Identities Involving Zeros of Ramanujan and Shanks Cubic Polynomials

Stefano Barbero, Umberto Cerruti, Nadir Murru Department of Mathematics University of Turin via Carlo Alberto 10 10123 Turin Italy

               stefano.barbero@unito.it
               umberto.cerruti@unito.it
                 nadir.murru@unito.it

Marco Abrate DIMEAS Polytechnic University of Turin Corso Duca degli Abruzzi 24 10129 Turin Italy

                 marco abrate@polito.it

Abstract

In this paper we highlight the connection between Ramanujan cubic polynomials (RCPs) and a class of polynomials, the Shanks cubic polynomials (SCPs), which generate cyclic cubic fields. In this way we provide a new characterization for RCPs and we express the zeros of any RCP in explicit form, using trigonometric functions. Moreover, we observe that a cyclic transform of period three permutes these zeros. As a consequence of these results we provide many new and beautiful identities. Finally we connect RCPs to Gaussian periods, finding a new identity, and we study some integer sequences related to SCPs.

1 Ramanujan and Shanks polynomials

Definition 1. A Ramanujan cubic polynomial (RCP) x^{3}+px^{2}+qx+r is a cubic polynomial, with p, q, r ∈ ℝ, r ≠ 0, which has real zeros and with coefficients satisfying the relation

pr^{\frac{1}{3}} + 3r^{\frac{2}{3}} + q = 0.  (1)

Shevelev [6] first introduced the definition of RCP, in honour of the great mathematician Srinivasa Ramanujan, since such a class of polynomials arises from a theorem proved by Ramanujan [4]. Witula [11] gave the characterization for the RCPs showed in the next theorem.

Theorem 2. All RCPs have the form:

x^{3} − \frac{P (γ − 1)}{(γ − 1) (γ − 2)} r^{\frac{1}{3}} x^{2} − \frac{P (2 − γ)}{(1 − γ) (2 − γ)} r^{\frac{2}{3}} x + r =

= (x − \frac{r^{\frac{1}{3}}}{2 − γ} ) (x − (γ − 1) r^{\frac{1}{3}} ) (x − \frac{2 − γ}{1 − γ} r^{\frac{1}{3}} ) ,

where r ∈ ℝ \ {0}, γ ∈ ℝ \ {1, 2}, and

P (γ) = γ^{3} − 3 γ + 1 = (γ − 2 cos \frac{2 π}{9} ) (γ − 2 cos \frac{4 π}{9} ) (γ − 2 cos \frac{8 π}{9} ) .

We provide a different representation for the RCPs in the following

Theorem 3. The polynomial x^{3} + px^{2} + qx + r is a RCP if and only if there exist h, s ∈ ℝ, s ≠ 0, such that

r = s^{3}, q = −(h + 3)s^{2}, p = hs.  (2)

Proof. If equations (2) hold, then it is straightforward to check equation (1). Vice-versa, if equation (1) holds, then for h = \frac{p}{r^{\frac{1}{3}}} and s = r^{\frac{1}{3}} we obtain equations (2).

Corollary 4. The polynomial x^{3} + px^{2} + qx + r is a RCP if and only if it has the form

ρ(h, s, x) = x^{3} + hsx^{2} − (h + 3)s^{2}x + s^{3},

where h, s ∈ ℝ, s ≠ 0.

Definition 5. We call Shanks cubic polynomials (SCPs) the polynomials

ρ(h, −1, x) = x^{3} − hx^{2} − (h + 3)x − 1,  (3)

since Shanks [5] deeply studied the cyclic cubic fields generated by the polynomials (3).

In the following theorem, we provide an explicit expression for the zeros of SCPs using trigonometric functions. Then we use this result and the connection between SCPs and RCPs in order to obtain a closed expression for the RCPs zeros.

Lemma 6. Let us consider the cubic polynomial Q(x) = ax^{3}+bx^{2}+cx+d and the polynomial P (w) = w^{2} + fx − \frac{e^{3}}{27}, where e = \frac{1}{a} (c − \frac{b^{2}}{3a}) , f = \frac{1}{a} (d + \frac{2b^{3}}{27a^{2}} − \frac{bc}{3a}). If the polynomial P (w) has two complex zeros α ± iβ, then the polynomial Q(x) has the zeros

−\frac{b}{3a} + 2\sqrt[3]{\sqrt{α^{2} + β^{2}}} cos (\frac{1}{3} (arctan (\frac{β}{α}) + kπ)) , k = 0, 2, 4, if α > 0

−\frac{b}{3a} − 2\sqrt[3]{\sqrt{α^{2} + β^{2}}} cos (\frac{1}{3} (arctan (\frac{β}{α}) + kπ)) , k = 0, 2, 4, if α < 0.

