sums of factors in GF(2)

[franktaw at netscape.net]

That is indeed impossible.  If p(x) in GF(2)[x] is irreducible (and not 
x+1), then in particular p(1) = 1, and then p(1) + 1 = 0, so p(x)+x is 
divisible by x+1.

Franklin T. Adams-Watters


-----Original Message-----
From: Antti Karttunen <antti.karttunen at gmail.com>

Marc LeBrun wrote: 
 
> Recently I've been trying to exhort the Sequence Phanatiques to > 
compute analogs of the sum-of-prime-factors (with and without > 
multiplicity) in other arithmetics, such as Gaussian integers, GF(2), > 
etc. 
> 
Don't worry, I will do the GF(2)[X] -thing, in my due time. (This is 
easy to do 
by using the existing the 
http://www.research.att.com/~njas/sequences/a091247.scm.txt 
Analogues for http://www.research.att.com/~njas/sequences/A000203 and 
similar sequences 
as well...) 
 
> I was wondering, specifically about GF(2), summing (ie XORing) the > 
prime factors of N with multiplicity: 
> 
> Noting that only the square-free part of N matters, since the square 
> parts sum to 0... 
> 
> A. Aside from the perfect squares (eg 5) are there any other N that > 
sum to 0? Can they be characterized? 
>
Antti:
 
If we could just find from 
http://www.research.att.com/~njas/sequences/A014580 
two successive terms, with first A014580(k) = 1 mod 4, and the second 
A014580(k+1) = A014580(k)+2, 
then by xoring them we could get 2, and the product A014580(k) x 
A014580(k+1) x 2 
(where x would be GF(2)[X] multiplication: 
http://www.research.att.com/~njas/sequences/A048720 ) 
would satisfy the condition. 
However, skimming cursorily over the first 100 terms of 
http://www.research.att.com/~njas/sequences/A058943 
I didn't find such a pair. (Maybe there's a reason for that? GF(2)[X] 
secrets elide me for a monent...) 
 
Yours, 
 
Antti