n consecutive integers / each integer has n distinct prime factors

 
n  consecutive integers such that each integer has  n  distinct prime factors
 

n = 2

14   =   2   *   7
15   =   3   *   5

n = 3

644   =   2^2   *   7   *   23
645   =   3   *   5   *   43
646   =   2   *   17   *   19

n = 4

134043   =   3   *   7   *   13   *   491
134044   =   2^2   *   23   *   31   *   47
134045   =   5   *   17   *   19   *   83
134046   =   2   *   3^2   *   11   *   677

n = 5

129963314   =   2   *   13   *   37   *   53   *   2549
129963315   =   3   *   5   *   31   *   269   *   1039
129963316   =   2^2   *   7   *   97   *   109   *   439
129963317   =   11^2   *   17   *   23   *   41   *   67
129963318   =   2   *   3   *   89   *   199   *   1223

n = 6

626804494291   =   23   *   29   *   83   *   97   *   151   *   773
626804494292   =   2^2   *   47   *   59   *   101   *   131   *   4271
626804494293   =   3   *   61   *   127   *   191   *   337   *   419
626804494294   =   2   *   11   *   19   *   71   *   197   *   107209
626804494295   =   5   *   7^2   *   41   *   79   *   443   *   1783
626804494296   =   2^3   *   3   *   31   *   37   *   107   *   212801

 
 
n = 7?   n = 8?

and so on.
 
 
 
 
 
 
 
 
 
 

Advertisements

About benvitalis

math grad - Interest: Number theory
This entry was posted in Math Beauty, Number Puzzles and tagged . Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s