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Characteristic value of Fibonacci Sequences 
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#1
Feb2409, 08:50 AM

P: 894

[tex] C(a,b) = a^2 + ab b^2[/tex]
The characteristic value of a Fibonacci sequence is an interesting property. 1) C(a,b) = C(a,ab) 2) C(a,b) * C(c,d) = C(ac+adbd,bcad) 3) C(a,b) * C(a,ab) = C(2a^22ab+b^2,2abb^2) =(C(a,b))^2 [tex]C(a,b)^n = C(A_{n},B_{n})[/tex] [tex]A_{n}=\sum_{i=0}^{n}F_{i1}nCia^{i}b^{(ni)}[/tex] [tex]B_{n}=\sum_{i=0}^{n}F_{i2}nCia^{i}b^{(ni)}[/tex] Opps the last two equations are sums as i goes from 0 to n [tex]F_i[/tex] ={1,1,0,1,1,2,3...} with [tex]F_{0}= 0[/tex] nCi are the binominal coefficients 


#2
Feb2809, 10:13 PM

P: 894




#3
Mar109, 01:46 AM

P: 813

Forgive me for this question but what is the:
Characteristic value of Fibonacci Sequences I wasn't a math major when I was in university. 


#4
Mar109, 09:09 AM

P: 894

Characteristic value of Fibonacci Sequences



#5
Mar109, 02:23 PM

P: 894

Let x and y be [tex]F_0[/tex] and [tex]F_1[/tex] respectively and [tex]F_{n}=F_{n2}+F_{n1}[/tex] Then [tex]C(a,b)^{n} = C(A_{n},B_{n})/C(x,y)[/tex] [tex]A_{n} = \sum_{i=0}^{n}F_{i}nCia^{ni}b^{i}[/tex] [tex]B_{n} = \sum_{i=0}^{n}F_{i+1}nCia^{ni}b^{i}[/tex] Now given that [tex]pCi = 0[/tex] mod p(prime) except for [tex]i=0[/tex] and [tex]i=p[/tex]; C(a,b)^(p)=C(a,b) mod p = C(xa +b* F_{p},ya + bF_{p+1}) /C(x,y) would seem to be a new property of Fibonacci series assuming that each of C(a,b), a and b <> 0 mod p. This property is the next candidate for proof. Eventually proof of the whole might be found. 


#6
Mar209, 12:07 AM

P: 894

My new hypothesis Let N Be odd and define the subscripts A,B,C,D to be the sequential integers from (N3)/2 to (N+3)/2. Then the following equivalence holds if and only if N is Prime If N = 1 mod 4 then [tex]C(F_{C},F_{B}) = C(F_{A},F_{D}) \mod N[/tex] If N = 1 mod 4 then [tex]C(F_{C},F_{B}) = C(F_{D},F_{A}) \mod N[/tex] I check this as far as my excel program would allow (up to prime = 53). This may or may not be useful but it is an improvement on Wilson's Theory. The only doubt I have is the "and only if" part but neither part stands proven. 


#7
Mar2109, 10:29 PM

P: 26

To: ramsey 2879
Quoting ramsey 2879 This is truly one of the most interesting results that I have ever seen, and I'm an "old timer"! The "if and only if part" does indeed stand a very good chance of being correct and from a purely theoretical standpoint, it is clearly of immense importance! Please continue posting the results of your reasearch, investigations and other work on this incredible discovery! Don. 


#8
Mar2209, 12:45 AM

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#9
Mar2409, 01:35 PM

P: 894

I have been limiting my investigation so far by chosing various other expresions in the first degree for a and b, and determining A(n) and B(n) as polynominals of degree n by the multiplication operation, relation 3 of my first post using alternative first order expresions (a,b) and (a,ab) as above and determining the relation between a and b such that the polynominals of degree n give all terms divisible by n if n is prime with the exception of one or both of the first and last terms. One way to speed the process is to determine the polynoninals for m = 2n by multiplying (A(n),B(n)) with (A(n),A(n)B(n)) then multiplying these polynominals together using multiple operations 3. This is how I deduced the result given earlier in my thread with respect to the coefficients all being a combination of Fibonacci numbers and the Binominal coefficients. 