Proof. It is well known that the polynomial Q(x) becomes the quadratic polynomial P (w) by means of the Vieta substitutions x = −\frac{b}{3a} + z + \frac{μ}{z}, z^{3} = w, where μ = −\frac{e}{3}. Thus from the zeros α ± iβ of P (w) we find the zeros of Q(x) using the relations

−\frac{b}{3a} + z_{i} + \frac{s}{z_{i}}, i = 1, …, 6  (4)

where

z_{1} = \sqrt[3]{α + iβ}, z_{2} = ω\sqrt[3]{α + iβ}, z_{3} = ω^{2}\sqrt[3]{α + iβ},
z_{4} = \sqrt[3]{α − iβ}, z_{5} = ω\sqrt[3]{α − iβ}, z_{6} = ω^{2}\sqrt[3]{α − iβ},

and the number ω represents one of the complex roots of unity. Obviously, in equations (4) some z_{i} determine the same values. Therefore we find exactly three different zeros for Q(x). Now ,without loss of generality, we focus on z_{1}, whose complex exponential form is

z_{1} = \sqrt[3]{ρ}e^{i \frac{θ}{3}} ,

where ρ = \sqrt{α^{2} + β^{2}} and θ = arctan (\frac{β}{α}). Furthermore, observing that s = \sqrt[3]{ρ^{2}}, from equations (4) we have

−\frac{b}{3a} + 2\sqrt[3]{ρ} cos (\frac{θ}{3}) = −\frac{b}{3a} ± 2\sqrt[3]{\sqrt{α^{2} + β^{2}}} cos (\frac{1}{3} (arctan (\frac{β}{α}))) ,

and plus or minus sign depends on whether α > 0 or α < 0, respectively. In a similar way we obtain the remaining zeros, proving the thesis.

Theorem 7. The zeros of the SCP ρ(h, −1, x) are

\frac{1}{3} (h + 2\sqrt{τ(h)} cos (\frac{1}{3} (arctan (\frac{3\sqrt{3}}{3 + 2h}) + kπ))) if h ≥ −\frac{3}{2},

\frac{1}{3} (h − 2\sqrt{τ(h)} cos (\frac{1}{3} (arctan (\frac{3\sqrt{3}}{3 + 2h}) + kπ))) if h ≤ −\frac{3}{2},

where τ(h) = h^{2}+3h+9 and k = 0, 2, 4. In particular, if h = −\frac{3}{2}, we have arctan(+∞) = \frac{π}{2} in the first formula and arctan(−∞) = −\frac{π}{2} in the second formula, retrieving the roots 1,−\frac{1}{2},−2 of the SCP ρ(−\frac{3}{2}, −1, x).

Proof. If h = −\frac{3}{2}, the proof is straightforward. If h > −\frac{3}{2}, we apply Lemma 6. In this case we have e = \frac{τ(h)}{3}, f = −\frac{(2h+3)τ(h)}{27}. The zeros of P (x) are

α ± iβ = \frac{1}{54}((3 + 2h)τ(h) ± 3i\sqrt{3}τ(h))

with α > 0 and

α^{2} + β^{2} = \frac{(3 + 2h)^{2}τ(h)^{2} + 27τ(h)^{2}}{54^{2}} = \frac{τ(h)^{3}}{3^{6}}, \frac{β}{α} = \frac{3\sqrt{3}}{3 + 2h}.

Similarly, using Lemma 6, we can prove the case h < −\frac{3}{2}, when α < 0.

