mersenneforum.org Formula for the term S_p-2 in Lucas-Lehmer primality test
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2023-01-18, 11:06   #23
kriesel

"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest

737510 Posts

Quote:
 Originally Posted by kriesel ... = 963 = 1024
Ack, edit fail, = 963 doesn't belong, time's expired.

 2023-01-23, 14:42 #24 User140242   Jul 2022 22·19 Posts 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 $$\sqrt{(S_5-2)/12}=408855776=4 \cdot 14 \cdot\sum \limits_{k=0}^{3}{(-1)^{3-k}\left(\begin{matrix} 3+k+1 \\ 2 \cdot k+1 \end{matrix}\right) \cdot (14^2)^k}=4 \cdot14 \cdot (-4+10 \cdot 14^2-6 \cdot 14^4+14^6)$$ Last fiddled with by User140242 on 2023-01-23 at 15:27
2023-01-23, 19:40   #25
User140242

Jul 2022

22·19 Posts

Quote:
 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:

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

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