mersenneforum.org  

Go Back   mersenneforum.org > Extra Stuff > Miscellaneous Math

Reply
 
Thread Tools
Old 2023-01-18, 11:06   #23
kriesel
 
kriesel's Avatar
 
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest

737510 Posts
Default

Quote:
Originally Posted by kriesel View Post
... = 963 = 1024
Ack, edit fail, = 963 doesn't belong, time's expired.
kriesel is offline   Reply With Quote
Old 2023-01-23, 14:42   #24
User140242
 
Jul 2022

22·19 Posts
Default

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
User140242 is offline   Reply With Quote
Old 2023-01-23, 19:40   #25
User140242
 
Jul 2022

22·19 Posts
Default

Quote:
Originally Posted by User140242 View Post
...

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:


\(S_{i+1} *S_{i+2}* \dots *S_{i+m} = \sum\limits_{k=0}^{2^m-1}{(-1)^{2^m-1-k}\left(\begin{matrix} 2^m+k \\ 2 \cdot k+1 \end{matrix}\right) \cdot (S_i^2)^k}\)
User140242 is offline   Reply With Quote
Reply

Thread Tools


Similar Threads
Thread Thread Starter Forum Replies Last Post
Lucas-Lehmer Primality Testing User140242 Programming 20 2022-09-10 22:44
Modifying the Lucas Lehmer Primality Test into a fast test of nothing Trilo Miscellaneous Math 25 2018-03-11 23:20
Lucas-Lehmer test Mathsgirl Information & Answers 23 2014-12-10 16:25
Lucas-Lehmer Test storm5510 Math 22 2009-09-24 22:32
Lucas-Lehmer primality test CMOSMIKE Math 5 2002-09-06 18:46

All times are UTC. The time now is 14:56.


Wed Feb 8 14:56:19 UTC 2023 up 174 days, 12:24, 1 user, load averages: 1.36, 1.09, 0.95

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2023, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.

≠ ± ∓ ÷ × · − √ ‰ ⊗ ⊕ ⊖ ⊘ ⊙ ≤ ≥ ≦ ≧ ≨ ≩ ≺ ≻ ≼ ≽ ⊏ ⊐ ⊑ ⊒ ² ³ °
∠ ∟ ° ≅ ~ ‖ ⟂ ⫛
≡ ≜ ≈ ∝ ∞ ≪ ≫ ⌊⌋ ⌈⌉ ∘ ∏ ∐ ∑ ∧ ∨ ∩ ∪ ⨀ ⊕ ⊗ 𝖕 𝖖 𝖗 ⊲ ⊳
∅ ∖ ∁ ↦ ↣ ∩ ∪ ⊆ ⊂ ⊄ ⊊ ⊇ ⊃ ⊅ ⊋ ⊖ ∈ ∉ ∋ ∌ ℕ ℤ ℚ ℝ ℂ ℵ ℶ ℷ ℸ 𝓟
¬ ∨ ∧ ⊕ → ← ⇒ ⇐ ⇔ ∀ ∃ ∄ ∴ ∵ ⊤ ⊥ ⊢ ⊨ ⫤ ⊣ … ⋯ ⋮ ⋰ ⋱
∫ ∬ ∭ ∮ ∯ ∰ ∇ ∆ δ ∂ ℱ ℒ ℓ
𝛢𝛼 𝛣𝛽 𝛤𝛾 𝛥𝛿 𝛦𝜀𝜖 𝛧𝜁 𝛨𝜂 𝛩𝜃𝜗 𝛪𝜄 𝛫𝜅 𝛬𝜆 𝛭𝜇 𝛮𝜈 𝛯𝜉 𝛰𝜊 𝛱𝜋 𝛲𝜌 𝛴𝜎𝜍 𝛵𝜏 𝛶𝜐 𝛷𝜙𝜑 𝛸𝜒 𝛹𝜓 𝛺𝜔