Theorem 8. If we consider the zeros set ζ(h, s) of the RCP ρ(h, s, x) and α is one of these zeros, then

ζ(h, s) = \{α, \frac{s^{2}}{s − α}, −\frac{s(s − α)}{α}\} = \{α, η_{s}(α), η_{s}^{2}(α)\},  (5)

where η_{s}(z) = \frac{s^{2}}{s − z}. Moreover, for any RCP ρ (\frac{p}{r^{\frac{1}{3}}} , r^{\frac{1}{3}} , x) = x^{3} + px^{2} + qx + r we have

ζ (\frac{p}{r^{\frac{1}{3}}} , r^{\frac{1}{3}} ) = −r^{\frac{1}{3}} ζ (\frac{p}{r^{\frac{1}{3}}} , −1) .  (6)

Proof. Let us observe that if α is a zero of ρ(h, s, x), then α ≠ 0, α ≠ s and the transform η_{s} is periodic with period 3, satisfying η_{s}^{3}(z) = z. We easily check that

ρ(h, s, η_{s}(z)) = −\frac{s^{3}ρ(h, s, z)}{(s − z)^{3}}, ρ(h, s, η_{s}^{2}(z)) = −\frac{s^{3}ρ(h, s, z)}{z^{3}}.

So we have

ρ(h, s, η_{s}(α)) = −\frac{s^{3}ρ(h, s, α)}{(s − α)^{3}} = 0, ρ(h, s, η_{s}^{2}(α)) = −\frac{s^{3}ρ(h, s, α)}{α^{3}} = 0.

We also point out that for any SCP

ζ(h, −1) = \{α, −\frac{1}{α + 1}, −\frac{(α + 1)}{α}\} , η_{−1}(z) = −\frac{1}{z + 1}.

The zeros of ρ (\frac{p}{r^{\frac{1}{3}}} , −1, x) = x^{3}−\frac{p}{r^{\frac{1}{3}}} x^{2}−(\frac{p}{r^{\frac{1}{3}}} + 3) x−1 are the elements of the set ζ (\frac{p}{r^{\frac{1}{3}}} , −1).

Thus, if ρ (\frac{p}{r^{\frac{1}{3}}} , −1, α) = 0, we have ζ (\frac{p}{r^{\frac{1}{3}}} , −1) = \{α, −\frac{1}{α + 1}, −\frac{α + 1}{α}\} . Furthermore we

easily prove that ρ (\frac{p}{r^{\frac{1}{3}}} , r^{\frac{1}{3}} , −r^{\frac{1}{3}} α) = 0, in fact

−rα^{3} + pr^{\frac{2}{3}} α^{2} − qr^{\frac{1}{3}} α + r = −r (α^{3} − \frac{p}{r^{\frac{1}{3}}} α^{2} + (\frac{p}{r^{\frac{1}{3}}} + 3) α − 1) = 0.

Therefore we obtain

ζ (\frac{p}{r^{\frac{1}{3}}} , r^{\frac{1}{3}} ) = \{−r^{\frac{1}{3}} α, η_{r^{\frac{1}{3}}} (−r^{\frac{1}{3}} α) , η_{r^{\frac{1}{3}}}^{2} (−r^{\frac{1}{3}} α)\} =

= \{−r^{\frac{1}{3}} α, \frac{r^{\frac{2}{3}}}{r^{\frac{1}{3}} + r^{\frac{1}{3}} α}, \frac{r^{\frac{1}{3}} (r^{\frac{1}{3}} + r^{\frac{1}{3}} α)}{−r^{\frac{1}{3}} α}\} =

= −r^{\frac{1}{3}} \{α, −\frac{1}{α + 1}, −\frac{α + 1}{α}\} = −r^{\frac{1}{3}} ζ (\frac{p}{r^{\frac{1}{3}}} , −1) .

In the next corollary we highlight some interesting consequences of the previous theorems.

Corollary 9. Recalling that h = \frac{p}{r^{\frac{1}{3}}} , s = r^{\frac{1}{3}} and τ(h) = h^{2} + 3h + 9,

1.

ζ(h, s) = −sζ(h, 1);  (7)

2. the zeros of any RCP ρ(h, s, x) have the following expressions in terms of cos and arctan functions

−\frac{s}{3} (h ± 2\sqrt{τ(h)} cos (\frac{1}{3} (arctan (\frac{3\sqrt{3}}{3 + 2h}) + kπ))) k = 0, 2, 4;  (8)

3. if we consider a zero α of the RCP ρ(h, s, x) ∈ ℚ[X], then the Galois group of the extension ℚ(α) over ℚ has three elements: the identity automorphism and the automorphisms η_{s} and η_{s}^{2}.

Proof. The equality (7) directly follows from relation (6) and allows us to prove equations (8), using the closed expression for the zeros of any SCP given by Theorem 7. Finally, the relation (5) clearly shows that the identity automorphism and the automorphisms η_{s} and η_{s}^{2} are the only elements of the Galois group of the extension ℚ(α), since the transform η_{s} is periodic with period 3.