#10
Mar2409, 08:17 PM

P: 894




#11
Mar2609, 12:38 AM

P: 26

To: Ramsey 2879,
You took a different approach than did Lucas, and that, in and of itself, makes your discovery unique. Thanks, Don. 


#12
Mar3109, 12:40 AM

P: 894

I found another Prime Test much stronger than the Lucas Test Input an odd number N not divisible by 5 greater than 1 A = 1 B = 0 P = 2 Do C = A D = AB A = C+D B = D P = P + 2 Loop Until P> N Now (A + (1)^( N^2 = 1 mod 10)) = 0 Mod N if N is prime Also (B +(N^2 = 1 mod 10) = 0 Mod N if N is prime This two prong test yields only three presudo primes in the first thousand primes. The first presudo prime is 4181 but I haven't located the other two. I think that an advantage to this test is that it only involves addition and subtraction. The number of recursions could be shortened by multiplication as per method 3 of my first post in combination with the addition and subtraction I would be glad to answer any questions concerning this test. 


#13
Mar3109, 01:02 AM

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P: 3,684

Your test takes n/2 steps, much worse than trial division.
I don't understand the final step in the test; can you give an example, perhaps for n = 101? I get (a, b, c, d) = (1500520536206896083277, 927372692193078999176, 573147844013817084101, 927372692193078999176). 


#14
Mar3109, 01:36 AM

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P: 16,091

I reverse engineered it  he's using the [.] operator where
[P] = 1 if P is true [P] = 0 if P is false except he's using parentheses for added confusion. 


#15
Mar3109, 01:40 AM

Emeritus
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P: 16,091

I suppose since I calculated it, I should post it in case someone else wants to know: set
[tex]u = (3 + \sqrt{5}) / 2[/tex] [tex]v = (3  \sqrt{5}) / 2[/tex] This test computes [tex]A = \frac{(1u) u^E  (1v) v^E}{v  u}[/tex] [tex]B = \frac{u^E  v^E}{v  u}[/tex] where E = (N  1) / 2 


#16
Mar3109, 03:16 AM

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Thanks for posting that. Now I can post the first few pseudoprimes. My list matches (but exceeds) ramsey2879's list.
4181, 5777, 6479, 6721, 10877, 13201, 15251, 27071, 34561, 44099, 47519, 51841, 54839, 64079, 64681, 65471, 67861, 68251, 72831, 75077, 78089, 90061, 96049, 97921, 100127, 109871, 113573, 118441, 139359, 146611, 157079, 161027, 162133, 163081, 168299, 186961, 196559, 197209, 219781, 231703, 233519, 252601, 254321, 257761, 268801, 272611, 283361, 300847, 302101, 303101, 327359, 330929, 399001, 417601, 430127, 433621, 438751, 451979, 486359, 489601, 510719, 512461, 520801, 530611, 544159, 545279, 553679, 553839, 556421, 575599, 618639, 620279, 635627, 636641, 638189, 641199, 655201, 670879, 689359, 701569, 722261, 737471, 741751, 809999, 850541, 851927, 852841, 853469, 881011, 925681, 954239, 993509, 999941 


#17
Mar3109, 06:18 AM

P: 894

If N = 3 or 7 mod 10 then (the expressions = 0,1) (1)^0 = 1 so we have A+1 and B+1 If N = 1 or 9 mod 10 then (the expressions = 1,0) ..........=1 so we have A1 and B+0 Interesting reverse engineering, but I actually based this on alternative type 2 multiplication by a = 1+x and b = 1+x or 0 alternatively, but I did only the constant x^(0) part. 


#18
Mar3109, 06:35 AM

P: 894




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