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 Register FAQ Search Today's Posts Mark Forums Read  2019-11-07, 20:25   #320
sweety439

Nov 2016

22×691 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
Attached Files Williams PP.txt (22.2 KB, 38 views)   2019-11-09, 01:28   #321
sweety439

Nov 2016

53148 Posts searched (b-1)*b^n-1 and (b-1)*b^n+1 for bases 2<=b<=1000, both up to n=1000

(b+1)*b^n-1 is still running ....
Attached Files Williams MM.txt (35.1 KB, 57 views) Williams MP.txt (34.7 KB, 37 views)

Last fiddled with by sweety439 on 2019-11-09 at 01:30   2019-11-11, 01:33   #322
sweety439

Nov 2016

22×691 Posts ............
Attached Files Williams PM.txt (43.6 KB, 91 views)   2019-11-17, 17:47   #323
sweety439

Nov 2016

53148 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
Attached Files conjectured smallest Sierpinski number base b.txt (15.7 KB, 40 views) conjectured smallest Riesel number base b.txt (15.7 KB, 34 views)   2019-11-20, 15:02   #324
sweety439

Nov 2016

276410 Posts .......
Attached Files prime gaps.txt (20.3 KB, 35 views)   2019-11-23, 19:19   #325
sweety439

Nov 2016

22×691 Posts Quote:
 Originally Posted by sweety439 Not including the single-digit primes, proof of that these sets are complete: b=2: we obtain the 2-digit primes 10 and 11, since for any prime p > 11, p must start and end with digit 1, and we have 11 <<< p, thus the 2-kernel {10, 11} is complete. b=3: we obtain the 2-digit primes 10, 12 and 21, for any prime p > 21, if p end with 2 and 12 !<<< p, then p must contain only 0 and 2, thus p is divisible by 2 and > 2, thus not prime, therefore, p must end with 1 (p cannot end with 0, or p is divisible by 10 and not prime), since 21 !<<< p, p must contain only 0 and 1, but since 10 !<<< p and p cannot have leading zeros, thus p can only have the digit 1, i.e. p is a repunit, and the smallest repunit prime is 111, thus completed the 3-kernel {10, 12, 21, 111}. b=4: we obtain the 2-digit primes 11, 13, 23 and 31, for any prime p > 31, if p end with 3 and 13 !<<< p and 23 !<<< p, then p must contain only 0 and 3, thus p is divisible by 3 and > 3, thus not prime, therefore, p must end with 1 (p cannot end with 0 or 2, or p is divisible by 2 and not prime), since 11 !<<< p and 31 !<<< p, thus p (before the final digit 1) must contain only 0 and 2, and we obtain the prime 221, since p cannot have leading zeros, the remain case is only 2{0}1, but all numbers of the form 2{0}1 are divisible by 3 and > 3, thus not prime, thus we completed the 4-kernel {11, 13, 23, 31, 221}. b=5: we obtain the 2-digit primes 10, 12, 21, 23, 32, 34 and 43, for any prime p>43: p end with 1 --> before this 1, p cannot contain 2 --> if p end with 11, then we find the prime 111, and all other primes contain at most two 1, ... p end with 2 --> before this 2, p cannot contain 1 or 3 --> p only contain 0, 2 and 4 --> p is divisible by 2 and > 2 --> p is not prime (thus, 12 and 32 are the only such primes end with 2) p end with 3 --> before this 3, p cannot contain 2 or 4 --> p only contain 0, 1 and 3 --> we obtain the primes 133 and 313, thus other primes p cannot contain both 1 and 3 (before the final digit 3), and since p cannot have leading zeros, if p begin with 1, then p is of the form 1{0,1}3 --> it must be of the form {1}3 (to avoid the prime 10) --> but 113 is not prime and all primes except 111 contain at most two 1 --> this way cannot find any primes, if p begin with 3, then p is of the form 3{0,3}3 --> p is divisible by 3 and > 3 --> p is not prime (thus, all such primes end with 3 are 23, 43, 133 and 313) p end with 4 --> before this 4, p cannot contain 3 --> since all primes > 2 are odd, p must contain at least one 1 --> we obtain the prime 414 and we know that no 2 can before this 1 (to avoid the prime 21) and no 0 or 2 can after this 1 (to avoid the primes 10 and 12) --> 1 must be the leading digit (since p cannot have leading zeros, and no 2, 3, 4 can before this 1 (to avoid the primes 21, 34 and 414, respectively) --> p must be of the form 1{4} or 11{4} (since all primes except 111 contain at most two 1) --> and we obtain the prime 14444 (thus, all such primes end with 4 are 34, 414 and 14444)
b=5, p end with 1: (assume p is a prime in the minimal set of the strings for primes with at least two digits in base 5 other than 10, 12, 21, 23, 32, 34, 43, 111, 131, 133, 313, 401, 414, 14444, 30301, 33001, 33331, 44441, 300031)

* 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 2019-11-24 at 15:23   2019-11-23, 19:46   #326
sweety439

Nov 2016

22×691 Posts Quote:
 Originally Posted by sweety439 b=5, p end with 1: (assume p is a prime in the minimal set of the strings for primes with at least two digits in base 5 other than 10, 12, 21, 23, 32, 34, 43, 111, 131, 133, 313, 401, 414, 14444, 30301, 33001, 33331, 44441, 300031) * 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) ** 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
Therefore, before this 1, p can only contain the digits 0, 3, and 4.