2 Identities

With the expressions for the RCPs zeros, we can derive some interesting and unexpected iden-

tities. Witula, Slota and Warzynski [9] showed that 2 cos (\frac{2π}{7}) , 2 cos (\frac{4π}{7}) and 2 cos (\frac{8π}{7})

are the zeros of the RCP ρ(−1, −1, x) = x^{3}+x^{2}−2x−1 (h = −1, s = −1). Thus, for example, we find the following identity

2 cos (\frac{2π}{7} ) = \frac{1}{3} (−1 + 2\sqrt{7} cos (\frac{1}{3} arctan (3\sqrt{3}))) .

Many other identities may result from previous theorems observing that a RCP with zero α is

ρ(h, s, x) = x^{3} + hsx^{2} − (h + 3)s^{2}x + s^{3} =
= (x − α) (x − \frac{s^{2}}{s − α}) (x + \frac{s(s − α)}{α}) =
= x^{3} + \frac{s^{3} − 3s^{2}α + α^{3}}{(s − α)α} x^{2} + \frac{−s^{4} + 3s^{2}α^{2} − sα^{3}}{(s − α)α} x + s^{3}.  (9)

Thanks to these equalities we can easily express any real number α in terms of cos and arctan functions. Indeed, choosing a nonzero real number s ≠ α and h = \frac{s^{3}−3s^{2}α+α^{3}}{s(s−α)α}, we create a RCP, with one of the zeros equal to α, on which we use the results of Theorems 7 and 8.

Example 10. We select α = \sqrt{2} and s = 3\sqrt{2}. In this way, h = \frac{1}{6} and \sqrt{2} is a zero of ρ(\frac{1}{6}, 3\sqrt{2}, x). By Theorem 7 we obtain the zeros of ρ(\frac{1}{6}, −1, x) and multiplying these zeros by −3\sqrt{2} we get the zeros of ρ(\frac{1}{6}, 3\sqrt{2}, x). We approximately evaluate these zeros (with any mathematical software) in order to make the correct one equal to \sqrt{2}. In this case we find

\sqrt{2} = −\frac{1 + 14\sqrt{7} cos (\frac{1}{3} (arctan (\frac{9\sqrt{3}}{10}) + 4π))}{3\sqrt{2}},

and with further simplifications we have the identity

1 = \sqrt{7} cos (\frac{1}{3} arctan (\frac{9\sqrt{3}}{10})) − \sqrt{21} sin (\frac{1}{3} arctan (\frac{9\sqrt{3}}{10})) .

Example 11. We consider α = π and s = 1. Hence, h = \frac{1−3π+π^{3}}{(1−π)π} < −\frac{3}{2} and π is a zero of ρ(h, 1, x). Using Theorem 7 we immediately have the zeros of ρ(h, −1, x) and multiplying them by -1 we obtain the zeros of ρ(h, 1, x). We approximately evaluate these zeros (with any mathematical software) in order to make the correct one equal to π. In this case, after some algebraic calculations, we find

\frac{1 − 3π + π^{3} + 2 (1 − π + π^{2})^{\frac{3}{2}} cos (\frac{1}{3} arctan (^{\frac{3\sqrt{3}(−1+π)π}{2−3π−3π^{2}+2π^{3}}} ))}{3(−1 + π)π} = π.

From the previous equality we derive the nontrivial identity

\frac{2π − 1}{2\sqrt{π^{2} − π + 1}} = cos (\frac{1}{3} arctan (\frac{3\sqrt{3}(−1 + π)π}{2 − 3π − 3π^{2} + 2π^{3}} )) .

A wonderful and famous Ramanujan identity on RCPs is

x1^{\frac{1}{3}} + x2^{\frac{1}{3}} + x3^{\frac{1}{3}} = (−p − 6r^{\frac{1}{3}} + 3(9r − pq)^{\frac{1}{3}} )^{\frac{1}{3}}  (10)

where x_{1}, x_{2}, x_{3} are the RCP zeros (see, e.g., Berndt [1, p. 22]). Using our expression (9) and Theorem 8, we rewrite identity (10) as

α^{\frac{1}{3}} = − (\frac{s^{2}}{s − α})^{\frac{1}{3}} − (−\frac{s(s − α)}{α})^{\frac{1}{3}} +

+ (\frac{s^{3} + 3s^{2}α − 6sα^{2} + α^{3}}{−sα + α^{2}} + 3 (s^{2} − sα + α^{2}) (\frac{s}{α^{2}(s − α)^{2}} )^{\frac{1}{3}} )^{\frac{1}{3}} . (11)

If we choose suitable real values for α and s, equation (11) generates infinite nontrivial identities.

