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Old 2020-09-18, 18:58   #980
sweety439
 
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Quote:
Originally Posted by sweety439 View Post
The case for R40 k=490, since all odd n have algebra factors, we only consider even n:

n-value : factors
2 : 3^3 · 9679
4 : 43 · 79 · 83 · 1483
6 : 881 · 759379493
8 : 3 · 356807111111111
10 : 31 · 67883 · 813864335521
12 : 53 · 51703370062893081761
18 : 163 · 68860007363271983640081799591
22 : 4801 · 23279 · 3561827 · 4036715519 · 17881240410679
28 : 210323 · 6302441 · 88788971627962097615055082730651231
30 : 38270136643 · 4920560231486977484668641122451121981831

and it does not appear to be any covering set of primes, so there must be a prime at some point.

R40 also has two special remain k: 520 and 11560, 520 = 13 * base, 11560 = 289 * base, and the further searching for k = 11560 is 289 with odd n > 1

Another base is R106, which has many k with algebra factors:

64 = 2^6 (thus, all n == 0 mod 2 and all n == 0 mod 3 have algebra factors)
81 = 3^4 (thus, all n == 0 mod 2 have algebra factors)
400 = 20^2 (thus, all n == 0 mod 2 have algebra factors)
676 = 26^2 (thus, all n == 0 mod 2 have algebra factors)
841 = 29^2 (thus, all n == 0 mod 2 have algebra factors)
1024 = 2^10 (thus, all n == 0 mod 2 and all n == 0 mod 5 have algebra factors)

We should check whether they have covering set for the n which do not have algebra factors, like the case for R88 k=400 and R30 k=1369
We consider (289*40^n-1)/3 (which is prime for n=1, but there may be covering set for n>1 (and the prime for n=1 (i.e. 3853) must be in the covering set), we should check it: (k=289 for odd n):

n-value : 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 55-digit prime)
47 : 207551 · 510199 · 2088787 · (a 60-digit prime)
49 : 15240209 · 10666161587 · 167148848268429277 · (a 47-digit 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 2020-09-18 at 20:39
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Old 2020-09-18, 19:14   #981
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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
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Old 2020-09-18, 20:38   #982
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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.

n-value : 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 89-digit composite with no known prime factor)
67 : 2843 · (a 134-digit composite with no known prime factor)
79 : 1129 · 32491 · (a 155-digit composite with no known prime factor)
91 : 105899 · (a 181-digit 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 k-value 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.

n-value : 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 55-digit prime)
33 : 311 · 1888306597 · 1129552782935923 · 2923571188269551 · 28251866661502752658291361
51 : 2843 · (a 101-digit composite with no known prime factor)
53 : 13456811 · 88286677 · 6437291630956799 · (a 78-digit prime)
59 : (a 121-digit composite with no known prime factor)
63 : 2371 · 6059059478263861 · (a 110-digit 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 5-th 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.

n-value : 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 67-digit prime)
37 : 7753 · 2460302303 · (a 65-digit prime)
49 : 97 · 839 · 25561 · 136811 · 45385621130173559982180883 · (a 62-digit prime)
57 : 1777 · 66191 · 10482163 · 4863222893 · (a 94-digit prime)
61 : (a 127-digit 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 2020-09-20 at 21:58
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Old 2020-09-20, 22:11   #983
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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 2020-09-20 at 22:16
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Old 2020-09-20, 22:12   #984
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For Riesel base b, a k-value has algebra factors if and only if there exists n such that k*b^n is perfect power

For Sierpinski base b, a k-value 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 2020-09-20 at 22:13
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