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[ Extra info pseudoprime to base 2 (Poulet Number) The smallest composite pseudoprime to base 2 is 341.
Distribution laws of the pseudoprimes Carlos Rivera found something interesting about the question e^(ln x^(5/14)) < P2(x) < x.e^((-ln x.lnlnln x)/2lnln x) (Pomerance) has a heuristic argument that the true estimate This estimative formula for the population of pseudoprimes According with the upper bound the probability that Well, when the number we are looking for in CYF1 is of Even numbers much smaller that 1000 digits are practically C. Caldwell's site is very informative regarding this topic. With this information the readers can themselves See A068216 for See A067845 for Gupta's largest solutions Similar to the above list, Shyam Sunder Gupta created More websites about pseudoprimes
Yes, I proved the following proposition: If p & 2p-1 are odd prime then p*(2p-1) is base-2 psp iff p = 1 (mod 12) So there exist many base-2 psp of the form p*(2p-1). Also it seems that there exist infinitely many base-2 psp numbers For finding a(17) (the first 17-digit base-2 psp) the steps were : 1. I found the smallest number m, where prime(m)*(2prime(m)-1) > 10^16. Hence the smallest base-2 psp of the form p*(2p-1) is As I remember there exists only one base-2 psp number r between a(17) and n Similarly for finding a(18) at first I found some upper bound for a(18) With the approach I get a candidate that is not necessarily the smallest.
Proof of the proposition:
If p & 2p-1 are odd prime then p*(2p-1) is base-2 psp iff p = 1 (mod 12) Since p is prime by Fermat's little theorem (FLT) we have, Hence from n is a 2-base pseudoprime iff 2^(n-1) = 1 (mod 2p-1) But 2p-1 is prime hence by FLT we have 2^(2p-1) = 2 (mod 2p-1) n is a 2-base pseudoprime iff 2^(p-1) = 1 (mod 2p-1) Now if q = 2p-1 according to Euler's criterion (2/q) = 2^((q-1)/2) (mod q) n is a 2-base pseudoprime iff (2/q) = 1 (mod q) A theorem says (one of the applications of Gauss's lemma)
Nov 17, 2008 ] !Smallest 2-psp with 17, 18, 19, 20 digits: My program agrees with the listed results below 17 digits. The page now lists Farideh Firoozbakht in 2007 for the first 2 numbers. The page says x.e^(-ln x.lnlnln x)/2lnln x) several times. I cannot explain the loss of this submission. Anyway my | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

A000125 Prime Curios! Prime Puzzle Wikipedia 125 Le nombre 125 |

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