Example 12. For α = π^{3} and s = 1, we obtain

π = \frac{1}{(−1 + π^{3})^{1/3}} − \frac{(−1 + π^{3})^{1/3}}{π} + (\frac{3 (1 − π^{3} + π^{6})}{π^{2} (−1 + π^{3})^{2/3}} + \frac{1 + 3π^{3} − 6π^{6} + π^{9}}{−π^{3} + π^{6}} )^{1/3} .

3 Gaussian periods and Ramanujan-Shanks polynomials

Definition 13. Let m be a positive integer and p a prime number satisfying the condition p ≡ 1 (mod m). We consider the multiplicative group G_{m,p} of the mth powers mod p whose order is \frac{p−1}{m}. The Gaussian periods are

η_{m,p}^{(k)} = ∑_{j∈C_{k}} e^{\frac{2jπi}{p}} ,

where C_{0} = G_{m,p} and for k ≥ 1 C_{k} are the cosets of G_{m,p} in ℤ_{p}^{∗}.

Example 14. If p = 13 and m = 3, we have G_{3,13} = C_{0} = {1, 5, 8, 12}, C_{1} = {2, 3, 10, 11} and C_{2} = {4, 6, 7, 9}. The cubic Gaussian periods are

η_{3,13}^{(0)} = e^{− \frac{2π}{13} i} + e^{\frac{2π}{13} i} − e^{− \frac{3π}{13} i} − e^{\frac{3π}{13} i} = 2 cos (\frac{2π}{13}) − 2 cos (\frac{3π}{13}) ,

η_{3,13}^{(1)} = e^{− \frac{4π}{13} i} + e^{\frac{4π}{13} i} + e^{− \frac{6π}{13} i} + e^{\frac{6π}{13} i} = 2 cos (\frac{4π}{13}) + 2 cos (\frac{6π}{13}) ,

η_{3,13}^{(2)} = e^{− \frac{8π}{13} i} + e^{\frac{8π}{13} i} + e^{− \frac{12π}{13} i} + e^{\frac{12π}{13} i} = 2 cos (\frac{8π}{13}) + 2 cos (\frac{12π}{13}) .

In the following, we focus on cubic Gaussian periods, i.e., we set m = 3 and we write η_{p}^{(k)} = η_{3,p}^{(k)}. Lehmer [3] connected the Shanks polynomials zeros with cubic Gaussian periods. In particular, when h ∈ ℤ and 3 |̸h, we call Shanks prime the value of τ(h) = h^{2} + 3h + 9 when it is a prime number. Remark 15. The sequence of Shanks primes in OEIS [7] is \underline{A005471}:

(7, 13, 19, 37, 79, 97, 139, …).

When τ(h) = p is a Shanks prime, the elements of ζ(h, −1) are fundamental units of ℚ(α),

differing from the cubic Gaussian periods by the integer number \frac{L − 1}{6}, where L = ±(2h+3).

Moreover, the minimal polynomial of η_{p}^{(0)}, η_{p}^{(1)}, η_{p}^{(2)} is

x^{3} + x^{2} − \frac{p − 1}{3}x − \frac{(L + 3)p − 1}{27}.  (12)

The sign of the integer L depends on the parameter h. If h ≡ 1 (mod 3), then L = −(2h+3)

and \frac{(L + 3)p − 1}{27} is an integer. If h ≡ 2 (mod 3), then L = 2h + 3. Thus, we can write the polynomial (12) as

G_{1}(h, x) = x^{3} + x^{2} − \frac{p − 1}{3}x + \frac{2hp + 1}{27}, if h ≡ 1 (mod 3),

G_{2}(h, x) = x^{3} + x^{2} − \frac{p − 1}{3}x − \frac{(6 + 2h)p + 1}{27}, if h ≡ 2 (mod 3).

