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 _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun --------------------------------- Do you Yahoo!? Yahoo! Shopping - Send Flowers for Valentine's Day