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• 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 =
Aug 2, 2001 1 of 6
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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

and these patterns seem to repeat often.

Milton L. Brown
miltbrown@...

[Non-text portions of this message have been removed]
• ... 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:
Aug 2, 2001 1 of 6
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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.
• ... These are algebraic factors. +---------------------------------------------------+ ... +---------------------------------------------------+
Aug 2, 2001 1 of 6
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At 08:14 PM 8/2/2001 -0700, 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

These are algebraic factors.
+---------------------------------------------------+
| "In order to think outside the box one first must |
| know what is inside the box." - Michael Shermer |
| |
| Jud McCranie |
+---------------------------------------------------+
• Jack Brennen wrote ... 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 -
Aug 2, 2001 1 of 6
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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
Aug 2, 2001 1 of 6
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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@...>
Sent: Thursday, August 02, 2001 8:26 PM

> 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 ... The latter is ID Number: A001562 (Formerly M3767 and N1537) Sequence: 5,7,19,31,53,67,293,641,2137,3011 Name: (10^p + 1)/11 is
Aug 3, 2001 1 of 6
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Milton Brown wrote

> 9901 (9909901,...) and 9091 (90901,...)
> which also quite often are prime. I think David
> indicated where these have been studied before.

The latter is

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
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