n+1)+1 == (a+1) * (a^2n - a^(2n-1) + ... + a^2 - a + 1)

For instance, set a=10^5 and n=3:

10^35+1 == (10^5+1) * (10^30-10^25+10^20-10^15+10^10-10^5+1)

Note that a^n+1 (with a >= 2, n >= 2) can only be prime if n is a power

of two, as a direct consequence of the above algebraic factorization;

if n has an odd non-trivial factor k, then a^n+1 has the non-trivial

factor a^(n/k)+1. - At 08:14 PM 8/2/2001 -0700, Milton Brown wrote:
>I think that the following is an interesting pattern.

These are algebraic factors.

>(It may have been studied before, I would appreciate

>a reference.)

>

>10^18+1 is divisble by 10^6+1 = 101*9901

>10^20+1 " " 10^4+1 = 73*137

>10^24+1 " " 10^8+1 = 17*5882353

>10^35+1 " " 10^5+1 = 11* 9091

+---------------------------------------------------+

| "In order to think outside the box one first must |

| know what is inside the box." - Michael Shermer |

| |

| Jud McCranie |

+---------------------------------------------------+ - Jack Brennen wrote

> 10^35+1 == (10^5+1) * (10^30-10^25+10^20-10^15+10^10-10^5+1)

and it has further algebraic factorizations because

x^n + 1 = product_{d|n,d is odd} Phi(2*n/d,x)

which with n=35 gives

(x + 1)*

(x^4 - x^3 + x^2 - x + 1)

(x^6 - x^5 + x^4 - x^3 + x^2 - x + 1)*

(x^24 + x^23 - x^19 - x^18 - x^17 - x^16 + x^14 + x^13 + x^12 + x^11

+ x^10 - x^8 - x^7 - x^6 - x^5 + x + 1)

With x=10, the last algebraic factor is composite.

The Cunningham project thus records:

> 35 (1,5,7) 4147571.265212793249617641

- Yes, yours is a nice analysis.

I thought that an interesting part of the patterns,

was also 9901 (9909901,...) and 9091 (90901,...)

which also quite often are prime. I think David

indicated where these have been studied before.

Milton L. Brown

miltbrown@...

----- Original Message -----

From: "Jack Brennen" <jack@...>

To: <primenumbers@yahoogroups.com>

Sent: Thursday, August 02, 2001 8:26 PM

Subject: Re: [PrimeNumbers] Interesting Pattern

> Milton Brown wrote:

> >

> > I think that the following is an interesting pattern.

> > (It may have been studied before, I would appreciate

> > a reference.)

> >

> > 10^18+1 is divisble by 10^6+1 = 101*9901

> > 10^20+1 " " 10^4+1 = 73*137

> > 10^24+1 " " 10^8+1 = 17*5882353

> > 10^35+1 " " 10^5+1 = 11* 9091

>

> All of these are algebraic factorizations. Specifically,

>

> a^(2n+1)+1 == (a+1) * (a^2n - a^(2n-1) + ... + a^2 - a + 1)

>

> For instance, set a=10^5 and n=3:

>

> 10^35+1 == (10^5+1) * (10^30-10^25+10^20-10^15+10^10-10^5+1)

>

> Note that a^n+1 (with a >= 2, n >= 2) can only be prime if n is a power

> of two, as a direct consequence of the above algebraic factorization;

> if n has an odd non-trivial factor k, then a^n+1 has the non-trivial

> factor a^(n/k)+1.

>

>

> Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com

> The Prime Pages : http://www.primepages.org

>

>

>

> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/

>

> - Milton Brown wrote

> 9901 (9909901,...) and 9091 (90901,...)

The latter is

> which also quite often are prime. I think David

> indicated where these have been studied before.

ID Number: A001562 (Formerly M3767 and N1537)

Sequence: 5,7,19,31,53,67,293,641,2137,3011

Name: (10^p + 1)/11 is prime.

References J. Brillhart et al., Factorizations of b^n +- 1.

in the EIS. But the former does not figure in the

Cunningham project tables for cyclotomic

factors of 10^n +/- 1.

Perhaps you meant

f(n)=(10^(6*n)+1)/(10^(2*n)+1)

f(1)= 9901 is prime

f(2)= 99990001 is prime

f(3)= 999999000001 is prime

f(4)= 9999999900000001 is prime

Puzzle: Find the next prime of this form :-)

David