A192004 Alternating row sums of array A187360: minimal polynomial of 2*cos(Pi/n) evaluated for x=-1.

1, -1, -2, -1, 1, -2, 1, -1, 1, 1, 1, -2, 1, 1, 1, -1, 1, 1, 1, 1, 1, 1, 1, -2, 1, 1, 1, 1, 1, 1, 1, -1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, -2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, -1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 1

Author

Wolfdieter Lang, Jul 14 2011

Keywords

• sign
• easy

Comments

It seems that after a(1) = 1, -1's occur only at the positions 2^k (with k >= 1) and -2's only at positions 3*2^k (with k >= 0, A007283), with everything else being 1. It would be nice to know whether this is true. - Antti Karttunen, May 27 2017

From Wolfdieter Lang, May 29 2017: (Start)

The preceding conjecture can be checked by using for even n Theorem 1A, eq. (41), and for odd n Theorem 2A, eq. (50) of the W. Lang arXiv link given in A187360 putting x = -1.

One uses for the polynomials that (A127672) and q (A130777) appearing there the result that(n, -1) = A099837(n+3), i.e., = 2 if n == 0 (mod 3), = -1 if n == 1 or 2 (mod 3), and q(n, -1) = A061347(n+2), i.e., = 1 if n == 0 or 2 (mod 3) and = -2 if n == 1 (mod 3).

E.g., n = 2^k, k >= 1: C(2^k, -1) = that(2^(k-1), -1) = -1 because 2^(k-1) == 1 or 2 (mod 3).

n = 3*2^k, k >= 1: C(2^k*3) = that(2^(k-1)*3, -1) / that(2^(k-1), -1) = 2/(-1) = -2 because 2^(k-1)*3 == 0 (mod 3), and the previous congruence. C(3, -1) = -2 also, by theorem 2A, see the next example.

n = 3^k, k >= 1: C(3^k, -1) = q((3^k-1)/2, -1) / q((3^(k-1)-1)/2, -1) = (-2)/1 = -2 if k = 1, and = (-2)/(-2) = +1 if k >= 2. (End)

Links

• R. J. Mathar, Table of n, a(n) for n = 1..220

Crossrefs

Cf. A055034, A192003.

Formula

a(n) = Sum_{m=0..A055034(n)} (-1)^m*A187360(n,m), n >= 1.

a(n) = C(n,x=-1), with the minimal (monic and integer) polynomial C(n,x) of 2*cos(Pi/n).