Furthermore, the SCP ρ(h, −1, x) has zeros

\frac{h − 1}{3} − η_{p}^{(k)}, k = 0, 1, 2 if h ≡ 1 (mod 3),  (13)

\frac{h + 1}{3} + η_{p}^{(k)}, k = 0, 1, 2 if h ≡ 2 (mod 3).  (14)

Remark 16. Using the expression of the SCP zeros provided by Theorem 7, we have a new explicit expression for cubic Gaussian periods involving cos and arctan functions.

Definition 17. We call Lehmer-Ramanujan cubic polynomials (LRCPs), a RCPs ρ(h, s, x) such that h ∈ ℤ, 3 |̸h and τ(h) is a prime number.

Considering the relationships between zeros of Ramanujan and Shanks polynomials showed in the previous section, we can express LRCPs zeros in terms of cubic Gaussian periods.

Theorem 18. Let ρ(h, s, x) be a LRCP and

θ_{1}(h, s, x) = \frac{x}{s} + \frac{h − 1}{3}, θ_{2}(h, s, x) = −\frac{x}{s} − \frac{h + 1}{3},

γ_{1}(h, s, x) = −\frac{1}{3}s(h − 1 − 3x), γ_{2}(h, s, x) = −\frac{1}{3}s(h + 1 + 3x).

The zeros of ρ(h, s, x) are

γ_{1}(h, s, η_{p}^{(k)}), k = 0, 1, 2 if h ≡ 1 (mod 3),  (15)
γ_{2}(h, s, η_{p}^{(k)}), k = 0, 1, 2 if h ≡ 2 (mod 3).  (16)

Therefore we have

ρ(h, s, x) = s^{3}G_{1}(h, θ_{1}(h, s, x)) if h ≡ 1 (mod 3),  (17)

ρ(h, s, x) = −s^{3}G_{2}(h, θ_{2}(h, s, x)) if h ≡ 2 (mod 3).  (18)

Proof. By Corollary 9, the set ζ(h, s) corresponds to the set −sζ(h, −1). Thus, by equations (13) and (14), we obtain the equalities (15) and (16). By comparison between the explicit expressions of ρ(h, s, x), G_{1}(h, x) and G_{2}(h, x) we easily prove the identities (17) and (18).

Now, the Ramanujan identity (10) provides us the following new relations involving the cubic Gaussian periods:

(\frac{h − 1}{3} − η_{p}^{(0)})^{\frac{1}{3}} + (\frac{h − 1}{3} − η_{p}^{(1)})^{\frac{1}{3}} + (\frac{h − 1}{3} − η_{p}^{(2)})^{\frac{1}{3}} = (6 + h − 3p^{\frac{1}{3}} )^{\frac{1}{3}}  (19)

and

(\frac{h + 1}{3} + η_{p}^{(0)})^{\frac{1}{3}} + (\frac{h + 1}{3} + η_{p}^{(1)})^{\frac{1}{3}} + (\frac{h + 1}{3} + η_{p}^{(2)})^{\frac{1}{3}} = (6 + h − 3p^{\frac{1}{3}} )^{\frac{1}{3}}  (20)

when h ≡ 1 (mod 3) and h ≡ 2 (mod 3), respectively. D. H. Lehmer and E. Lehmer [2] studied differences of Gaussian periods, proving that their minimal polynomial is x^{3} −px+p. Let us consider

δ_{p}^{(0)} = η_{p}^{(0)} − η_{p}^{(1)}, δ_{p}^{(1)} = η_{p}^{(1)} − η_{p}^{(2)}, δ_{p}^{(2)} = η_{p}^{(2)} − η_{p}^{(0)}.

Using previous results, we provide a closed formula for these differences.

Proposition 19. If p = τ(h) is a prime number and θ_{h} = \frac{3\sqrt{3}}{3 + 2h}, then

δ_{p}^{(0)} = ±2\sqrt{\frac{p}{3}} cos (\frac{1}{3} arctan (\frac{1}{θ_{h}} )) , δ_{p}^{(1)} = ±2\sqrt{\frac{p}{3}} cos (\frac{1}{3} (arctan (\frac{1}{θ_{h}} ) + 2π)) ,

δ_{p}^{(2)} = ±2\sqrt{\frac{p}{3}} cos (\frac{1}{3} (arctan (\frac{1}{θ_{h}} ) + 4π)) .

