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Count with me (Posted on 2005-05-26) Difficulty: 2 of 5

See The Solution Submitted by Tristan    
Rating: 4.2500 (4 votes)

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No Subject | Comment 9 of 16 |

With the given terms of sequence, the solution proposed by Jer and the solution of some other mails (I've proposed it also in a precedent mail = 7,1,7,4,8,3,9) are both possible.

Perhaps (probably) the intended solution was that of Jer, based on prime numbers, which is better related to the title of the sequence, but in my opinion the sequence need some more terms to have uniqueness.

Edited on May 26, 2005, 10:26 pm
  Posted by armando on 2005-05-26 22:20:41

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