 mersenneforum.org > Math Probable primality test for numbers of the form (10^n-1)/9-2 and (10^n+1)/11-2 ?
 Register FAQ Search Today's Posts Mark Forums Read 2022-05-13, 20:32 #1 kijinSeija   Mar 2021 France 2×3×5 Posts Probable primality test for numbers of the form (10^n-1)/9-2 and (10^n+1)/11-2 ? Here is what I observed : For (10^n−1)/9 - 2 : Let the sequence Si=S^10(i-1)−10*S^8(i−1)+35*S^6(i−1)−50*S^4(i−1)+25*S^2(i−1)−2 with S0=123. Then N is prime if and only if Sn−1≡710647 (modN). I choose 123 for S0 because this is the 10th Lucas number L10. For the sequence, I choose the Lucas' polynomial L10(x) and alternate + and − for each part as shown in the sequence or 2^(-10)*[(S - sqrt(S^2 - 4))^10 + (S + sqrt(S^2 - 4))^10]. Interestingly, 710647 is the 28th Lucas Number For example with (10^9-1)/9-2 Code: Mod(123, 111111109) Mod(51829516, 111111109) Mod(44205291, 111111109) Mod(61285299, 111111109) Mod(10762512, 111111109) Mod(7042864, 111111109) Mod(86600637, 111111109) Mod(105924187, 111111109) Mod(710647, 111111109) And 111111109 is indeed prime. Now for (10^n+1)/11-2 : Let the sequence Si=S^10(i-1)−10*S^8(i−1)+35*S^6(i−1)−50*S^4(i−1)+25*S^2(i−1)−2 with S0=123. Then N is prime if and only if Sn-1≡4870847 (modN). I choose 123 for S0 because this is the 10th Lucas number L10. For the sequence, I choose the Lucas' polynomial L10(x) and alternate + and − for each part as shown in the sequence or 2^(-10)*[(S - sqrt(S^2 - 4))^10 + (S + sqrt(S^2 - 4))^10]. Interestingly, 4870847 is the 32th Lucas Number For example with (10^23+1)/11-2 : Code: Mod(123, 9090909090909090909089) Mod(792070839848372253127, 9090909090909090909089) Mod(7013172659588921898968, 9090909090909090909089) Mod(5376631768234272959764, 9090909090909090909089) Mod(5567064784309146947242, 9090909090909090909089) Mod(4454192417285686785591, 9090909090909090909089) Mod(3887232900008722268298, 9090909090909090909089) Mod(5857621439229082751516, 9090909090909090909089) Mod(8657130616128185619457, 9090909090909090909089) Mod(5342237526653964258324, 9090909090909090909089) Mod(3638170586991402690760, 9090909090909090909089) Mod(3794383004766300228342, 9090909090909090909089) Mod(227556717345808697556, 9090909090909090909089) Mod(6483026222090999362884, 9090909090909090909089) Mod(3594924984182372500551, 9090909090909090909089) Mod(8063944666165464610737, 9090909090909090909089) Mod(4327641896999818513305, 9090909090909090909089) Mod(8276896390005679891227, 9090909090909090909089) Mod(1921266966330415715489, 9090909090909090909089) Mod(7582518327012313526239, 9090909090909090909089) Mod(6553601200532570675906, 9090909090909090909089) Mod(8580008981791729157791, 9090909090909090909089) Mod(4870847, 9090909090909090909089) And 9090909090909090909089 is indeed a prime number. I tested some extensions with (20^n−1)/19-2 ,(40^n−1)/39-2 and (80p+1)/81-2 etc. and it seems the primality test works for these too. For ((10*2^n)^m−1)/(10*2^n−1)-2 Let the sequence Si=2^(-10*2^n)*[(S(i-1) - sqrt(S(i-1)^2 - 4))^(10*2^n) + (S(i-1) + sqrt(S(i-1)^2 - 4))^(10*2^n)] with S0=L(10*2^n) Then N is prime if and only if Sn-1≡L(30*2^n-2)(modN). For ((10*2^n)^m+1)/(10*2^n+1)-2 Let the sequence Si=2^(-10*2^n)*[(S(i-1) - sqrt(S(i-1)^2 - 4))^(10*2^n) + (S(i-1) + sqrt(S(i-1)^2 - 4))^(10*2^n)] with S0=L(10*2^n) Then N is prime if and only if Sn-1≡L(30*2^n+2)(modN). It seems it works if (10^m±1)/(10±1)-2 > L(30*2^n±2) I found some new prime and PRP with that. Do you think it is useful ?   Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post sweety439 sweety439 7 2020-02-11 19:49 tapion64 Miscellaneous Math 40 2014-04-20 05:43 princeps Math 15 2012-04-02 21:49 Arkadiusz Math 6 2011-04-05 19:39 T.Rex Math 0 2004-10-26 21:37

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