Further Combinatorial Identities deriving from the $n$-th power of a $2   \times 2$ matrix

James Mc Laughlin; Nancy J. Wyshinski

arXiv:1901.00476·math.CO·January 3, 2019

Further Combinatorial Identities deriving from the $n$-th power of a $2 \times 2$ matrix

James Mc Laughlin, Nancy J. Wyshinski

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TL;DR

This paper derives new combinatorial identities from the powers of 2x2 matrices, explores Fibonacci polynomials, and characterizes a class of multiplicative functions using binomial sums.

Contribution

It introduces novel combinatorial identities based on matrix powers, provides a new functional equation for Fibonacci polynomials, and characterizes a class of multiplicative functions with binomial coefficient formulas.

Findings

01

Derived a complex combinatorial identity involving binomial coefficients and matrix powers.

02

Established a functional equation for generalized Fibonacci polynomials.

03

Provided a binomial sum formula for specially multiplicative functions.

Abstract

In this paper we use a formula for the $n$ -th power of a $2 \times 2$ matrix $A$ (in terms of the entries in $A$ ) to derive various combinatorial identities. Three examples of our results follow. 1) We show that if $m$ and $n$ are positive integers and $s \in {0, 1, 2, \dots,$ $⌊(mn - 1) /2 ⌋}$ , then \begin{multline*} \sum_{i,j,k,t}2^{1+2t-mn+n} \frac{(-1)^{nk+i(n+1)}}{1+\delta_{(m-1)/2,\,i+k}} \binom{m-1-i}{i} \binom{m-1-2i}{k}\times\\ \binom{n(m-1-2(i+k))}{2j}\binom{j}{t-n(i+k)} \binom{n-1-s+t}{s-t}\\ =\binom{mn-1-s}{s}. \end{multline*} 2) The generalized Fibonacci polynomial $f_{m} (x, s)$ can be expressed as \[ f_{m}(x,s)= \sum_{k=0}^{\lfloor (m-1)/2 \rfloor}\binom{m-k-1}{k}x^{m-2k-1}s^{k}. \] We prove that the following functional equation holds: \begin{equation*} f_{mn}(x,s)=f_{m}(x,s)\times f_{n}\left (\,f_{m+1}(x,s)+sf_{m-1}(x,s), \,-(-s)^{m}\right) . \end{equation*} 3) If…

Equations112

i, j, k, t \sum 2^{1 + 2 t - mn + n} \frac{( - 1 ) ^{nk + i (n + 1)}}{1 + δ _{(m - 1) /2, i + k}} (i m - 1 - i) (k m - 1 - 2 i) \times (2 j n ( m - 1 - 2 ( i + k ))) (t - n ( i + k ) j) (s - t n - 1 - s + t) = (s mn - 1 - s) .

i, j, k, t \sum 2^{1 + 2 t - mn + n} \frac{( - 1 ) ^{nk + i (n + 1)}}{1 + δ _{(m - 1) /2, i + k}} (i m - 1 - i) (k m - 1 - 2 i) \times (2 j n ( m - 1 - 2 ( i + k ))) (t - n ( i + k ) j) (s - t n - 1 - s + t) = (s mn - 1 - s) .

f_{m} (x, s) = k = 0 \sum ⌊(m - 1) /2 ⌋ (k m - k - 1) x^{m - 2 k - 1} s^{k} .

f_{m} (x, s) = k = 0 \sum ⌊(m - 1) /2 ⌋ (k m - k - 1) x^{m - 2 k - 1} s^{k} .

f_{mn} (x, s) = f_{m} (x, s) \times f_{n} (f_{m + 1} (x, s) + s f_{m - 1} (x, s), - (- s)^{m}) .

f_{mn} (x, s) = f_{m} (x, s) \times f_{n} (f_{m + 1} (x, s) + s f_{m - 1} (x, s), - (- s)^{m}) .

f (p^{n + 1}) = f (p) f (p^{n}) - g (p) f (p^{n - 1}), n \geq 1,

f (p^{n + 1}) = f (p) f (p^{n}) - g (p) f (p^{n - 1}), n \geq 1,

f (p^{k}) = j = 0 \sum ⌊ k /2 ⌋ (- 1)^{j} (j k - j) f (p)^{k - 2 j} g (p)^{j} .

f (p^{k}) = j = 0 \sum ⌊ k /2 ⌋ (- 1)^{j} (j k - j) f (p)^{k - 2 j} g (p)^{j} .

