T.Rex
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Let's say: L(x)=x^2-2 , L^1 = L, L^m = L \circ L^{m-1} = L \circ L \circ L \ldots \circ L.
Where L(x) is the polynomial used in the Lucas-Lehmer Test (LLT) :
S_0=4 \ , \ S_{i+1}=S_i^2-2=L(S_i) \ ; \ M_q \text{ is prime } \Longleftrightarrow \ S_{q-2} \equiv 0 modulo M_q .
We have:
L^2(x)=x^4-4x^2+2
L^3(x)=x^8-8x^6+20x^4-16x^2+2
L^4(x)=x^{16}-16x^{14}+104x^{12}-352x^{10}+660x^8-672x^6+336x^4-64x^2+2
Let's call C_m^+ the sum of the positive coefficients of the polynomial L^m(x).
We call C_m^+ a "LLT number": C_1^+ = 1 , C_2^+ = 3 , \ C_3^+ = 23 , \ C_4^+ = 1103 , \ C_5^+ = 2435423 .
It seems that we have the formula: C_m^+ = 2^m \prod_{i=1}^{m-1} C_{i}^+ - 1 \ \ \text{ for: } m>1.
How to prove it ? (I have no idea ...)
T.
Where L(x) is the polynomial used in the Lucas-Lehmer Test (LLT) :
S_0=4 \ , \ S_{i+1}=S_i^2-2=L(S_i) \ ; \ M_q \text{ is prime } \Longleftrightarrow \ S_{q-2} \equiv 0 modulo M_q .
We have:
L^2(x)=x^4-4x^2+2
L^3(x)=x^8-8x^6+20x^4-16x^2+2
L^4(x)=x^{16}-16x^{14}+104x^{12}-352x^{10}+660x^8-672x^6+336x^4-64x^2+2
Let's call C_m^+ the sum of the positive coefficients of the polynomial L^m(x).
We call C_m^+ a "LLT number": C_1^+ = 1 , C_2^+ = 3 , \ C_3^+ = 23 , \ C_4^+ = 1103 , \ C_5^+ = 2435423 .
It seems that we have the formula: C_m^+ = 2^m \prod_{i=1}^{m-1} C_{i}^+ - 1 \ \ \text{ for: } m>1.
How to prove it ? (I have no idea ...)
T.
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