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 5-kernel is complete!!!   2019-11-23, 19:58   #327
sweety439

Nov 2016

22×691 Posts Quote:
 Originally Posted by sweety439 Not including the single-digit primes, proof of that these sets are complete: b=2: we obtain the 2-digit primes 10 and 11, since for any prime p > 11, p must start and end with digit 1, and we have 11 <<< p, thus the 2-kernel {10, 11} is complete. b=3: we obtain the 2-digit primes 10, 12 and 21, for any prime p > 21, if p end with 2 and 12 !<<< p, then p must contain only 0 and 2, thus p is divisible by 2 and > 2, thus not prime, therefore, p must end with 1 (p cannot end with 0, or p is divisible by 10 and not prime), since 21 !<<< p, p must contain only 0 and 1, but since 10 !<<< p and p cannot have leading zeros, thus p can only have the digit 1, i.e. p is a repunit, and the smallest repunit prime is 111, thus completed the 3-kernel {10, 12, 21, 111}. b=4: we obtain the 2-digit primes 11, 13, 23 and 31, for any prime p > 31, if p end with 3 and 13 !<<< p and 23 !<<< p, then p must contain only 0 and 3, thus p is divisible by 3 and > 3, thus not prime, therefore, p must end with 1 (p cannot end with 0 or 2, or p is divisible by 2 and not prime), since 11 !<<< p and 31 !<<< p, thus p (before the final digit 1) must contain only 0 and 2, and we obtain the prime 221, since p cannot have leading zeros, the remain case is only 2{0}1, but all numbers of the form 2{0}1 are divisible by 3 and > 3, thus not prime, thus we completed the 4-kernel {11, 13, 23, 31, 221}. b=5: we obtain the 2-digit primes 10, 12, 21, 23, 32, 34 and 43, for any prime p>43: p end with 1 --> before this 1, p cannot contain 2 --> if p end with 11, then we find the prime 111, and all other primes contain at most two 1, ... p end with 2 --> before this 2, p cannot contain 1 or 3 --> p only contain 0, 2 and 4 --> p is divisible by 2 and > 2 --> p is not prime (thus, 12 and 32 are the only such primes end with 2) p end with 3 --> before this 3, p cannot contain 2 or 4 --> p only contain 0, 1 and 3 --> we obtain the primes 133 and 313, thus other primes p cannot contain both 1 and 3 (before the final digit 3), and since p cannot have leading zeros, if p begin with 1, then p is of the form 1{0,1}3 --> it must be of the form {1}3 (to avoid the prime 10) --> but 113 is not prime and all primes except 111 contain at most two 1 --> this way cannot find any primes, if p begin with 3, then p is of the form 3{0,3}3 --> p is divisible by 3 and > 3 --> p is not prime (thus, all such primes end with 3 are 23, 43, 133 and 313) p end with 4 --> before this 4, p cannot contain 3 --> since all primes > 2 are odd, p must contain at least one 1 --> we obtain the prime 414 and we know that no 2 can before this 1 (to avoid the prime 21) and no 0 or 2 can after this 1 (to avoid the primes 10 and 12) --> 1 must be the leading digit (since p cannot have leading zeros, and no 2, 3, 4 can before this 1 (to avoid the primes 21, 34 and 414, respectively) --> p must be of the form 1{4} or 11{4} (since all primes except 111 contain at most two 1) --> and we obtain the prime 14444 (thus, all such primes end with 4 are 34, 414 and 14444)
b=6: we obtain the 2-digit primes 11, 15, 21, 25, 31, 35, 45 and 51, for any prime p > 51:

* 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 6-kernel is complete!!!   2019-11-23, 22:11   #328
sweety439

Nov 2016

ACC16 Posts ..........
Attached Files divisors of googolplex - googol.txt (19.7 KB, 37 views) nondivisors of googolplex - googol.txt (5.0 KB, 48 views)   2019-12-27, 17:14   #329
sweety439

Nov 2016

22·691 Posts ........
Attached Files (k-1)k(k+1) has 4 distince prime factors.txt (3.4 KB, 38 views)   2019-12-27, 18:20   #330
sweety439

Nov 2016

276410 Posts ...........
Attached Files (k-1)k(k+1) has 4 distinct prime factors.txt (3.4 KB, 35 views)   Thread Tools Show Printable Version Email this Page

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