Dickson Polynomials, Chebyshev Polynomials, and Some Conjectures of Jeffery

Stefano Barbero Department of Mathematics University of Turin via Carlo Alberto 10 10123 Turin Italy

                   stefano.barbero@unito.it

Abstract By using Dickson polynomials in several variables and Chebyshev polynomials of the second kind, we derive the explicit expression of the entries in the array defining the sequence \underline{A185095} . As a result, we obtain a straightforward proof of some conjectures of Jeffery concerning this sequence and other related ones.

1 Proof of main conjectures

As we can find in the Encyclopedia of Integer Sequences [1], the sequence \underline{A185095} is a rectangular array read by antidiagonals, where the n-th row has generating function

F_{n}(z) = \begin{matrix}∑ & {n \atop r=0} & (n + 1 − r)(−1)^{r}\binom{2(n+1)−r}{r}z^{r} \\ ∑ & {n+1 \atop r=0} & (−1)^{r}\binom{2(n+1)−r}{r}z^{r} \end{matrix}  (1)

for n = 0, 1, 2, . . .. In this section we find the explicit expression for the entries of this rectangular array and we determine the ordinary generating functions for the columns.

Let us consider the Dickson polynomials of the first kind in several indeterminates, as defined in the book of Lidl, Mullen, and Turnwald [2].

Definition 1. The Dickson polynomials of the first kind in n indeterminates of total degree k

D_{k}^{(i)}(x_{1}, …, x_{n}, a), i = 1, …, n

satisfy the functional equations

D_{k}^{(i)}(x_{1}, …, x_{n}, a) = S_{i}(u_{1}^{k}, …, u_{n+1}^{k}), i = 1, …, n  (2)

where x_{i} = S_{i}(u_{1}, …, u_{n+1}), S_{i}(y_{1}, …, y_{n+1}) is the i-th symmetric function of y_{1}, …, y_{n+1}, and u_{1} · . . . · u_{n+1} = a.

In particular when i = 1 we have D_{k}^{(1)}(x_{1}, …, x_{n}, a) = S_{1}(u_{1}^{k}, …, u_{n+1}^{k}) = ∑_{j=1}^{n+1} u_{j}^{k}. Moreover if we pose x_{0} = 1 and x_{n+1} = a, the Dickson polynomials of the first kind D_{k}^{(1)}(x_{1}, …, x_{n}, a) have the following generating function (see Lidl, Mullen, and Turnwald [2, Lemma 2.23] ):

∑_{k=0}^{+∞} D_{k}^{(1)}(x_{1}, …, x_{n}, a)z^{k} = \frac{∑_{r=0}^{n}(n + 1 − r)(−1)^{r}x_{r}z^{r}}{∑_{r=0}^{n+1}(−1)^{r}x_{r}z^{r}} , k ≥ 0.  (3)

Proposition 2. The k-th entry in the n-th row R_{n,k}, k, n ≥ 0, of the rectangular array which defines the sequence \underline{A185095}, corresponds to D_{k}^{(1)}(x_{1}, …, x_{n}, 1) where x_{r} = \binom{2(n+1)−r}{r} for r = 0, …, n + 1. Furthermore we have that

R_{n,k} = 2^{2k} ∑_{j=1}^{n+1} cos^{2k} (\frac{jπ}{2n + 3}) .  (4)

Proof. After a comparison between the two generating functions (1) and (3), we immediately find that R_{n,k} = D_{k}^{(1)}(x_{1}, …, x_{n}, 1). Moreover, by definition (1), the role played by symmetric functions when x_{r} = \binom{2(n+1)−r}{r} for r = 0 …, n + 1, allow us to prove that

R_{n,k} = D_{k}^{(1)}(x_{1}, …, x_{n}, 1) = ∑_{j=1}^{n+1} α_{j}^{k},

where α_{j}, for j = 1, …, n + 1 are the zeros of the polynomial

P_{n}(x) = ∑_{j=0}^{n+1}(−1)^{j}\binom{2(n + 1) − j}{j}x^{n+1−j}.  (5)

Now let us recall the definition of the Chebyshev polynomials of the second kind U_{h} (\frac{x}{2}) (see for references the books of Rivlin [3] and of Mason and Hascomb [4]):

U_{h} (\frac{x}{2}) = ∑_{j=0}^{⌊ \frac{h}{2} ⌋}(−1)^{j}\binom{h − j}{j}x^{h−2j}.  (6)

It is clear that P_{n}(x) = U_{2n+2} (\frac{\sqrt{x}}{2}). So α_{j} is a zero of P_{n}(x) if and only if \frac{\sqrt{α_{j}}}{2} is a positive

zero for U_{2n+2}(x). Since the zeros of U_{2n+2}(x) are x_{j} = cos (\frac{jπ}{2n+3}), j = 1, …, 2n+3, a simple trigonometric consideration allow us to find the positive ones when j = 1, …, n + 1, showing that α_{j} = 2^{2} cos^{2} (\frac{jπ}{2n+3}). The thesis immediately follows.

