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Old 2020-09-03, 08:33   #969
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2 (probable) primes found for R70:

(376*70^6484-1)/3
(496*70^4934-1)/3

k=811 still remains ....
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Old 2020-09-03, 21:37   #970
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https://docs.google.com/document/d/e...Ns5LlL_FA0/pub

Update newest file for Sierpinski problems to include the top 10 (probable) primes for S78, S96, and S126
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Old 2020-09-04, 22:43   #971
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https://docs.google.com/document/d/e...MpCV9458Wi/pub

Update newest file for Sierpinski conjectures, according to GFN2 and GFN10, the test limit of S512 k=2 is (2^54-1)/9-1 = 2001599834386886, and the test limit of S10 k=100 is 2^31-3 = 2147483645
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Old 2020-09-04, 23:03   #972
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Quote:
Originally Posted by sweety439 View Post
S117:

k=11
k=47
k=67
k=75
k=77
k=81 (proven by N+1-method)

S256: (k=11 is only probable prime)

k=23

S1024:

k=14
k=41
k=44
R4:

k=106

R7: (k=197 and 367 are only probable primes)

k=79
k=139 (proven by N-1-method) (certificate for large prime factor for N-1)
k=159
k=299
k=313
k=391
k=419
k=429
k=437
k=451

R10:

k=121

R12:

k=298

R17:

k=13
k=29

R26:

k=121

R31:

k=21
k=39 (proven by N-1-method) (certificate for large prime factor for N-1)
k=49
k=113
k=115
k=123
k=124

R33:

k=213

R35:

k=1 (proven by N-1-method)

R37:

k=5 (proven by N-1-method)

R39:

k=1 (proven by N-1-method)

R43:

k=4 (proven by N-1-method)

R45:

k=53

Last fiddled with by sweety439 on 2020-09-05 at 12:03
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Old 2020-09-05, 20:17   #973
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Quote:
Originally Posted by sweety439 View Post
Update files.
Update files
Attached Files
File Type: txt Sierp k1.txt (7.6 KB, 15 views)
File Type: txt Sierp k10.txt (7.6 KB, 16 views)
File Type: txt Sierp k12.txt (7.4 KB, 18 views)
File Type: txt Riesel k9.txt (12.1 KB, 18 views)
File Type: txt Riesel k10.txt (7.5 KB, 15 views)

Last fiddled with by sweety439 on 2020-09-05 at 20:18
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Old 2020-09-06, 19:49   #974
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Also reserved R88 and found these (probable) primes:

(49*88^2223-1)/3
(79*88^7665-1)/3
(235*88^1330-1)/3
(346*88^2969-1)/3
(541*88^1187-1)/3
(544*88^8904-1)/3

k = 46, 94, 277, 508 are still remaining.
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Old 2020-09-06, 20:13   #975
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Still no (probable) prime found for R70 k=811

If a (probable) prime for R70 k=811 were found, then will produce a group of 11 consecutive proven Riesel conjectures: R67 to R77
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Old 2020-09-06, 23:53   #976
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Quote:
Originally Posted by sweety439 View Post
R4:

k=106

R7: (k=197 and 367 are only probable primes)

k=79
k=139 (proven by N-1-method) (certificate for large prime factor for N-1)
k=159
k=299
k=313
k=391
k=419
k=429
k=437
k=451

R10:

k=121

R12:

k=298

R17:

k=13
k=29

R26:

k=121

R31:

k=21
k=39 (proven by N-1-method) (certificate for large prime factor for N-1)
k=49
k=113
k=115
k=123
k=124

R33:

k=213

R35:

k=1 (proven by N-1-method)

R37:

k=5 (proven by N-1-method)

R39:

k=1 (proven by N-1-method)

R43:

k=4 (proven by N-1-method)

