A079279

[Jon Perry]

The author asks for a proof of his sequence:

Assume the sequence opens 1,2,3,4,8,9,10.

The first step is to prove that only numbers of the form 6k+2,6k+3,6k+4 are
in the sequence.

As 6k+5 is odd, as is 6k+7, and neither are divisible by 3, they will not
share any common factors with 6k+2,6k+3,6k+4, except if 6k+2 and 6k+7 are
divisible by 5.

However, if 6k+2 is divisible by 5, so is 6k-3, and so 6k+2 (it is also
even, as are 6k-2,6k-4)will not be in the sequence for 6k+2=0mod5, i.e.
k=3mod5, i.e. 20mod30 are not present.

6k+6 is divisible by 6, and will hence always share common factors with 3 of
the previous terms, either as all three of 6k+2,6k+3,6k+4 are present, or we
have the pattern 6k-2,6k+3,6k+4.

Jon Perry
perry at globalnet.co.uk
http://www.users.globalnet.co.uk/~perry/maths/
http://www.users.globalnet.co.uk/~perry/DIVMenu/
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