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T.Rex

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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.

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.

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