Corollary 3. The entry R_{n,k} for n ≥ 1 has the alternative expression

\begin{cases}R_{n,k} = \binom{2k}{k} (n + \frac{3}{2}) − 2^{2k−1}, 1 ≤ k < 2n + 3 \\ R_{n,k} = \binom{2k}{k} (n + \frac{3}{2}) − 2^{2k−1} + (2n + 3) ∑_{i=1}^{⌊ \frac{k}{2n+3} ⌋} \binom{2k}{k−i(2n+3)}, k ≥ 2n + 3 \end{cases}  (7)

where R_{n,0} = n + 1, R_{0,k} = 1.

Proof. Obviously if k = 0 we obtain R_{n,0} = n + 1 and if n = 0 we have R_{0,k} = 1. When k ≥ 1, by (4) and thanks to the summation formulas (see Gradshteyn and Ryzhik [5]):

cos^{2k}(x) = \frac{1}{2^{2k}} \{\binom{2k}{k} + 2 ∑_{h=0}^{k−1} \binom{2k}{h} cos(2(k − h)x)\}

and

∑_{j=0}^{n} cos(jx) = \frac{1}{2} [1 + \begin{matrix}sin & (\frac{(2n+1)x}{2}) \\ sin & (\frac{x}{2}) \end{matrix}] ,

we get

R_{n,k} = ∑_{j=1}^{n+1} (\binom{2k}{k} + 2 ∑_{h=0}^{k−1} \binom{2k}{h} cos (2(k − h)\frac{jπ}{2n + 3})) =

= \binom{2k}{k}(n + 1) + 2 ∑_{h=0}^{k−1} \binom{2k}{h} (∑_{j=0}^{n+1} cos (2(k − h)\frac{jπ}{2n + 3}) − 1) =

= \binom{2k}{k}(n + 1) + 2 ∑_{h=0}^{k−1} \binom{2k}{h} (∑_{j=0}^{n+1} cos (2(k − h)\frac{jπ}{2n + 3})) − 2 ∑_{h=0}^{k−1} \binom{2k}{h}

Now we have

\begin{cases}∑_{j=0}^{n+1} cos (2(k − h)\frac{jπ}{2n+3}) = \frac{1}{2} [1 + \begin{matrix}sin & (\frac{2n+3}{2} \frac{2(k−h)π}{2n+3}) \\ sin & (\frac{(k−h)π}{2n+3}) \end{matrix}] = \frac{1}{2} if k ≠ i(2n + 3) + h \\ ∑_{j=0}^{n+1} cos (2(k − h)\frac{jπ}{2n+3}) = n + 2 if k = i(2n + 3) + h \end{cases}  (8)

for a certain positive integer i.

Recalling the following well known identity

∑_{h=0}^{k−1} \binom{2k}{h} = 2^{2k−1} − \frac{1}{2}\binom{2k}{k},

if 1 ≤ k < 2n + 3 we clearly obtain

R_{n,k} = \binom{2k}{k} (n + \frac{3}{2}) − 2^{2k−1},

on the other hand, when k ≥ 2n + 3 we find

R_{n,k} = \binom{2k}{k}(n+1)+2 (∑_{h=0}^{k−1} \binom{2k}{h} (−\frac{1}{2}) − \frac{1}{2} ∑_{i=1}^{⌊ \frac{k}{2n+3} ⌋} \binom{2k}{k − i(2n + 3)} + (n + 2) ∑_{i=1}^{⌊ \frac{k}{2n+3} ⌋} \binom{2k}{k − i(2n + 3)}) =

= \binom{2k}{k} (n + \frac{3}{2}) − 2^{2k−1} + (2n + 3) ∑_{i=1}^{⌊ \frac{k}{2n+3} ⌋} \binom{2k}{k − i(2n + 3)}.

Considering the rectangular array in the definition of sequence \underline{A 185095} we can also study its columns, finding their generating functions. Thanks to the previous result we have the following

Proposition 4. The ordinary generating function G_{k}(z) for the k-th column of the rectangular array in the definition of \underline{A 185095} is

G_{k}(z) = \frac{(2^{2k−1} − \frac{3}{2} \binom{2k}{k} + 1) z^{2} + (\frac{5}{2} \binom{2k}{k} − 2^{2k−1} − 2) z + 1}{(1 − z)^{2}} + P_{n_{0}}(z), k ≥ 0, (9)

where n_{0} = ⌊\frac{k−3}{2}⌋ and

P_{n_{0}}(z) = ∑_{n=1}^{n_{0}} ((2n + 3) ∑_{i=1}^{⌊ \frac{k}{2n+3} ⌋} \binom{2k}{k − i(2n + 3)}) z^{n}  (10)

with the convention that the summation in (10) is equal to 0 when n_{0} < 1.

