View Single Post
Old 2007-01-10, 17:36   #4
Citrix
 
Citrix's Avatar
 
Jun 2003

24×97 Posts
Default

Quote:
Originally Posted by thommy View Post
b=4, as 2*1^1+1 and 2*2^1+1 and 2*3^1+1 are prime
For all n : 2*4^n+1=2*1^n+1=2*1+1=3=0 (mod 3), so every number divisible by 3.
Those questions seem not that interesting.
Thommy.
2 is not considered a sierpinski number for base 4, since the solution is trivial and no covering set is involved.
Citrix is offline   Reply With Quote