A218664 Coefficients of cubic polynomials p(x+n), where p(x) = x^3 + x^2 - 2*x - 1.

1, 1, -2, -1, 1, 4, 3, -1, 1, 7, 14, 7, 1, 10, 31, 29, 1, 13, 54, 71, 1, 16, 83, 139, 1, 19, 118, 239, 1, 22, 159, 377, 1, 25, 206, 559, 1, 28, 259, 791, 1, 31, 318, 1079, 1, 34, 383, 1429, 1, 37, 454, 1847, 1, 40, 531, 2339, 1, 43, 614, 2911, 1, 46, 703, 3569, 1, 49, 798, 4319

Offset: 0

Author

Roman Witula, Nov 04 2012

Keywords

• sign

Comments

We have p(x) = (x - c(1))*(x - c(2))*(x - c(4)), where c(j) := 2*cos(2*Pi*j/7). We note that c(4) = c(3) = -c(1/2), c(1) = s(3) and c(2) = -s(1), where s(j) := 2*sin(Pi*j/14). Moreover we obtain -p(-x) = x^3 - x^2 - 2*x + 1 = (x + c(1))*(x + c(2))*(x + c(4)), q(x) := -x^3*p(1/x) = x^3 + 2*x^2 + x - 1 = (x - c(1)^(-1))*(x - c(2)^(-1))*(x - c(4)^(-1)), and -q(-x) = x^3 - 2*x^2 + x + 1 = (x + c(1)^(-1))*(x + c(2)^(-1))*(x + c(4)^(-1)).

We also have p(x+2) = x^3 + 7*x^2 + 14*x + 7 = (x + s(2)^2)*(x + s(4)^2)*(x + s(6)^2). The polynomial -p(-x-2) = x^3 - 7*x^2 + 14*x - 7 = (x - s(2)^2)*(x - s(4)^2)*(x - s(6)^2) is known as Johannes Kepler's cubic polynomial (see Witula's book).

Let us set r(x) := p(x+1). It can be verified that -x^3*r(1/x) = x^3 - 3*x^2 - 4*x - 1 = (x - c(1)/c(4))*(x - c(4)/c(2))*(x - c(2)/c(1)); for example, we have c(1)^3 + c(1)^2 - 2*c(1) - 1 = 0 which implies that c(1)^2 + 2*c(1) = 1/(c(1) - 1), and then c(1)^2 + 2*c(1) = c(4)/c(2) since c(4)/c(2) = (c(1)^4 - 4*c(1)^2 + 2)/(c(1)^2 - 2).

The polynomials p(x+n) and the ones obtained as above (i.e., after simple algebraic transformations) are the characteristic polynomials of many sequences in the OEIS; see crossrefs.

References

• R. Witula, Complex Numbers, Polynomials and Partial Fraction Decomposition, Part 3, Wydawnictwo Politechniki Slaskiej, Gliwice 2010 (Silesian Technical University publishers).

Crossrefs

Cf. A094648, A096975, A033304, A214683, A006053, A215076, A215100, A215007, A215008, A215143, A215493, A215494, A215510.

Formula

We have a(4*k) = 1, a(4*k + 1) = 3*k + 1, a(4*k + 2) = 3*k^2 + 2*k - 2, a(4*k + 3) = k^3 + k^2 - 2*k - 1. Further, the following relations hold true: b(k+1) = b(k) + 3, c(k+1) = 2*b(k) -2*c(k) + 3, d(k+1) = b(k) - 2*c(k) - d(k) + 1, where p(x + k) = x^3 + b(k)*x^2 + c(k)*x + d(k).

Empirical g.f.: -(x^15 - x^14 - 2*x^13 + x^12 - 5*x^11 + 10*x^10 + 3*x^9 - 3*x^8 - 3*x^7 - 11*x^6 + 3*x^4 + x^3 + 2*x^2 - x - 1) / ((x-1)^4*(x+1)^4*(x^2+1)^4). - Colin Barker, May 17 2013