Proof. Thanks to the relation between Gaussian periods and Shanks polynomials, we know that δ_{p}^{(0)} = α − η_{−1}(α), where α is a zero of ρ(h, −1, x). Moreover, by Theorem 7 we have

δ_{p}^{(0)} = ±\frac{2}{3}\sqrt{p} cos (\frac{1}{3} arctan(θ_{h})) ∓ \frac{2}{3}\sqrt{p} cos (\frac{1}{3} (arctan(θ_{h}) + 2π)) =

= \frac{\sqrt{p}}{3} (±2 cos (\frac{1}{3} arctan(θ_{h})) ± cos (\frac{1}{3} arctan(θ_{h})) ± \sqrt{3} sin (\frac{1}{3} arctan(θ_{h}))) =

= ±\sqrt{p} cos (\frac{1}{3} arctan(θ_{h})) ± \sqrt{\frac{p}{3}} sin (\frac{1}{3} arctan(θ_{h})) .

Recalling that, arctan x + arctan \frac{1}{x} = \frac{π}{2} if x > 0 and arctan x + arctan \frac{1}{x} = −\frac{π}{2} if x < 0, we have

δ_{p}^{(0)} = ±\frac{\sqrt{3}}{2}\sqrt{p} cos (\frac{1}{3} arctan (\frac{1}{θ_{h}} )) ± \frac{\sqrt{p}}{2} sin (\frac{1}{3} arctan (\frac{1}{θ_{h}} )) ±
± \frac{1}{2}\sqrt{\frac{p}{3}} cos (\frac{1}{3} arctan (\frac{1}{θ_{h}} )) \frac{\sqrt{3}}{2}\sqrt{\frac{p}{3}} sin (\frac{1}{3} arctan (\frac{1}{θ_{h}} )) =
= ±2\sqrt{\frac{p}{3}} cos (\frac{1}{3} arctan (\frac{1}{θ_{h}} )) .

Similarly, we obtain the thesis for δ_{p}^{(1)} and δ_{p}^{(2)}.

4 On a conjecture of Jeffery

In this section we use our results on SCPs zeros to give a proof that the sequence \underline{A198636} = (3, 5, 13, 38, 117, 370, 1186, …) in OEIS [7] is a linear recurrent sequence. We also prove a related conjecture due to L. Edson Jeffery (January 21 2012) mentioned in \underline{A198636}.

Let α, β, and γ be the zeros of the SCP ρ(h, −1, x). We consider the integer sequence (A(k, n))_{n=0}^{+∞}, where k is a fixed integer and

A(k, n) = α^{kn} + β^{kn} + γ^{kn}.  (21)

Moreover, the companion matrix M of the SCP is

M = (\begin{matrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 3 + h & h \end{matrix}) .

Clearly, we have

A(1, n) = Tr(M^{n}),

and (A(1, n))_{n=1}^{+∞} is a linear recurrent sequence with characteristic polynomial ρ(h, −1, x). Similarly,

A(k, n) = Tr(M^{kn}),

and (A(k, n))_{n=0}^{+∞} is a linear recurrent sequence whose characteristic polynomial is

x^{3} − A(k, 1)x^{2} + B(k, 1)x − 1,

with zeros α^{k}, β^{k} and γ^{k}. When h = −1, the SCP ρ(−1, −1, x) has zeros

ᾱ = 2 cos (\frac{2π}{7}) , β̄ = 2 cos (\frac{4π}{7}) , γ̄ = 2 cos (\frac{8π}{7})

and the integer sequences arising from this polynomial are particularly interesting. These sequences have been deeply studied in some recent works of Witula and Slota [8, 9, 10].

Lemma 20. Let (a_{n})_{n=0}^{+∞} be the linear recurrent sequence with characteristic polynomial x^{3} − 5x^{2} + 6x − 1 and initial conditions a_{0} = 3, a_{1} = 5 and a_{2} = 13, then

a_{n} = 2^{2n} (cos^{2n} (\frac{2π}{7}) + cos^{2n} (\frac{4π}{7}) + cos^{2n} (\frac{8π}{7})) ,

for n = 0, 1, 2, . . ..

Proof. The matrix

M^{2} = (\begin{matrix}0 & 0 & 1 \\ 1 & 2 & −1 \\ −1 & −1 & 3 \end{matrix})

has the same characteristic polynomial x^{3} − 5x^{2} + 6x − 1 of the sequence (A(2, n))_{n=1}^{+∞}. Moreover, evaluating the traces of the matrices M^{0}, M^{2} and M^{4}, we observe that A(2, 0) = 3, A(2, 1) = 5 and A(2, 2) = 13, i.e., a_{n} = A(2, n), for every n = 0, 1, 2, . . .. Thus, by equation (21), we have the thesis.

