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LLT numbers |
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| Nov19-06, 04:57 PM | #1 |
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LLT numbers
Let's say: [tex]L(x)=x^2-2[/tex] , [tex]L^1 = L[/tex], [tex]L^m = L \circ L^{m-1} = L \circ L \circ L \ldots \circ L[/tex].
Where [tex]L(x)[/tex] is the polynomial used in the Lucas-Lehmer Test (LLT) : [tex]S_0=4 \ , \ S_{i+1}=S_i^2-2=L(S_i) \ ; \ M_q \text{ is prime } \Longleftrightarrow \ S_{q-2} \equiv 0 [/tex] modulo [tex]M_q[/tex] . We have: [tex]L^2(x)=x^4-4x^2+2[/tex] [tex]L^3(x)=x^8-8x^6+20x^4-16x^2+2[/tex] [tex]L^4(x)=x^{16}-16x^{14}+104x^{12}-352x^{10}+660x^8-672x^6+336x^4-64x^2+2[/tex] Let's call [tex]C_m^+[/tex] the sum of the positive coefficients of the polynomial [tex]L^m(x)[/tex]. We call [tex]C_m^+[/tex] a "LLT number": [tex]C_1^+ = 1 , C_2^+ = 3 , \ C_3^+ = 23 , \ C_4^+ = 1103 , \ C_5^+ = 2435423 [/tex] . It seems that we have the formula: [tex]C_m^+ = 2^m \prod_{i=1}^{m-1} C_{i}^+ - 1 \ \ \text{ for: } m>1 [/tex]. How to prove it ? (I have no idea ...) T. |
| Nov21-06, 01:13 AM | #2 |
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Recognitions:
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| Nov21-06, 01:38 AM | #3 |
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If that pattern of alternating signs keeps up, oberve that
[tex] C_m^+ = \frac{L^m(i) + L^m(1)}{2} [/tex] |
| Nov21-06, 10:20 AM | #4 |
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LLT numbers
OK. I've got a proof. Thanks for the hints! Not so difficult once you think using "i" !
T. |
| Nov21-06, 10:25 AM | #5 |
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Now, more difficult, I think.
Can you prove: [tex] \prod_{i=1}^{2^n-1}C_i^{+} \equiv 1 \pmod{F_n}\ \ [/tex] iff [tex]\ \ F_n=2^{2^n}+1[/tex] is prime. T. |
| Nov21-06, 11:53 AM | #6 |
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Can you prove:
[tex]C_{2^n}^{+} \equiv -2 \pmod{F_n} \ \Longleftrightarrow \ F_n=2^{2^n}+1 \text{ is prime.}[/tex]. T. |
| Nov21-06, 12:33 PM | #7 |
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| Nov21-06, 01:56 PM | #8 |
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Hello,
As examples, look at: Generalized Pell Numbers: Zhi-Wei SUN helped me. New properties of Mersenne numbers: an anonymous reviewer provided a nice proof. Cycles under the LLT modulo a Mersenne prime: 2 guys helped me, including ZetaX from this forum. Regards, Tony |
| Nov23-06, 04:17 PM | #9 |
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It appears that we have: [tex]L^m(i) = V_{2^m}(1,-1)[/tex], where i is the square root of -1, m is greater than 1, and [tex]V_n(1,-1)[/tex] is a Lucas number defined by: [tex]V_0=2 , V_1=1, V_{n+1}=V_n+V_{n-1}[/tex]. Look at "The Little Book of BIGGER primes" by Paulo Ribenboim, 2nd edition, page 59. So, what I called LLT numbers are: [tex]C_m^+ = \frac{V_{2^m}-1}{2}[/tex]. I did not know this relationship. Now, in order to prove the theorem, I think I could use either the method used by Lehmer or the one used by Ribenboim. I'll see. Any more ideas ? Regards, Tony |
| Nov23-06, 04:20 PM | #10 |
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There's a U sequence too, I think. There's a nice formula not just for increasing the indices by 1, but also for doubling the indices.
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| Nov23-06, 04:33 PM | #11 |
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When n is even and Q=-1 or 1, we have: [tex]V_{2^n}=V_{2^{n-1}}^2-2[/tex] which is the LLT basic formula. Have I found something useful or is it simply another way to look at an old result ? (Probably second one !) T. |
| Nov23-06, 04:46 PM | #12 |
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IIRC, there's a more general one. If you know [itex]U_m, U_{m+1}, V_m, V_{m+1}[/itex], you can leap directly to [itex]U_{2m}, U_{2m+1}, V_{2m}, V_{2m+1}[/itex]. I don't remember the details; I just wanted to throw it out there, in case the idea was useful. The U and V sequences have a ton of useful properties!
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