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 2020-09-18, 19:14 #981 sweety439     Nov 2016 3·5·132 Posts For the case for R106: k = 64: since 64 is square and cube, all even n and all n divisible by 3 have algebra factors, and we only want to know whether it has a covering set of primes for all n == 1 or 5 (mod 6), if so, then this k makes a full covering set with algebraic factors and be excluded from the conjecture; if not, then this k does not make a full covering set with algebraic factors and be included from the conjecture. n-value : factors 1 : 17 · 19 5 : 7 · 17 · 13669 · 25073 7 : 19 · 739 · 32636508923 11 : 105137 · 710341 · 774645021719 13 : 17 · 19 · 2012493124713603631831681 17 : 17 · 16036907 · 301016884615451673389616697 19 : 7 · 19 · 81929 · 1441051 · 1392403219 · 42173384412226351 23 : 4691 · 240422191 · 359534531 · 287087966317907212195482133 35 : 241 · 389 · 39161 · 3351132509456839 · (a 47-digit prime) 47 : 7 · 421 · 17069162801611 · 14667444266312619953 · (a 60-digit prime) 59 : 487 · (a 118-digit composite without known prime factor) 71 : 4289 · 10093 · (a 137-digit composite without known prime factor) Although this number is divisible by 17 for all n == 1 mod 4 and by 19 for all n == 1 mod 6 (which makes this k-value very low weight, since only n == 11 mod 12 can be prime), but it does not appear to be any covering set of primes, so there must be a prime at some point. k = 81: since 81 is square, all even n have algebra factors, and we only want to know whether it has a covering set of primes for all odd n, if so, then this k makes a full covering set with algebraic factors and be excluded from the conjecture; if not, then this k does not make a full covering set with algebraic factors and be included from the conjecture. n-value : factors 1 : 17 · 101 3 : 67 · 287977 5 : 17 · 431 · 727 · 40699 7 : 857 · 2842334911979 11 : 883 · 347963521 · 1000887146689 15 : 47 · 1359940313999 · 607414685128749427 19 : 5 · 1049 · 3331 · 1861172051723 · 150736978974366072719 23 : 6637 · 74623 · 45940781149 · 27196124333848915407481172821 27 : 2135773 · 2196601133149 · 16652026043310698243659019628892454299 31 : 367 · 3894307 · (a 55-digit prime) 35 : 12589419042703 · 73042126655937895819733 · 1354070261224865451982856575186891049 and it does not appear to be any covering set of primes, so there must be a prime at some point. k = 400: since 400 is square, all even n have algebra factors, and we only want to know whether it has a covering set of primes for all odd n, if so, then this k makes a full covering set with algebraic factors and be excluded from the conjecture; if not, then this k does not make a full covering set with algebraic factors and be included from the conjecture. n-value : factors 1 : 3 · 673 3 : 19 · 743 · 1607 5 : 179 · 1424022961 7 : 3 · 4657 · 23917 · 8571317 9 : 19 · 1693713242107962001 11 : 47^2 · 19991 · 8187946182350101 17 : 3362709722608729 · 152528509553573862011 23 : 10889 · 66817096529447428049947387228178558168776171 29 : 67 · 2445989705956469367060937 · 6297691198803985156665528870701561 35 : 34352269373675266693 · 889339893798719344479307 · 47920658139709491455114469269 47 : 607 · (a 94-digit composite without known prime factor) Although this number is divisible by 3 for all n == 1 mod 6 and by 19 for all n == 3 mod 6 (which makes this k-value low weight, since only n == 5 mod 12 can be prime), but it does not appear to be any covering set of primes, so there must be a prime at some point. Last fiddled with by sweety439 on 2020-09-18 at 19:15