20200918, 18:58  #980  
Nov 2016
2·1,117 Posts 
Quote:
nvalue : factors 3 : 3^2 · 317 · 2161 5 : 37 · 601 · 443609 7 : 71 · 222299342723 9 : 3 · 8417735111111111 11 : 521 · 77553029814459373 13 : 1093 · 135966569 · 435014942249 17 : 173 · 1201 · 796539523771295275773721 19 : 199 · 827 · 125878441037<12> · 12782225695980733 23 : 31 · 37 · 4493 · 131539610664636811448698039308523 25 : 1693 · 14071 · 83071 · 2786867 · 196665766270295693879723 29 : 43 · 15523495249 · 366735559693 · 11342410093643652930353483 31 : 271 · 1471 · 11144340056387535855201380021957935418919111013 35 : 1289 · (a 55digit prime) 47 : 207551 · 510199 · 2088787 · (a 60digit prime) 49 : 15240209 · 10666161587 · 167148848268429277 · (a 47digit prime) and 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 20200918 at 20:39 

20200918, 19:14  #981 
Nov 2016
2234_{10} 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. nvalue : 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 47digit prime) 47 : 7 · 421 · 17069162801611 · 14667444266312619953 · (a 60digit prime) 59 : 487 · (a 118digit composite without known prime factor) 71 : 4289 · 10093 · (a 137digit 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 kvalue 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. nvalue : 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 55digit 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. nvalue : 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 94digit 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 kvalue 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 20200918 at 19:15 
20200918, 20:38  #982 
Nov 2016
2×1,117 Posts 
k = 676:
since 676 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. nvalue : factors 1 : 17 · 281 3 : 3 · 277 · 64591 5 : 5 · 17 · 19 · 67 · 5573621 7 : 1949 · 3476839593221 9 : 3^3 · 17 · 3271 · 50712496951637 11 : 19 · 42937 · 2432147 · 431166327217 13 : 17 · 1373 · 6351547249 · 64838460350149 17 : 17 · 19 · 61 · 61591784776543827671882518345783 19 : 6299 · 756585273193 · 2861128642099661938794059 23 : 19 · 98443 · 920347627017000007051307391604325416676033 43 : (a 89digit composite with no known prime factor) 67 : 2843 · (a 134digit composite with no known prime factor) 79 : 1129 · 32491 · (a 155digit composite with no known prime factor) 91 : 105899 · (a 181digit composite with no known prime factor) Although this number is divisible by 3 for all n == 3 mod 6 and by 19 for all n == 5 mod 6 and by 17 for all n == 1 mod 4 (which makes this kvalue very low weight, since only n == 7 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 = 841: since 841 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. nvalue : factors 1 : 3 · 283 3 : 271 · 35201 5 : 61 · 677 · 2595479 7 : 3^2 · 5^2 · 5352605493383 9 : 4679 · 8663 · 333839809991 11 : 67 · 11440889 · 198352025576693 15 : 359487408541 · 53396278847280064403 17 : 5 · 19927 · 140909 · 15362282538731494849528849 21 : 271 · 7457 · 663563 · 20305527277370848392217057350779 23 : 94547 · 534824108672537 · 6050383020924045192372407269 29 : 84737 · (a 55digit prime) 33 : 311 · 1888306597 · 1129552782935923 · 2923571188269551 · 28251866661502752658291361 51 : 2843 · (a 101digit composite with no known prime factor) 53 : 13456811 · 88286677 · 6437291630956799 · (a 78digit prime) 59 : (a 121digit composite with no known prime factor) 63 : 2371 · 6059059478263861 · (a 110digit composite with no known prime factor) and it does not appear to be any covering set of primes, so there must be a prime at some point. k = 1024: since 1024 is square and 5th power, all even n and all n divisible by 5 have algebra factors, and we only want to know whether it has a covering set of primes for all n == 1, 3, 7, 9 (mod 10), 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. nvalue : factors 1 : 97 · 373 3 : 17 · 3407 · 7019 7 : 17 · 3019053696484613 9 : 647 · 3581827 · 248841380929 11 : 3 · 17^2 · 7473501436891484179943 13 : 449 · 1447 · 112057280449127255045987 17 : 3^2 · 406591 · 2126171 · 1181353712721405831409129 19 : 17 · 283 · 258373 · 9179867 · 9050472811960369021895401 21 : 1831 · 972605267 · 1597539586927967 · 407873305308400559 33 : 1223 · (a 67digit prime) 37 : 7753 · 2460302303 · (a 65digit prime) 49 : 97 · 839 · 25561 · 136811 · 45385621130173559982180883 · (a 62digit prime) 57 : 1777 · 66191 · 10482163 · 4863222893 · (a 94digit prime) 61 : (a 127digit composite with no known prime factor) and 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 20200920 at 21:58 
20200920, 22:11  #983 
Nov 2016
2·1,117 Posts 
For R40, these remain k's have algebra factors:
* k=490 (=10*square), all odd n are algebraic * k=11560 (=289*base, 289 itself was eliminated at n=1), all odd n are algebraic * k=12250 (=10*square), all odd n are algebraic * k=12544 (=square), all even n are algebraic * k=15376 (=square), all even n are algebraic For R52, these remain k's have algebra factors: * k=21316 (=square), all even n are algebraic For R78, these remain k's have algebra factors: * k=4489 (=square), all even n are algebraic * k=7800 (=100*base, 100 itself was eliminated at n=1), all odd n are algebraic * k=8649 (=square), all even n are algebraic * k=12167 (=cube), all n divisible by 3 are algebraic * k=13824 (=cube), all n divisible by 3 are algebraic * k=59536 (=square), all even n are algebraic For R96, these remain k's have algebra factors: * k=1681 (=square), all even n are algebraic * k=5046 (=6*square), all odd n are algebraic * k=9216 (=1*base^2, 1 itself was eliminated at n=2), all even n are algebraic * k=16641 (=square), all even n are algebraic For R106, these remain k's have algebra factors: * k=64 (=square and cube), all even n and all n divisible by 3 are algebraic * k=81 (=square), all even n are algebraic * k=400 (=square), all even n are algebraic * k=676 (=square), all even n are algebraic * k=841 (=square), all even n are algebraic * k=1024 (=square and 5th power), all even n and all n divisible by 5 are algebraic * k=2116 (=square), all even n are algebraic * k=3136 (=square), all even n are algebraic * k=3481 (=square), all even n are algebraic * k=4096 (=square and cube), all even n and all n divisible by 3 are algebraic * k=5776 (=square), all even n are algebraic * k=7744 (=square), all even n are algebraic * k=10816 (=square), all even n are algebraic * k=12321 (=square), all even n are algebraic For R124, these remain k's have algebra factors: * k=441 (=square), all even n are algebraic * k=1156 (=square), all even n are algebraic * k=1519 (=31*square), all odd n are algebraic * k=4096 (=square and cube), all even n and all n divisible by 3 are algebraic * k=7396 (=square), all even n are algebraic Last fiddled with by sweety439 on 20200920 at 22:16 
20200920, 22:12  #984 
Nov 2016
2×1,117 Posts 
For Riesel base b, a kvalue has algebra factors if and only if there exists n such that k*b^n is perfect power
For Sierpinski base b, a kvalue has algebra factors if and only if there exists n such that k*b^n is either perfect odd power or of the form 4*m^4 Last fiddled with by sweety439 on 20200920 at 22:13 
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