20120311, 14:23  #23 
"Jonathan"
Jul 2010
In a tangled web...
11010110_{2} Posts 
Well, with surprisingly little effort the Aurifeuillian LM calculator is now online! (the system() command being the culprit).
So, presenting with minimalist style this page: (Near the bottom should be the said calculator) 
20120311, 14:50  #24  
"Forget I exist"
Jul 2009
Dumbassville
8,369 Posts 
Quote:


20120311, 22:24  #25 
"Jonathan"
Jul 2010
In a tangled web...
214_{10} Posts 
Yes, random values will likely produce that result. I really should put some educational type info on there.
In brief, your base will have an L,M factorization at every exponent which is an odd multiple of your base (with square parts removed). It will be a '1' if the squarefree part has a remainder of 1 when dividing by 4 else it will be a '+1'. eg. 52^{39}1 is an LM since, squarefree base = 52 / (2^{2}) = 13 exponent is 3 (an odd number) x squarefree base = 13 1 is correct since 13 / 4 = 3 remainder 1 (I'm sure the math types here are having a heart attack looking at my primitive analysis) If that doesn't work, then there really is a problem. 
20120311, 23:01  #26 
"Jonathan"
Jul 2010
In a tangled web...
2×107 Posts 
Oh, I forgot to mention that those rules don't apply to bases that are perfect powers.
eg. base 1000 = 10^{3} is a perfect power. The solution is of course to just convert your number to a nonperfect power. 1000^{50}+1 = (10^{3})^{50}+ 1 = 10^{3x50}+1 = 10^{150}+1 Cheers [Edit: Actually that was a bad example. But if your base was say 100 = 10^{2} then I think you can see there would be a problem.] Last fiddled with by jcrombie on 20120311 at 23:11 Reason: logic error 
20120426, 20:04  #27  
"Jonathan"
Jul 2010
In a tangled web...
2·107 Posts 
Quote:
myfactors.mooo.com Enjoy! 

20120427, 03:32  #28  
Romulan Interpreter
Jun 2011
Thailand
8824_{10} Posts 
mooo dot com blocked by barracuda:
Quote:
Last fiddled with by LaurV on 20120427 at 03:33 

20120427, 03:59  #29 
"Jonathan"
Jul 2010
In a tangled web...
2×107 Posts 
mooo.com is a domain registered to freedns.afraid.org. Anyone is allowed to add their
subdomain onto it for free. (My guess, some malware was also put there on a different subdomain). 
20120711, 03:02  #30 
Noodles
"Mr. Tuch"
Dec 2007
Chennai, India
2351_{8} Posts 
I realize right now that the Aurifeuillian factorizations come out from the polynomial factorization for the values for
For the prime bases b N^{N}x^{2N}1, if b = 1 mod 4 (Totally the forms b = k^{2}p, for the primes p = 1 mod 4?) N^{N}x^{2N}+1, if b = 2, 3 mod 4 (Case elsewhere?) By the way, what are being the best known fastest algorithms for the polynomial factorization, as such? Berlekamp Algorithm? Or which otherwise that works out only inside the finite fields, off actually? This way, away Question For the Cunningham numbers, for the bases b below 12, why does the base 12 alone have the Aurifeuillian coefficient's power to be 6n3, while whereas the rest of the bases have got their respective powers to be precisely 2bnb ? What are the other bases that are likely being to have such a property as this, the left hand side coefficient's power is not being equivalent to the value for 2bnb at all? If so, then what is being the dividing factor? Is it always being likely to be an integer value, again? Is it being the largest square value that divides the base b ? Is it the case that all the square free bases b have got their left hand side powers to be precisely 2bnb to being likely to be the true case? Such as the cases b = 12, 18 where the bases are not being the square free themselves at all. Thinking for it, I hope that it is not! For this example, consider the observation for the base value system for b = 15. It unusually splits out into three parts for the degree 8 coefficient parts. Does that mean that the base 15 has got the Aurifeuillian factors away into the three parts 15A, 15B, 15C as such?! Or otherwise not the case at all!? Last fiddled with by Raman on 20120711 at 04:02 
20120711, 05:45  #31  
Dec 2011
10001111_{2} Posts 
Quote:
I think the previous comments in this thread also should answer many of your questions. But, here are some examples based on some of your specific questions... Among the bases in the Cunningham Tables (2, 3, 5, 6, 7, 10, 11, 12), 12 is the only base that is not square free. If you exclude the "squarefull" part of the base, what is left determines which exponents will have Aurifeuillean Factors. (I.e., will have Aurifeuillean Factors at the same exponents where would have Aurifeuillean Factors. So, if has Aurifeuillean Factors, then also has Aurifeuillean Factors.) Similarly, so have Aurifeuillean Factors, just as and have Aurifeuillean Factors. If a number has Aurifeuillean Factors, by convention, it has exactly two Aurifeuillean Factors. However, for a number that has algebraic factors as well as Aurifeuillean Factors, some of the algebraic factors may interact with the Aurifeuillean Factors. For example, where and But is divisible by which are its Aurifeuillean factors. But, is also divisible by which are its Aurifeuillean factors. But and are all divisible by which are its Aurifeuillean factors. [Thanks to jcrombie's Web Site at http://myfactors.mooo.com/ for serving these Aurifeuillean factors up on a silver platter.] By combining the Aurifeuillean factors of the algebraic factors of the complete factorization is obtained. In general, a number that has two Aurifeuillean Factors will also have a nonAurifeuillean component. Base 2 is a special case where the nonAurifeuillean factor is unity. Your question about is a case that always combines Aurifeuillean Factors with algebraic factors. is always divisible by and Your computergenerated(?) factorization appears to be providing a combination of Aurifeuillean and algebraic factors. Last fiddled with by rcv on 20120711 at 06:00 

20191004, 16:20  #32 
Feb 2017
Nowhere
2^{2}×887 Posts 
Aurifeuillian factorizations of Lucastype numbers...
I have been going over the basics of generalized Lucas and Fibonacci sequences based on the quadratic polynomials
x^2  a*x  1 (a = positive integer) and have gotten to the generalization of the wellknown Aurifeuillian factorization for a = 1, and odd n, As mentioned, e.g. here, it is based on the algebraic identity The hypothesis that n is odd manifests itself with xy being 1 rather than +1. I am sure the following has been well known for a long time, but it proved easier to derive it from scratch than to dig up a reference. If anyone knows a reference for the generalized version, I'd like to know it. If the coefficient a is odd, there will be a similar algebraic identity, with the coefficient 5 above replaced by N = a^{2} + 4, based on the algebraic factorization (x^{2N} + 1)/(x^{2} + 1) over the quadratic field defined by t^{2} + N. The situation is simplest when N is prime. For the case a = 3, N = 13 the appropriate identity for L_{13n}/L_{n}, n odd, is 
20191004, 17:46  #33  
Nov 2003
7232_{10} Posts 
Quote:
in Riesel's book. I believe that Brent wrote a paper on an algorithm to derive the coefficients. 

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