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 Flawless Series? (Posted on 2003-11-29)
What are the next three numbers in this sequence? 6, 28, 496, 8128, ... ? Please explain how you determined these three numbers.

 See The Solution Submitted by SilverKnight Rating: 2.8889 (9 votes)

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 my solution | Comment 66 of 67 |

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
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