mersenneforum.org > Math Chebyshev polynomials and higher order Lucas Lehmer algorithm by Kok Seng Chu
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 2022-01-10, 11:51 #1 T.Rex     Feb 2004 France 929 Posts Chebyshev polynomials and higher order Lucas Lehmer algorithm by Kok Seng Chu Hi, I've found this recent (2021, October 3rd) paper named "CHEBYSHEV POLYNOMIALS AND HIGHER ORDER LUCAS LEHMER ALGORITHM", by KOK SENG CHUA, based on previous work by Pedja Terzi´c, and talking about the necessity part. This looks very interesting to me, since it provides a search about generalized Mersennes and the LLT (x^2-2). However, I've not spent yet enough time to read it, and it's not easy for me to understand it. So, I'd like to get comments from true mathematicians. https://arxiv.org/pdf/2010.02677.pdf Regards
 2022-01-11, 21:40 #2 T.Rex     Feb 2004 France 929 Posts The formula for Wagstaff numbers seems OK. With Pari/gp . Code: t(q)={w=(2^q+1)/3;S0=4;print("w: ",w);S=S0;for(i=1,q-1,S=Mod(S^2-2,w));s1=lift(Mod(S-5-9,w));s2=lift(Mod(S-5+9,w));print(s1," ",s2)} ? t(11) w: 683 665 0 ? t(13) w: 2731 0 18 ? t(17) w: 43691 43673 0 ? t(19) w: 174763 0 18 ? t(23) w: 2796203 2796185 0 ? t(29) w: 178956971 Not prime 59834419 59834437 ? t(31) w: 715827883 0 18 ? t(37) w: 45812984491 Not prime 24875527143 24875527161 ? t(41) w: 733007751851 Not prime 634893124730 634893124748 ? t(43) w: 2932031007403 0 18 ? t(61) w: 768614336404564651 0 18
 2022-01-15, 14:44 #3 Dr Sardonicus     Feb 2017 Nowhere 10111000110112 Posts The sufficiency result (Lemma 2.1) requires a complete factorization of Q + 1 or Q - 1 in order to prove that Q is prime.
 2022-06-12, 20:37 #4 kijinSeija   Mar 2021 France 468 Posts I don't know if it's related to this topic but I made some probable primality test for numbers of the forum (a^p-1)/(a-1) and (a^p+1)/(a+1) using Chebyshev polynomials : but the test isn't perfect and there are some conditions to apply : - a must not be a perfect power otherwise you can get false positive. - p must be a prime number >2 otherwise you can "break" the primality test and get false positive Let $N = \frac{a^p-1}{a-1}$ Let the sequence $S_i = 2 \cdot T_{a}(S_{i-1}/2)$ where $T_{n}(x)$ is the Chebyshev's polynomial of the first kind with $S_0 = p$ $N$ is prime if $S_{p} \equiv 2 \cdot T_{a}(p/2) (mod N)$ or $S_{p} \equiv 2 \cdot T_{a-2}(p/2) (mod N)$ You can found the test here : https://sagecell.sagemath.org/?z=eJx...yLjgUAARUAuQ== For $N = \frac{a^p+1}{a+1}$ Let the sequence $S_i = 2 \cdot T_{a}(S_{i-1}/2)$ where $T_{n}(x)$ is the Chebyshev's polynomial of the first kind with $S_0 = p$ $N$ is prime if $S_{p} \equiv 2 \cdot T_{a}(p/2) (mod N)$ or $S_{p} \equiv 2 \cdot T_{a+2}(p/2) (mod N)$ You can found the test here : https://sagecell.sagemath.org/?z=eJx...yLjgUAARUAuQ== For the moment I didn't get a counterexample If you found one please tell me. EDIT : (Second test fails for (a^3+1)/(a+1) apparently where a = 5 and 7) Last fiddled with by kijinSeija on 2022-06-12 at 21:20

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