Theorem 21. The sequence (A_{n})_{n=0}^{+∞} =\underline{A198636} in OEIS [7] is a linear recurrent sequence corresponding to the sequence (A(2, n))_{n=1}^{+∞}. In particular

A_{0} = 3
A_{1} = 5
A_{2} = 13
A_{n+3} = 5A_{n+2} − 6A_{n+1} + A_{n}.

Moreover

A_{n} = 2^{2n} (cos^{2n} (\frac{π}{7} ) + cos^{2n} (\frac{2π}{7} ) + cos^{2n} (\frac{3π}{7} )) ,

i.e., the Jeffery conjecture is true.

Proof. We recall that A_{n} = \frac{w(6,2n)}{2} for all n ≥ 0, where w(6, l) is the total number of closed walks of length l on the graph P_{6}, the simple path with 6 points (vertices) and 5 lines (or edges). As explained in the comments related to the sequence \underline{A198632} in OEIS [7], if we consider the adjacency matrix J_{N} of the graph P_{N}, we have

w(N, l) = Tr(J_{N}^{l}) = ∑_{j=1}^{N} λ_{j}^{l},

where λ_{j}, j = 1, …, N, are the eigenvalues of J_{N}. Since the matrix J_{N} is a Jacobi matrix, its characteristic polynomial is S_{N}(x) = U_{N}(\frac{x}{2}), where U_{N}(x) is the second-kind Chebyshev polynomial of degree N (see, e.g., Mason and Handscomb [12] for a survey on Chebyshev polynomials). The zeros of S_{N}(x) are λ_{j} = 2 cos (\frac{jπ}{N+1}), j = 1, …, N and in paricular we have

w(6, 2n) = ∑_{j=1}^{6} (2 cos (\frac{jπ}{N + 1}))^{2n} = 2^{(2n+1)} (cos^{2n} (\frac{2π}{7} ) + cos^{2n} (\frac{4π}{7} ) + cos^{2n} (\frac{8π}{7} )) .

Therefore, using the results of Lemma 20, we clearly obtain

A_{n} = \frac{w(6, 2n)}{2} = A(2, n) n = 0, 1, …

and the thesis immediately follows.

5 Acknowledgements

The authors thank sincerely the referee for his appreciations, useful suggestions and comments about this paper.

References

[1] B. C. Berndt, Ramanujan Notebooks, Part IV, Springer-Verlag, 1994.

[2] D. H. Lehmer and E. Lehmer, The Lehmer project, Math. Comp. 61 (1993), 313–317.

[3] E. Lehmer, Connection between Gaussian periods and cyclic units, Math. Comp. 50 (1988), 535–541.

[4] S. Ramanujan, Notebooks, Tata Institute of Fundamental Research, 1957.

[5] D. Shanks, The simplest cubic fields, Math. Comp. 28 (1974), 1137–1152.

[6] V. Shevelev, On Ramanujan cubic polynomials, preprint, 2007. Available at

   http://arxiv.org/abs/0711.3420v1.

[7] N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences, http://oeis.org.

[8] R. Witula, D. Slota, and A. Warzynski, Quasi-Fibonacci numbers of the seventh order, J. Integer Seq. 9 (2006), Article 06.4.3.

[9] R. Witula and D. Slota, New Ramanujan-Type Formulas and Quasi-Fibonacci Numbers of Order 7, J. Integer Seq. 10 (2007), Article 07.5.6.

[10] R. Witula, Ramanujan type trigonometric formulas: the general form for the argument \frac{2π}{7}, J. Integer Seq. 12 (2009), Article 09.8.5.

[11] R. Witula, Full description of Ramanujan cubic polynomials, J. Integer Seq. 13 (2010), Article 10.5.9.

[12] J. C. Mason and D. C. Handscomb, Chebyshev Polynomials, CRC Press, 2003.

2010 Mathematics Subject Classification: Primary 11C08; Secondary 11B83. Keywords: Ramanujan cubic polynomial.

(Concerned with sequences \underline{A005471}, \underline{A198632}, and \underline{A198636}.)

Received May 6 2013; revised version received August 26 2013. Published in Journal of Integer Sequences, October 12 2013.

Return to Journal of Integer Sequences home page.