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-   -   A Sierpinski/Riesel-like problem (https://www.mersenneforum.org/showthread.php?t=21839)

 sweety439 2016-12-24 19:07

1 Attachment(s)
The correct text file for extended R17 is here. (the above file only lists the k's < 44, we should lists all k's < 49)

 Batalov 2016-12-24 19:45

2 Attachment(s)
You can surely post 10 messages an hour. You have proven your capacity to type.
You come across as a graphomaniac.

But do you think anybody is reading your inane stream of conscience?
Why would they -- if you are appear to be deaf to other people's messages?

 sweety439 2016-12-26 19:26

The (probable) primes with n > 1000 for the extended Sierpinski/Riesel problems (with bases b <= 32, except b = 2, 3, 6, 15, 22, 24, 28, 30) are:

S4:

186*4^10458+1

S7:

(141*7^1044+1)/2

S10:

804*10^5470+1

S12:

404*12^714558+1
378*12^2388+1

S16:

(23*16^1074+1)/3

S17:

10*17^1356+1

S18:

122*18^292318+1
381*18^24108+1
291*18^2415+1

S25:

(61*25^3104+1)/2

S26:

32*26^318071+1
217*26^11454+1
95*26^1683+1
178*26^1154+1

R4:

(106*4^4553-1)/3
74*4^1276-1
296*4^1275-1 ( = 74*4^1276-1, the two primes are the same, since 296 = 74 * 4)

R7:

(367*7^15118-1)/6
(313*7^5907-1)/6
(159*7^4896-1)/2
(429*7^3815-1)/2
(419*7^1052-1)/2

R12:

(298*12^1676-1)/11

R17:

44*17^6488-1
(29*17^4904-1)/4
(13*17^1123-1)/4

R25:

86*25^1029-1

R26:

115*26^520277-1
32*26^9812-1
(121*26^1509-1)/5

 sweety439 2016-12-26 19:34

[QUOTE=sweety439;449149]
Extend Sierpinski problem base b: Finding and proving smallest k such that (k*b^n+1)/gcd(k+1, b-1) is composite for all integer n>1.

Extend Riesel problem base b: Finding and proving smallest k such that (k*b^n-1)/gcd(k-1, b-1) is composite for all integer n>1.
[/QUOTE]

It should not be "n>1", it should be "n>=1", we allow n=1 for (k*b^n+1)/gcd(k+1, b-1) and (k*b^n-1)/gcd(k-1, b-1), but we do not allow n=0.

 sweety439 2016-12-26 19:41

[QUOTE=gd_barnes;449175]
First suggestion: When showing which k's need a prime use the k-value in this form:
k*3^n+(3^n-1)/2

Reasoning for the suggestion. For a more complex base/append combination, such as base 9 append 7, it would not be clear what k is being searched unless you use the form:
k*9^n+(9^n-1)*7/8

Second suggestion: Do not show k=3m as remaining. It is redundant.
[/QUOTE]

This is another problem, different from the "[URL="http://mersenneforum.org/showthread.php?t=21832"]Add repeated digits after a number until it is prime[/URL]" problem. First, although for the Riesel case this problem are special cases (when d divides b-1) of that problem, but for the Sierpinski case it is not. Second, in that problem, if d does not divide b-1, then it cannot be transfer to this problem.

Thus, in this problem, we say "k=1617 is remaining for the extended Riesel base 3 problem", but in that problem, we say "k=806 is remaining for the base 3 d=1 problem".

 sweety439 2016-12-27 16:48

These problems are just my extending for the original Sierpinski/Riesel problems.

 sweety439 2016-12-27 16:49

For (269*10^n+1)/9, tested to n=6000, still no (probable) prime found.

 sweety439 2016-12-28 16:34

For the two bases that have only one k remaining:

S10, k=269:

(269*10^n+1)/9 seems to be tested to n=10000 (I ran the program of 25 hours and 38 minutes!!!), still no (probable) prime found, base released.

R7, k=197:

See the link: [URL]https://www.rose-hulman.edu/~rickert/Compositeseq/[/URL], (197*7^n-1)/2 is already tested to n=15000 with no (probable) prime found.

 sweety439 2016-12-28 16:43

3 Attachment(s)
Also tested extended S15, S22 and S24 to k=500.

For extended S15, there are 4 k's <= 500 remain: 219, 225, 341, 343.
For extended S22, there are 5 k's <= 500 (excluding the GFNs: i.e. k=22 and k=484) remain: 173, 346, 383, 461, 464.
For extended S24, there are 5 k's <= 500 remain: 319, 346, 381, 461, 486. However, all of them are correspond to CRUS primes, since all these k's satisfy that gcd(k+1,24-1)=1, and according to the CRUS page, the smallest remain k for the original S24 problem is k=656. Thus, there are in fact no k's <= 500 remain for extended S24.

 sweety439 2016-12-28 16:51

If the k satisfies that gcd(k+-1, b-1)=1, then the correspond prime of the extended Sierpinski/Riesel problem is the same as the correspond prime of the original Sierpinski/Riesel problem.

The extended Sierpinski/Riesel problem is just my extending of the original Sierpinski/Riesel problem to the k's such that gcd(k+-1, b-1) is not 1. Since k*b^n+-1 is always divisible by gcd(k+-1, b-1), the division is necessary for there to be any chance of finding primes.

 sweety439 2016-12-28 17:20

3 Attachment(s)
Also tested extended R15, R22 and R24 to k=500.

For extended R15, there are 6 k's <= 500 remain: 47, 203, 239, 407, 437, 451.
For extended R22, there are 4 k's <= 500 (k=185 prime is given by the CRUS page) remain: 208, 211, 355, 436.
For extended R24, there are 7 k's <= 500 remain: 69, 201, 339, 346, 364, 389, 461. However, the k's != 1 mod 23 are correspond to CRUS primes, according to the CRUS page, k=389 is the only k <= 500 remain for the original R24 problem.

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