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%I #15 Feb 03 2022 02:36:05
%S 0,0,0,2,0,2,0,0,4,4,0,4,4,6,0,6,0,6,8,6,8,8,0,16,8,0,8,10,12,10,0,12,
%T 12,32,0,12,12,12,16,14,12,14,16,18,16,16,0,28,32,18,16,18,0,54,16,18,
%U 20,20,24,20,20,18,0,62,24,22,24,24,64,24,0,24,24
%N Number of unimodular roots of the equation z^n + z^k - 1 for all 1 <= k <= n-1.
%C A complex root is called unimodular if it lies on the unit circle.
%C While there is technically 1 unimodular root for z-1, the offset of this sequence is 2 so that the polynomial in question has three terms.
%C Let g=gcd(n,k). If 6 divides n/g + k/g, then z^n + z^k - 1 has exactly 2*g unimodular roots each of the form exp(i(Pi/3g + 2*Pi*m/g)), or its conjugate, where 0 <= m <= g-1 (see Theorem 2 from College Math Journal reference).
%H Michael A. Brilleslyper and Lisbeth E. Schaubroeck, <a href="http://dx.doi.org/10.4169/college.math.j.45.3.162">Locating Unimodular Roots</a>, College Mathematics Journal, Volume 45, Number 3, May 2014, pp. 162-168(7).
%F a(n) = Sum_{k=1..n-1; 6 divides (n/gcd(n,k) + k/gcd(n,k))} 2*gcd(n,k).
%e The polynomial z^20 + z^4 - 1 has 8 roots (of the 20 possible) lying on the unit circle; moreover, z^20 + z^k - 1 has no roots lying on the unit circle when 1 <= k <= 19 and k != 4. Thus a(20) = 8.
%o (Sage)
%o [sum(2*gcd(n,k) for k in [1..n-1] if Integer(n/gcd(n,k)+k/gcd(n,k))%6==0) for n in [2..100]]
%Y Cf. A120963.
%K nonn
%O 2,4
%A _Tom Edgar_, May 23 2014
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