R45:

k=53
R46: (k=86, 576, and 561 are only probable primes)

k=100
k=121
k=142
k=256
k=386

R49:

k=79

R51:

k=1

R57:

k=87 (proven by N-1-method) (certificate for large prime factor for N-1)

R58: (k=382, 400, and 421 are only probable primes)

k=4 (proven by N-1-method)
k=103
k=109
k=142
k=163
k=217
k=271
k=334
k=361
k=379
k=445
k=457
k=487

R61: (k=13 is only probable prime)

k=10
k=41
k=77

Last fiddled with by sweety439 on 2020-09-06 at 23:56
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Old 2020-09-07, 00:27   #977
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Update newest file for Riesel problems to include recent status for R70 and R88
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Old 2020-09-13, 12:37   #978
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I conjectured that:

If (k,b,c) is integer triple, k>=1, b>=2, c != 0, gcd(k,c)=1, gcd(b,c)=1, if (k*b^n+c)/gcd(k+c,b-1) does not have covering set of primes for the n such that: (the set of the n satisfying these conditions must be nonempty, or (k*b^n+c)/gcd(k+c,b-1) proven composite by full algebra factors)

* if c != +-1, (let r be the largest integer such that (-c) is perfect r-th power) k*b^n is not perfect r-th power

* if c = 1, k*b^n is not perfect odd power (of the form m^r with odd r>1), except the case: b = q^m, k = q^r, where q is not of the form t^s with odd s>1, and m and r have no common odd prime factor, and the exponent of highest power of 2 dividing r < the exponent of highest power of 2 dividing m

* if c = -1, k*b^n is not perfect power (of the form m^r with r>1), except the case: b = q^m, k = q^r, where q is not of the form t^s with odd s>1, and m and r have no common odd prime factor

* 4*k*b^n*c is not perfect 4-th power

Then there are infinitely many primes of the form (k*b^n+c)/gcd(k+c,b-1), except the case: b = q^m, k = q^r, where q is not of the form t^s with odd s>1, and m and r have no common odd prime factor, and the exponent of highest power of 2 dividing r >= the exponent of highest power of 2 dividing m, and the equation 2^x = r (mod m) has no solution.

Last fiddled with by sweety439 on 2020-09-13 at 12:39
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Old 2020-09-17, 19:28   #979
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Quote:
Originally Posted by sweety439 View Post
Also these cases:

S15 k=343:

since 343 is cube, 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 not divisible by 3, 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 : 31 · 83
2 : 2^2 · 11 · 877
4 : 2^2 · 809 · 2683
5 : 811 · 160583
7 : 11^2 · 242168453
11 : 31 · 101 · 25357 · 18684739
13 : 397 · 1281101 · 656261029
17 : 11 · 27479311 · 55900668804553
29 : 53 · 197741 · 209188613429183386499227445981
35 : 1337724923 · 18667724069720862256321575167267431
43 : 20943991 · 3055827403675875709696160949928034201885723243
61 : 23539 · (a 61-digit prime)

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

S61 k=324:

since 324 is of the form 4*m^4, all n divisible by 4 have algebra factors, and we only want to know whether it has a covering set of primes for all n not divisible by 4, 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 : 59 · 67
2 : 41 · 5881
3 : 13 · 1131413
5 : 5 · 7 · 1563709723
6 : 13 · 256809250661
7 : 23 · 1255679 · 7051433
13 : 191 · 7860337 · 27268229 · 256289843
14 : 1540873 · 1698953 · 244480646906833
31 : 1888149043321 · 441337391577139 · 1721840403480692512106884569347
34 : 10601 · 174221 · (a 54-digit prime)

and it does not appear to be any covering set of primes, so there must be a prime at some point.
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 k = 289 with odd n > 1 (since 289 is square, all even n for k = 289 have algebra factors)

Another base is R106, which has many k with algebra factors (these k are all squares):

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 R30 k=1369 and R88 k=400

Last fiddled with by sweety439 on 2020-09-18 at 19:18
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