cp's OEIS Frontend

This is half of a two sided sequences.

The other half is A274664. - Kai Wang, Aug 02 2016

a(n) is x1^n + x2^n + x3^n, where x1, x2, x3 are the roots of the polynomial x^3 + 4*x^2 - 11*x - 1.

x1 = (cos(Pi/7))^2/(cos(2*Pi/7)*cos(4*Pi/7)),

x2 = -(cos(2*Pi/7))^2/(cos(4*Pi/7)*cos(Pi/7)),

x3 = -(cos(4*Pi/7))^2/(cos(Pi/7) *cos(2*Pi/7)).

In general, for integer h, k let

X = (sin^(h+k)(2*Pi/7))/(sin^(h)(4*Pi/7)*sin^(k)(8*Pi/7)),

Y = (sin^(h+k)(4*Pi/7))/(sin^(h)(8*Pi/7)*sin^(k)(2*Pi/7)),

Z = (sin^(h+k)(8*Pi/7))/(sin^(h)(2*Pi/7)*sin^(k)(4*Pi/7)).

then X, Y, Z are the roots of a monic equation

t^3 + a*t^2 + b*t + c = 0

where a, b, c are integers and c = 1 or -1.

Then X^n + Y^n + Z^n, n = 0, 1, 2, ... is an integer sequence.

Instances of such sequences with (h,k) values:

(-3,0), (0,3), (3,-3): gives A274663;

(-3,3), (0,-3): give A274664;

(-2,0), (0,2), (2,-2): give A198636;

(-2,-3), (-1,-2), (2,-1), (3,-1): give A274032;

(-1,-1), (-1,2): give A215076;

(-1,0), (0,1), (1,-1): give A094648;

(-1,1), (0,-1), (1,0): give A274975;

(1,1), (-2,1), (1,-2): give A274220;

(1,2), (-3,1), (2,-3): give A274075;

(1,3): this sequence.

The Ramanujan-type sequence the number 9 for the argument 2*Pi/7. The sequence is connecting with the following decomposition: (s(4)/s(1))^(1/3)*s(1)^n + (s(1)/s(2))^(1/3)*s(2)^n + (s(2)/s(4))^(1/3)*s(4)^n = x(n)*(4-3*7^(1/3))^(1/3) + y(n)*(11-3*49^(1/3))^(1/3), where s(j) := sin(2*Pi*j/7), x(0)=1, x(1)=-7^(1/6)/2, x(2)=y(0)=y(1)=0, y(2)=7^(1/3)/4 and X(n)=sqrt(7)*(X(n-1)-X(n-3)) for every n=3,4,..., and X=x or X=y. It could be deduced the formula 4*y(n) = a(n)*7^(1/3 + (3+(-1)^n)/4), which implies a(0)=0, a(1)= 0, a(2)= 1/7, a(3)=1/7, a(4)=1, a(5)=6/7, i.e., A163260(n)=7*a(n) for every n=0,1,...,5. The sequence a(n) is discussed in third Witula paper.

a(n) = x1^n + x2^n + x3^n, where x1, x2, x3 are the roots of x^3 - 35*x^2 + 147*x - 49, x1 = 7*(cot(1*Pi/7))^2, x2 = 7*(cot(2*Pi/7))^2, x3 = 7*(cot(4*Pi/7))^2.

Let A = sin(2*Pi/7), B = sin(4*Pi/7), C = sin(8*Pi/7).

In general, for integer h, k let

X = (B^h*C^k)/A^(h+k),

Y = (C^h*A^k)/B^(h+k),

Z = (A^h*B^k)/C^(h+k).

then X, Y, Z are the roots of a monic equation

t^3 + a*t^2 + b*t + c = 0

where a, b, c are integers and c = 1 or -1.

Then X^n + Y^n + Z^n, n = 0, 1, 2, ... is an integer sequence.

This sequence has (h,k) = (1,3) and its other half is A320918.

Let A = cos(2*Pi/7), B = cos(4*Pi/7), C = cos(8*Pi/7).

In general, for integer h, k let

X = A^(h+k)/(B^h*C^k),

Y = B^(h+k)/(C^h*A^k),

Z = C^(h+k)/(A^h*B^k).

then X, Y, Z are the roots of a monic equation

t^3 + a*t^2 + b*t + c = 0

where a, b, c are integers and c = 1 or -1.

Then X^n + Y^n + Z^n , n = 0, 1, 2, ... is an integer sequence.

This sequence has (h,k) = (1,3).

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.