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 Register FAQ Search Today's Posts Mark Forums Read 2019-11-06, 15:02 #1 enzocreti   Mar 2018 2·269 Posts Is a plausible conjecture? pg(69660), pg(92020),pg(174968),pg(176006) and pg(541456) are probable primes 69660,92020,174968,176006 and 541456 are even numbers 69660,92020,174968,176006 and 541456 are congruent to 2^k-1 mod 43 for k some non-negative integer I mean in all these cases the binary form of 69660,92020,174968,176006 and 541456 contains a number of 1's which is a multiple of 5. I mean this is the vector leading to a probabble prime pg(s) 2, 3, 4, 7, 8, 12, 19, 22, 36, 46, 51, 67, 79, 215, 359, 394, 451, 1323, 2131, 3336, 3371, 6231, 19179, 39699, 51456, 56238, 69660, 75894, 79798, 92020, 174968, 176006, 181015, 285019, 331259, 360787, 366770,...541456] for s>394 the even exponents s such that pg(s) is probable prime are s=3336, 51456, 56238, 69660, 75894, 79798, 92020, 174968, 176006,366770, 541456 when s is congruent to 2^k-1 for some k nonnegative then the binary form of s contains a number of 1's multiple of 5 i think that the conjecture can be adjusted: for s>394 if s is even or s is an odd number not multiple of 3 and if s is congruent to 2^k-1 mod 43, then the binary expansion of s contains a number of 1's multiple of 5 for s>394 infact we have that s= 69660,92020,174968,176006, 285019 (285019 is not a multiple of 3), 541456 are congruent to 2^k-1 mod 43 and the binary expansion of 69660, 92020,174968, 176006 and 541456 have a number of 1's multiple of 5 For s>394, if s is even or odd but not multiple of 3 and if s is congruent to 2^k-1 mod 43, with k some nonnegative integer, then the binary expansion of s contains a number of 1's which is a multiple of 5. Note that the only odd number multiple of 3 which is congruent to 2^k-1 mod 43 is s=19179=3^2*2131 19179 is a multiple of a prime 2131 such that pg(2131) is probable prime in particular the odd exponents leading to a probable prime s, such that s is congruent to 2^k-1 mod 43 are 7,359 and 285019 7 and 359 are prime and the binary expansion of 7 and 359 contain a numnber of 1's multiple of 3 285019 is not prime and the binary expansion of 285019 contain a number of 1's multiple of 5 consider the s such that pg(s) is probable prime and s is congruent to 2^k-1 mod 43 with k=0 or k an even integer i mean pg(359) pg(285019) pg(69660) pg(92020) pg(541456) if s is prime as in the case of 359 then the binary expansion of 359 contains 4 1's if s is not prime as in the cases 285019, 69660, 92020, 541456 then the binary expansion of 285019, 69660, 92020, 541456 contain five 1's or a multiple of five 1's example 359 is prime and is congruent to 2^4-1 mod 43 so it contains 4 1's 285019 instead is not prime and is congruent to 2^4-1 mod 43 so the binary expansion contains a number of 1's which is multiple of 5 Last fiddled with by enzocreti on 2019-11-06 at 17:54  Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post MathDoggy Miscellaneous Math 11 2019-04-17 07:05 reddwarf2956 Prime Gap Searches 2 2016-03-01 22:41 devarajkandadai Math 13 2012-05-27 07:38 sascha77 Math 15 2010-05-08 00:33 AntonVrba Math 19 2005-07-26 12:49

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