20100106, 14:44  #12  
Oct 2004
Austria
7×353 Posts 
Quote:
Code:
c6: 1 c5: 0 c4: 15 c3: 20 c2: 30 c1: 18 c0: 5 Y0: ? Y1: ? How do I find the rational poly? (P.S.: No, I don't plan to factor this fibonacci number) 

20100106, 14:45  #13 
(loop (#_fork))
Feb 2006
Cambridge, England
2^{4}×5×79 Posts 
Y0=u and Y1=v as defined in the code above, I think. Perhaps you have to swap them round and fiddle with the signs.

20100106, 14:47  #14 
Oct 2004
Austria
7×353 Posts 

20100106, 18:47  #15  
Noodles
"Mr. Tuch"
Dec 2007
Chennai, India
3×419 Posts 
Quote:
For example, take up with . For a fifth degree polynomial for this number only: we multiply by 12: in order to get Dividing by to get Multiplying with , Now that I can take up with in order to get the algebraic polynomial as and then the rational polynomial as Otherwise, it can be also taken up as well as: with too This is for the Homogeneous Cunningham numbers part only.  For Lucas Numbers & Fibonacci Numbers, becomes . Since itself is only , which contains the surd , which cannot be written as a polynomial with integral coefficients at all purely, for as a whole, such as . Thus, it is not close to any power of integer at all. Thanks, that your post gave me up with the idea. Quote:
I am not going to factor this Fibonacci number right now anyway, however that I should learn about it up actually, properly. Fibonacci/Lucas numbers, Homogeneous Cunningham numbers, even Aliquot sequences have plenty of small numbers, with small factors that pop up so frequently that it is too difficult to manage up with my larger resources itself. A waste of computing power. Most of the time, I stay away from my resources only. Automation of each job takes up a lot of time to set up. Cunningham Numbers have the higher priority among me, harder numbers, larger factors, ECM being eliminated away, only a handful of numbers. Why two variables? SNFS polynomials, both rational side as well as algebraic side allow only one variable? I realized that dividing throughout by only, we can substitute up with , such that the rational polynomial becomes . I realized about that before your subsequent post itself. I will look up at Perrin number at Wikipedia, try out writing up larger Lucas/Fibonacci number as closer to power of smaller Lucas/Fibonacci numbers respectively. Then, that all becomes clear. How to write up in terms of and then only, etc. Quote:
Thanks to you. Please, answer to questions within the post #8 as well. By the way, @ fivemack: How much is the linear algebra for M941 completed up till now? What is the status so far? What is the expected completion date, if any predictions only... Last fiddled with by Raman on 20100106 at 19:29 

20100106, 19:10  #16  
"Robert Gerbicz"
Oct 2005
Hungary
2601_{8} Posts 
Quote:


20100106, 19:27  #17 
Noodles
"Mr. Tuch"
Dec 2007
Chennai, India
3×419 Posts 
I have understood on how to derive the Aurifeuillian factorizations for any base, by just simply glancing up through the Cunningham book only...
Substituting up with , the right hand side is a difference of two squares: we can make use of the identity In this case, besides , this works out for as well. Since , since 4 is a perfect square, thus remains as a perfect square even if only. Thus, I noticed up that the interval between Aurifeuillian factorizations is retained up as only, for any base For example, in that above example, the interval between Aurifeuillian factorizations of 3 is 6, for 12 it is 6 as well, but for all other bases N other than 12, it is only 2N. Similar to what you said out for Aurifeuillian factorizations for minus side: if the base p = 1 (mod 4), other bases of the form as well. Here, is it necessary that value of p is prime, or any value of p with no factor of any perfect squares at all? I tried out to derive up with the Aurifeuillian factorization of base 5 as well, only: Substituting up with within that above equation, I get up that Here, that the expression is of the form only, with which is not a perfect square at all. The other terms within the box brackets are perfect squares only, hence the terms within the box brackets cannot be written as difference of two perfect squares at all. How to write it up in that way, then factorize up with only, accordingly, thus? Note that, for the term , in order to be a perfect square, that should be replaced up with , even then the entire expression within the box brackets becomes the sum of two squares, but not the difference of two squares at all. Where did I go wrong? 
20100106, 20:39  #18 
(loop (#_fork))
Feb 2006
Cambridge, England
18B0_{16} Posts 
(5^5*x^101) is divisible by (5*x^21), naturally
The ratio is 625*x^8 + 125*x^6 + 25*x^4 + 5*x^2 + 1 which it turns out is but I did this backwards  I got pari to factor 5^5*x^101 into irreducibles, and noticed that the two quartic factors had a nice plusandminus form. Last fiddled with by fivemack on 20100106 at 20:40 
20100327, 02:49  #19 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
21643_{8} Posts 
an excercise in triviality
There are fairly many 2L and 2M, 3L and 3M prime numbers. They can only be prime for prime 2k1, because if 2k1=pq (with odd p), then they have algebraic factorizations. The situation is different for 5L/M, 7L/M, 11L/M. They also have algebraic factorizations when 2k1=pq but unlike 2L/M, 3L/M, this leads to a Fermatlike restriction 2k1=b^{N} ("their exponent already has a factor of b so they can only stack them up", one can say simplistically). Indeed, 11,11L is prime, and (for example) 5,25L. Very few. When they are composite, their factors also follow a Fermatesque form (like 2*k*11^{s}+1 for 11L/Ms). And just like Fermat primes, primes of each of these sets can be argued to be finite. Similar for 12L/M and 6L/M (with b=3). Example of a larger prime: 12,81M. Did I get this right? What is the simplest explanation for the Mersenneesque to Fermatesque restriction transition? (I guess, the absence of algebraic factorization for the exponents for the factor of b. Did I actually overlook some other forbidden algebraic factorizations?) What could be the largest known prime Aurifeuillian factor (excluding 2 and 3LM)? And I guess there can be no other than 2 and 3LM of the first kind (where the search space is in primes)? << (2^{p}+1)/3 search space is of the first kind, too. This line of thinking probably segues into Lucas Aurifeuillian primitive parts, for which there's a special category at UTM. >> Last fiddled with by Batalov on 20100327 at 03:18 Reason: hopelessly cleaning up the language; but some silliness will stay, no doubt 
20100327, 06:33  #20 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3×3,041 Posts 
Silly me! Actually, 3,27L/M numbers are prime.
Because 3,(3n)L/M don't have an algebraic factorization, they may be prime for n prime or a power of 3. 
20100410, 06:34  #21 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
9123_{10} Posts 
... I found all the answers for the 3LM primes (after seeing the similarly "converging" sequences A007670+A007671 to a supersequence A057429 for the 2LM primes). Thank you for not bashing my skull in and letting me find the answers in a leisurely fashion:
the A066408 sequence with a great reference set; Oakes: http://tech.groups.yahoo.com/group/p...s/message/4607 and Posting to the Number Theory list; 3^{534827}3^{267414}+1 (in disguise) 2^{1203793}2^{601897}+1 
20120304, 19:53  #22 
"Jonathan"
Jul 2010
In a tangled web...
2·107 Posts 
Utility program
Here is a little program that should show the L & M values for a^n +/ 1.
http://myfactors.mooo.com/source/cyclo.cpp Enjoy! 
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