What are the next three numbers in this sequence? 6, 28, 496, 8128, ... ? Please explain how you determined these three numbers.

The formula is 2^(n-1)*(2^n-1).  The n's in the sequence are the prime numbers:

2^(2-1)*(2^2-1)=2*3=6

2^(3-1)*(2^3-1)=4*7=28

2^(5-1)*(2^5-1)=16*31=496

2^(7-1)*(2^7-1)=64*127=8128

2^(11-1)*(2^11-1)=1024*2047=2096128

2^(13-1)*(2^13-1)=4096*8191=33550336

2^(17-1)*(2^17-1)=65536*131071=8589869056

Posted by Ed on 2007-10-19 19:56:46