20191107, 20:25  #320 
Nov 2016
2^{2}×19×31 Posts 
searched (b+1)*b^n+1 for bases 2<=b<=1000 (not including b end with 1, 4, 7 or X, since for these b, all such numbers are divisible by 3), up to n=1000

20191109, 01:28  #321 
Nov 2016
2^{2}·19·31 Posts 
searched (b1)*b^n1 and (b1)*b^n+1 for bases 2<=b<=1000, both up to n=1000
(b+1)*b^n1 is still running .... Last fiddled with by sweety439 on 20191109 at 01:30 
20191111, 01:33  #322 
Nov 2016
2^{2}×19×31 Posts 
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20191117, 17:47  #323 
Nov 2016
2^{2}×19×31 Posts 
The CK of the Sierpinski problems and the Riesel problems for these bases:
* all bases <= 1000 * all square bases <= 40^2 * all power of 2 bases <= 2^10 
20191120, 15:02  #324 
Nov 2016
934_{16} Posts 
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20191123, 19:19  #325  
Nov 2016
2^{2}×19×31 Posts 
Quote:
* before this 1, p cannot contain 2 (because of 21) * if before this 1, p contain 1 > assume p is {xxx}1{yyy}1 > y cannot contain 0, 1, 2, or 3 (because of 10, 111, 12, and 131) ** if y is not empty > y contain only the digits 4 > x cannot contain 1, 2, 3, or 4 (because of 111, 21, 34, and 414) > x contain only the digits 0 (a contradiction, since a number cannot have leading zeros, and all numbers of the form 1{4}1 is divisible by 2 and cannot be primes) ** thus y is empty > p is {xxx}11 > x cannot contain 1 or 2 (because of 111 and 21), but x cannot contain both 3 and 4 (because of 34 and 43) > x contain either only 0 and 3, or only 0 and 4, however, {0,3}11 is divisible by 3, and {0,4}11 is divisible by 2, and neither can be primes, a contradiction!!! > thus, before this 1, p cannot contain 1 Last fiddled with by sweety439 on 20191124 at 15:23 

20191123, 19:46  #326  
Nov 2016
2^{2}×19×31 Posts 
Quote:
However, p cannot contain both 3 and 4 (because of 34 and 43) thus, before this 1, p contain either only 0 and 3, or only 0 and 4 For the primes contain only 0 and 3: Since the first digit cannot be 0, it can only be 3 thus p is 3{xxx}1 x should contain at least one 3 (or p is of the form 3{0}1, and 3{0}1 is divisible by 2) thus we can assume p is 3{xxx}3{yyy}1 * if both x and y contain 0, then we have the prime 30301 * if x does not contain 0, then x contain only 3 ** if x contain at least two 3, then we have the prime 33331 ** if x contain exactly one 3, then p is 3{0}3{0}3{yyy}1, and y can contain only the digits 0 (y cannot contain 3, because of 33331) > p is of the form 3{0}3{0}3{0}1 and divisible by 2, a contradiction ** if x is empty, then p is 33{yyy}1, and y can contain only the digits 0 (y cannot contain at least two 3 because of 33331, and if y contain exactly one 3, then p is divisible by 2 and cannot be prime), then we have the prime 33001 * if y does not contain 0, then y contain only 3 ** if y contain at least two 3, then we have the prime 33331 ** if y contain exactly one 3, then p is 3{xxx}3{0}3{0}1, and x can contain only the digits 0 (x cannot contain 3, because of 33331) > p is of the form 3{0}3{0}3{0}1 and divisible by 2, a contradiction ** if y is empty, then p is 3{xxx}31, and x can contain only the digits 0 (x cannot contain at least two 3 because of 33331, and if x contain exactly one 3, then p is divisible by 2 and cannot be prime), then we have the prime 300031 For the primes contain only 0 and 4: Since the first digit cannot be 0, it can only be 4 thus p is 4{xxx}1 if x contain 0, then we have the prime 401 if x does not contain 0 (thus contain only the digit 4), then we have the prime 44441 Therefore, no such prime p can exist!!! The 5kernel is complete!!! 

20191123, 19:58  #327  
Nov 2016
4464_{8} Posts 
Quote:
* if p end with 5 > before this 5, p cannot contain 1, 2, 3, or 4, however, all numbers of the form {0, 5}5 is divisible by 5 and cannot be prime * thus, p can only end with 1 > before this 1, p cannot contain 1, 2, 3, or 5 > before this 1, p can only contain the digits 0 and 4 > since the first digit cannot be 0, it can only be 4 > p is of the form 4{xxx}1 > x should contain at least one 4 (or p is of the form 4{0}1, and 4{0}1 is divisible by 5) > thus we can assume p is 4{xxx}4{yyy}1 * if y contain at least one 0, then we have the prime 4401 * if y contain at least one 4, then we have the prime 4441 * if y is empty: ** if x contain at least one 4, then we have the prime 4441 ** if x does not contain 4 (thus contain only the digit 0), then we have the prime 40041 The 6kernel is complete!!! 

20191123, 22:11  #328 
Nov 2016
2^{2}×19×31 Posts 
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20191227, 17:14  #329 
Nov 2016
2^{2}·19·31 Posts 
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20191227, 18:20  #330 
Nov 2016
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