Formula for the term S_p-2 in Lucas-Lehmer primality test

[kriesel]

Originally Posted by kriesel

... = 963 = 1024

Ack, edit fail, = 963 doesn't belong, time's expired.

[User140242]

I'm just looking into the different ways to derive the sequence used in Lucas-Lehmer primality test in hopes of finding a simplified alternative method.

In these examples, that I have not tested higher values, ​​are used Triangle of coefficients of Chebyshev's OEIS A053122

\({(-1)^{n-k}\left(\begin{matrix} n+k+1 \\ 2 \cdot k+1 \end{matrix}\right)}\)

\(\sqrt{(S_3-2)/12}=\sqrt{(37634-2)/12}=56=4 \cdot \sum \limits_{k=0}^{1}{(-1)^{1-k}\left(\begin{matrix} 1+k+1 \\ 2 \cdot k+1 \end{matrix}\right) \cdot (4^2)^k}=4 \cdot(-2+4^2)\)

\(\sqrt{(S_4-2)/12}=10864=4 \cdot \sum \limits_{k=0}^{3}{(-1)^{3-k}\left(\begin{matrix} 3+k+1 \\ 2 \cdot k+1 \end{matrix}\right) \cdot (4^2)^k}=4 \cdot(-4+10 \cdot 4^2-6 \cdot 4^4+4^6)\)

\(\sqrt{(S_5-2)/12}=408855776=4 \cdot \sum \limits_{k=0}^{7}{(-1)^{7-k}\left(\begin{matrix} 7+k+1 \\ 2 \cdot k+1 \end{matrix}\right) \cdot (4^2)^k}\)

Furthermore 4 can be replaced example with S₁=14 to obtain the same result with fewer terms

\(\sqrt{(S_5-2)/12}=408855776=4 \cdot 14 \cdot\sum \limits_{k=0}^{3}{(-1)^{3-k}\left(\begin{matrix} 3+k+1 \\ 2 \cdot k+1 \end{matrix}\right) \cdot (14^2)^k}=4 \cdot14 \cdot (-4+10 \cdot 14^2-6 \cdot 14^4+14^6)\)

[User140242]

Originally Posted by User140242

...

Furthermore 4 can be replaced example with S₁=14 to obtain the same result with fewer terms

\(\sqrt{(S_5-2)/12}=408855776=4 \cdot 14 \cdot\sum \limits_{k=0}^{3}{(-1)^{3-k}\left(\begin{matrix} 3+k+1 \\ 2 \cdot k+1 \end{matrix}\right) \cdot (14^2)^k}=4 \cdot14 \cdot (-4+10 \cdot 14^2-6 \cdot 14^4+14^6)\)

I haven't checked but I speculate it could be:


\(S_{i+1} *S_{i+2}* \dots *S_{i+m} = \sum\limits_{k=0}^{2^m-1}{(-1)^{2^m-1-k}\left(\begin{matrix} 2^m+k \\ 2 \cdot k+1 \end{matrix}\right) \cdot (S_i^2)^k}\)