20200918, 18:58  #980  
Nov 2016
2^{2}×691 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
2^{2}·691 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^{2}×691 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^{2}×691 Posts 
For R40, these remain k's have algebra factors:
* k=490 (=square*base/4), all odd n are algebraic * k=11560 (=289*base, 289 itself was eliminated at n=1), all odd n are algebraic * k=12250 (=square*base/4), 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 (=square*base/16), all odd n are algebraic * k=9216 (=1*base^2, 1 itself was eliminated at n=2), all n such that n+2 is composite 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 (=square*base/4), 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 20200923 at 19:58 
20200920, 22:12  #984 
Nov 2016
2^{2}×691 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 
20200922, 20:52  #985 
Nov 2016
2^{2}×691 Posts 
Riesel case: (k*b^n1)/gcd(k1,b1)
Sierpinski case: (k*b^n+1)/gcd(k+1,b1) If k is not rational power of b, then: * In Riesel case, (k*b^n1)/gcd(k1,b1) has algebra factors if and only if k*b^n is perfect power (of the form m^r with r>1) * In Sierpinski case, (k*b^n+1)/gcd(k+1,b1) has algebra factors if and only if k*b^n is either perfect odd power (of the form m^r with odd r>1) or of the form 4*m^4 If k is rational power of b (let k = m^r, b = m^s): * In Riesel case, (k*b^n1)/gcd(k1,b1) has algebra factors if and only if n*s+r is composite * In Sierpinski case, (k*b^n+1)/gcd(k+1,b1) has algebra factors if and only if n*s+r is (not power of 2, if valuation(r,2) >= valuation(s,2)) (not of the form p*2^valuation(r,2) with p prime, if valuation(r,2) < valuation(s,2)) Last fiddled with by sweety439 on 20200923 at 19:56 
20200922, 22:42  #986  
Nov 2016
2764_{10} Posts 
Quote:


20200922, 22:56  #987 
Nov 2016
2764_{10} Posts 
Conjecture: For integer triple (k,b,c), k>=1, b>=2, c != 0, gcd(k,c)=1, gcd(b,c)=1, there is a prime of the form (k*b^n+c)/gcd(k+c,b1) with integer n>=1 if and only if (k*b^n+c)/gcd(k+c,b1) has no covering set (including: covering set of fixed prime factors or covering set of all algebra factors or full covering set of partial algebra factors and partial fixed prime factors).
Last fiddled with by sweety439 on 20200923 at 01:45 
20200923, 20:02  #988 
Nov 2016
2^{2}×691 Posts 
See https://github.com/xayahrainie4793/S...variablebase for the status of 1<=k<=12 and 2<=b<=1024, all searched to n>=6000 (n>=100000 for gcd(k+1,b1) (+ for Sierpinski,  for Riesel) = 1, n>=100000 for Riesel k=1, n>=8388607 for Sierpinski k=1 and b even, n>=524287 for Sierpinski k=1 and b odd)
Remain bases: Riesel k=1: {185, 269, 281, 380, 384, 385, 394, 452, 465, 511, 574, 601, 631, 632, 636, 711, 713, 759, 771, 795, 861, 866, 881, 938, 948, 951, 956, 963, 1005, 1015} (totally 30 bases) Riesel k=2: {581, 992, 1019} (totally 3 bases) Riesel k=3: {347, 575, 587, 588, 595, 659, 699, 711, 731, 751, 763, 783, 795, 823, 972} (totally 15 bases) Riesel k=4: {178, 223, 271, 275, 310, 373, 412, 438, 475, 535, 647, 650, 653, 655, 718, 727, 742, 751, 778, 790, 812, 862, 868, 871, 898, 927, 940, 968, 970, 997, 1003} (totally 31 bases) Last fiddled with by sweety439 on 20200924 at 00:47 
20200923, 21:16  #989 
Nov 2016
2^{2}·691 Posts 
Riesel k=5: {31, 117, 181, 338, 411, 429, 489, 499, 535, 581, 583, 631, 717, 757, 998} (totally 15 bases)
Riesel k=6: {234, 412, 549, 553, 573, 619, 750, 878, 894, 954, 986} (totally 11 bases) Riesel k=7: {202, 233, 308, 373, 392, 398, 437, 463, 518, 548, 638, 662, 713, 807, 821, 848, 878, 893, 895, 953, 1015} (totally 21 bases) Riesel k=8: {321, 328, 372, 374, 407, 432, 477, 575, 665, 680, 697, 710, 721, 722, 727, 728, 752, 800, 815, 836, 867, 957, 958, 972, 974} (totally 25 bases) Last fiddled with by sweety439 on 20200924 at 00:49 
20200923, 23:13  #990 
Nov 2016
2^{2}·691 Posts 
Riesel k=9: {107, 207, 237, 325, 347, 378, 438, 483, 536, 566, 570, 592, 636, 688, 705, 711, 718, 823, 830, 835, 852, 893, 907, 926, 927, 995, 1010} (totally 27 bases)
Riesel k=10: {80, 233, 262, 284, 307, 505, 530, 551, 611, 691, 712, 724, 883, 899, 912, 980} (totally 16 bases) Riesel k=11: {65, 123, 137, 163, 173, 207, 214, 221, 227, 235, 247, 263, 283, 293, 317, 331, 375, 377, 422, 444, 452, 458, 471, 487, 533, 542, 555, 603, 627, 638, 663, 668, 691, 723, 752, 793, 804, 823, 843, 857, 863, 872, 907, 911, 923, 949, 950, 962, 987} (totally 49 bases) Riesel k=12: {263, 615, 912, 978} (totally 4 bases) Last fiddled with by sweety439 on 20200924 at 00:50 
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