Let a[n] = 2^(n/2+1) sin(n t)/sqrt(7),

where t = atan(sqrt(7)).

Then sin(t) = sqrt(7)/(2 sqrt(2)), and

cos(t) = 1/(2 sqrt(2)).

To start with, a[0] = 0, a[1] = 1.

One can verify the identity

sin(n x) = 2 cos(x) sin((n-1) x) - sin((n-2) x).

Then we see that

a[n] = 2 sqrt(2) cos(t) a[n-1] - 2 a[n-2]

       = a[n-1] - 2 a[n-2],

and so a[n] is integral for all n.

 "R. William Gosper" <rwg@spnet.com> wrote:

For how many n in {0,1,...,999999} is

n/2 + 1
2 sin(atan(sqrt(7)) n)
-----------------------------
sqrt(7)

an integer? An odd integer?
--rwg

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