[math-fun] sums of factors in GF(2)
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. 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? B. If some sum S occurs for any N then it occurs infinitely, for all the square multiples of N. Does every value of S occur?
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?
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
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@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
A. There are other sums to zero. For example, (1+x+x^2) + (1+x^2+x^3) + (1+x+x^4) + (1+x^3+x^4) = 0, and the parenthesized polynomials are all irreducible. I doubt that these can be usefully characterized, but a few points can be noted: 1) Such a term can include x only if it also includes x+1. The terms other than x all have 1 for their constant term, so there must be an even number of them. Except for x+1, all have an odd number of non-zero coefficients, so if x was present and not x+1, the total number of non-zero coefficients would be odd, making the sum non-zero. 2) Likewise, if x+1 is present, x must be. (Substitute x+1 for x in each polynomial.) 3) If x and x+1 are not present, the number of terms must be even; if they are present, it must be odd. At least 4 terms are required. B. Yes. We can get every polynomial of degree <= 1: 0 from the empty set (sum of the prime divisors of 1), 1 from (x) + (1+x), and x and (1+x) are irreducible. Since there is at least one irreducible polynomial of each degree n, which can be added to the polynomials of lower degree to get all polynomials of degree n, by induction every polynomial of degree n occurs. Franklin T. Adams-Watters -----Original Message----- From: Marc LeBrun <mlb@well.com> 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. 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? B. If some sum S occurs for any N then it occurs infinitely, for all the square multiples of N. Does every value of S occur? _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
participants (3)
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Antti Karttunen -
franktaw@netscape.net -
Marc LeBrun