 12:30 PM
Hey @Peter what problem are you working on today ? 12:46 PM
Mainly, the factorization of 6^129+3^129+2^129 , the first hole in the 2^n+3^n+6^n-list. aka (6^{43})^3+(3^{43})^3+(2^{43})^3 1:21 PM
19 hours ago, by Mathphile
For every \$n\$ is there a semiprime that is \$n\$ digits long?
@Peter what do you think about the above question? This can surely be proven by the bounds of Dusart for the number of primes below some x. If you have found an example for every n<10, I think we can safely assume that we have a semiprime with every possible digit-length.
The Cunningham question is wide open, if I understood it right.
To prove the semiprime statement, we only need two primes in the range \$[10^n,2*10^n]\$ and two primes in the range \$[9*10^n,10*10^n]\$ for every integer n>1. This should not be too difficult to be proven. 1:55 PM
@Peter how did you up with \$n \lt 10\$? 2:07 PM
*come up 2:21 PM
As said, two primes in the ranges do the job.
101 * 103 , 991 * 997 , 1009 * 1013 , 9967 * 9973 and so on
For C101, I seem to need the quadratic sieve hmm
@Peter can we find a semiprime that is the square of a prime for any \$n\$ digits 2:50 PM
This works as well, for 1 and 2 digits we have 4 and 25 , for three digits 121 , for 4 digits 1369. And we can continue with one prime in my ranges and square it for the other cases. 3:37 PM
probably as it follows from Bertrand that a prime square occurs between n and 4n.