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].(adsbygoogle = window.adsbygoogle || []).push({});

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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# LLT numbers

**Physics Forums | Science Articles, Homework Help, Discussion**