phi(n!!)

[Leroy Quet]

Oh sorry, of course (since -2 = 1 IS a square mod 3...
this gives b.t.w. the term 9 in the sequence, for m=1).
So the interesting connection with A029547 applies only to the
subsequence of squares c(m=3n).
Maximilian

On 5/26/07, Don Reble <djr at nk.ca> wrote:
> > since 3 | (m²+2) is impossible
>    Eh?  3 | (1^2+2), 3 | (2^2+2), 3 | (4^2+2), ...
>    3 divides, unless m is a multiple of 3.
> --
> Don Reble  djr at nk.ca
>
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