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Old 2017-05-20, 15:46   #12
sweety439
 
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These problems are to find a prime of the form (k*b^n+c)/gcd(k+c,b-1) with integer n>=1 for fixed integers k, b and c, k>=1, b>=2, gcd(k,c)=1 and gcd(b,c)=1.

For some (k,b,c), there cannot be any prime because of covering set (e.g. (k,b,c) = (78557,2,1), (334,10,-1) or (84687,6,-1)) or full algebra factors (e.g. (k,b,c) = (9,4,-1), (2500,16,1) or (9,4,-25) (the case (9,4,-25) can produce prime only for n=1)) or partial algebra factors (e.g. (k,b,c) = (144,28,-1), (25,17,-9) or (1369,30,-1)). It is conjectured that for every (k,b,c) which cannot be proven that they do not have any prime, there are infinitely primes of the form (k*b^n+c)/gcd(k+c,b-1). (Notice the special case: (k,b,c) = (8,128,1), it cannot have any prime but have neither covering set nor algebra factors)

However, there are many such cases even not have a single known prime, like (21181,2,1), (2293,2,-1), (4,53,1), (1,185,-1), (1,38,1), (269,10,1), (197,7,-1), (4105,17,-9), (16,21,335), (5,36,821), but not all case will produce a minimal prime to base b, e.g. the form (197*7^n-1)/2 is the form 200{3} in base 7, but since 2 is already prime, the smallest prime of this form (if exists) will not be a minimal prime in base 7.

The c=1 and gcd(k+c,b-1)=1 case is the Sierpinski problem base b, and the c=-1 and gcd(k+c,b-1)=1 case is the Riesel problem base b.
gcd(k+c,b-1) is the largest number that divides k*b^n+c for all n.

Note: gcd(0, m) = m for all integer m, and gcd(1, m) = 1 for all integer m.

Last fiddled with by sweety439 on 2017-05-20 at 15:47
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Old 2017-05-20, 17:11   #13
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Note: gcd(0, m) = m for all integer m, and gcd(1, m) = 1 for all integer m.
Thank you, Captain Obvious!
In other news today, light travels faster than sound, and a minute contains 60 seconds.
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Old 2017-05-21, 03:45   #14
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…and a minute contains 60 seconds.
Is that always true?

https://en.wikipedia.org/wiki/Leap_second

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Old 2017-05-21, 05:15   #15
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Is that always true?
It is just as true as "gcd(0, m) = m for all integer m".
For m=0, gcd(0, 0) = 24. 24 is a greatest common divisor of 0 and 0, because it divides both 0 and 0, and there is no higher number: see goo.gl/ASN4Ov ... and 24 ≠ 0
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Old 2017-05-21, 17:43   #16
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It is just as true as "gcd(0, m) = m for all integer m".
For m=0, gcd(0, 0) = 24. 24 is a greatest common divisor of 0 and 0, because it divides both 0 and 0, and there is no higher number: see goo.gl/ASN4Ov ... and 24 ≠ 0
No, gcd(0, m) = m is true only for positive integer m.

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Old 2017-05-21, 22:03   #17
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No, gcd(0, m) = m is true only for positive integer m.
gcd(0, 0) = 0 so the rule holds in all cases except when you want to take 24 as maximal instead of 0 for humor.
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Old 2017-05-22, 14:29   #18
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light travels faster than sound
Depends on the medium. Through a vacuum, it certainly does. Also through the air we breathe. But it takes a long, long time for the EM energy produced in the solar core to make its way through the interior of the sun, and out as sunshine. Sound waves travel through the interior of the sun much more quickly.

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Old 2017-05-22, 18:00   #19
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Depends on the medium. Through a vacuum, it certainly does. Also through the air we breathe. But it takes a long, long time for the EM energy produced in the solar core to make its way through the interior of the sun, and out as sunshine. Sound waves travel through the interior of the sun much more quickly.
If the medium is not transparent, the the speed of (visible light) is zero, thus it is lower then that of sound. Besides, if the medium is vacuum, then the speed of sound is zero, since sound needs medium to spread.
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Old 2017-05-22, 18:05   #20
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Thank you, Captain Obvious!
In other news today, light travels faster than sound, and a minute contains 60 seconds.
A minute does not always contain 60 seconds, since the definition of second is from the cesium atomic, it is not always 1/60 minute = 1/86400 day, since the definition of day is form earth. Besides, in Alaska and in Amazon forest, the length of "second" is not the same, since the distance of them to geocentric is different XDDD...

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Old 2019-11-27, 09:36   #21
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Base 36 has only two unsolved family:

(4428*36^n+67)/5
(6480*36^n+821)/7

Base 40 has only two unsolved family:

(13998*40^n+29)/13
(86*40^n+37)/3
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Old 2019-11-27, 09:39   #22
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Base 36 has only two unsolved family:

(4428*36^n+67)/5
(6480*36^n+821)/7

Base 40 has only two unsolved family:

(13998*40^n+29)/13
(86*40^n+37)/3
A (probable) prime was found:

(13998*40^12381+29)/13

Written in base 40, this number is Qa{U12380}X

This number is likely the second-largest "base 40 minimal prime"
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