Proof. Using the alternative expression determined in Corollary 7, and the well–known formulas ∑_{n=0}^{+∞} z^{n} = \frac{1}{1−z} and ∑_{n=0}^{+∞}(n + 1)z^{n} = \frac{1}{(1−z)^{2}} , we can compute the ordinary generating function G_{k}(z) as follows:

G_{k}(z) = ∑_{n=0}^{+∞} R_{n,k}z^{n} = 1 + \binom{2k}{k} (∑_{n=0}^{+∞}(n + 1)z^{n} − 1) − (2^{2k−1} − \frac{1}{2}\binom{2k}{k}) (∑_{n=0}^{+∞} z^{n} − 1) + P_{n_{0}}(z) =

= 1 + \binom{2k}{k}\frac{(2z − z^{2})}{(1 − z)^{2}} − \frac{z}{1 − z} (2^{2k−1} − \frac{1}{2}\binom{2k}{k}) + P_{n_{0}}(z) =

= \frac{(2^{2k−1} − \frac{3}{2} \binom{2k}{k} + 1) z^{2} + (\frac{5}{2} \binom{2k}{k} − 2^{2k−1} − 2) z + 1}{(1 − z)^{2}} + P_{n_{0}}(z),

where the term P_{n_{0}}(z) described in equation (10) appears because when 2n + 3 ≤ k, or equivalently n ≤ ⌊\frac{k−3}{2}⌋ = n_{0}, we use the second expression of R_{n,k} in (7).

2 Concluding remarks

In this section we discuss some consequences of the previous results concerning the sequence \underline{A185095}. We show that the proof of the remaining conjectures and the relations with other sequences are immediate. First of all we observe that the k-th column of the array defining the sequence \underline{A185095} satisfies the recurrence relation

R_{n+1,k} = 2R_{n,k} − R_{n−1,k}, for n ≥ ⌊\frac{k − 3}{2}⌋ + 1,

which is a straightforward consequence of (7). The rectangular array which generates \underline{A186740} clearly is the transpose of the rectangular array that we have considered. In fact, if we start by numbering by 0 the columns of the rectangular array which defines \underline{A186740}, we find that the entry in the k-th row and n-th column corresponds to R_{n,k} by definition. Moreover by (7) and using our numeration for the columns we have

R_{n,0} = n + 1, n ≥ 0: the column 0 corresponds to the natural numbers \underline{A000027} ;

R_{n,1} = 2n + 1, n ≥ 0 : the column 1 is the sequence of odd integers \underline{A005408} ;

R_{n,2} = 6n + 1, n ≥ 0 : the column 2 is \underline{A016921} ;

R_{n,3} = 20n − 2, n ≥ 1 : the column 3 is \underline{A114698} ;

R_{n,4} = 70n − 93, n ≥ 2: the column 4 is \underline{A114646} ,

and so on .

Finally another interesting consequence of the formula for R_{n,k} concerns the sequence \underline{A198632}. By considering the array w(h, 2k), as defined in the page related to \underline{A198632} (see OEIS [1]), we have immediately w(2(n + 1), 2k) = 2R_{n,k}. In fact

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

where λ_{j}, j = 1, …, h, are the eigenvalues of the adjacency matrix J_{h}, a Jacobi matrix, whose characteristic polynomial is U_{h}(\frac{x}{2}). Thus

w(2(n + 1), 2k) = ∑_{j=1}^{2n+2} (2 cos (\frac{jπ}{2n + 3}))^{2k} = 2^{2k+1} ∑_{j=1}^{n+1} (cos (\frac{jπ}{2n + 3}))^{2k} = 2R_{n,k}.

3 Acknowledgements

In this paper we point out some relations between many sequences and we think that a lot of interesting formulas will also rise, considering the role played by Dickson and Chebyshev polynomials, their zeros and the summation identities involved. The author sincerely thanks L. Edson Jeffery for all his interesting work and for his invitation to try a proof of his conjectures.

References

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

[2] R. Lidl, G. L. Mullen, and G. Turnwald, Dickson Polynomials, John Wiley and Sons, 1993.

[3] T. J. Rivlin, The Chebyshev Polynomials, John Wiley and Sons, 1974.

[4] J. C. Mason and D. C. Handscomb, Chebyshev Polynomials, Chapman and Hall, 2002.

[5] I. S. Gradshteyn and I. M. Ryzhik, Tables of Integrals, Series and Products, Academic Press, 2007.

2010 Mathematics Subject Classification: Primary 11C08; Secondary 11B83. Keywords: Dickson polynomial, Chebyshev polynomial.

(Concerned with sequences \underline{A000027}, \underline{A005408}, \underline{A016921}, \underline{A114646}, \underline{A114698}, \underline{A114968}, \underline{A185095}, \underline{A186740}, \underline{A198632}, and \underline{A198636}.)

Received December 7 2013; revised version received January 25 2014. Published in Journal of Integer Sequences, February 16 2014. Corollary 3 and Proposition 4 corrected, May 21 2015.

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