A242848 Number of unimodular roots of the equation z^n + z^k - 1 for all 1 <= k <= n-1.

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, 12, 32, 0, 12, 12, 12, 16, 14, 12, 14, 16, 18, 16, 16, 0, 28, 32, 18, 16, 18, 0, 54, 16, 18, 20, 20, 24, 20, 20, 18, 0, 62, 24, 22, 24, 24, 64, 24, 0, 24, 24

Offset: 2

Author

Tom Edgar, May 23 2014

Keywords

• nonn

Comments

A complex root is called unimodular if it lies on the unit circle.

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.

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).

Examples

			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.
		

Links

• Michael A. Brilleslyper and Lisbeth E. Schaubroeck, Locating Unimodular Roots, College Mathematics Journal, Volume 45, Number 3, May 2014, pp. 162-168(7).

Crossrefs

Cf. A120963.

Programs

Sage

[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]]

Formula

a(n) = Sum_{k=1..n-1; 6 divides (n/gcd(n,k) + k/gcd(n,k))} 2*gcd(n,k).