A = (a c b d)

A = (a c b d)

y_{n} = i = 0 \sum ⌊ n /2 ⌋ (i n - i) T^{n - 2 i} (- D)^{i} .

y_{n} = i = 0 \sum ⌊ n /2 ⌋ (i n - i) T^{n - 2 i} (- D)^{i} .

A^{n} = (y_{n} - d y_{n - 1} c y_{n - 1} b y_{n - 1} y_{n} - a y_{n - 1}) .

A^{n} = (y_{n} - d y_{n - 1} c y_{n - 1} b y_{n - 1} y_{n} - a y_{n - 1}) .

y_{k + 1} = (a + d) y_{k} + (b c - a d) y_{k - 1} .

y_{k + 1} = (a + d) y_{k} + (b c - a d) y_{k - 1} .

k = 0 \sum n - 1 (k n - 1 - k) (w + k n) (m - k - w k + w) (- 1)^{k} = k = - 2 w - n + m + 1 \sum m - w (k + w n) (n + k + w - m n) (k k + n + 2 w - m - 1) (- 1)^{k} .

k = 0 \sum n - 1 (k n - 1 - k) (w + k n) (m - k - w k + w) (- 1)^{k} = k = - 2 w - n + m + 1 \sum m - w (k + w n) (n + k + w - m n) (k k + n + 2 w - m - 1) (- 1)^{k} .

i, j, k, t \sum 2^{1 + 2 t - mn + n} \frac{( - 1 ) ^{nk + i (n + 1)}}{1 + δ _{(m - 1) /2, i + k}} (i m - 1 - i) (k m - 1 - 2 i) \times (2 j n ( m - 1 - 2 ( i + k ))) (t - n ( i + k ) j) (s - t n - 1 - s + t) = (s mn - 1 - s),

i, j, k, t \sum 2^{1 + 2 t - mn + n} \frac{( - 1 ) ^{nk + i (n + 1)}}{1 + δ _{(m - 1) /2, i + k}} (i m - 1 - i) (k m - 1 - 2 i) \times (2 j n ( m - 1 - 2 ( i + k ))) (t - n ( i + k ) j) (s - t n - 1 - s + t) = (s mn - 1 - s),

δ_{p, q} = {1, 0, p = q, p \neq = q .

δ_{p, q} = {1, 0, p = q, p \neq = q .

A = (1 x 10) .

A = (1 x 10) .

A^{n} = (y_{n} x y_{n - 1} y_{n - 1} y_{n} - y_{n - 1}) = (y_{n} x y_{n - 1} y_{n - 1} x y_{n - 2}),

A^{n} = (y_{n} x y_{n - 1} y_{n - 1} y_{n} - y_{n - 1}) = (y_{n} x y_{n - 1} y_{n - 1} x y_{n - 2}),

y_{k} = i = 0 \sum ⌊ k /2 ⌋ (i k - i) x^{i} = y_{k - 1} + x y_{k - 2} .

y_{k} = i = 0 \sum ⌊ k /2 ⌋ (i k - i) x^{i} = y_{k - 1} + x y_{k - 2} .

T_{n} = y_{n} + x y_{n - 2} .

T_{n} = y_{n} + x y_{n - 2} .

T_{n + 1} = T_{n} + x T_{n - 1} .

T_{n + 1} = T_{n} + x T_{n - 1} .

T_{n}

T_{n}

= \frac{1}{2 ^{n + 1}} j = 0 \sum ⌊ n /2 ⌋ k = j \sum ⌊ n /2 ⌋ (2 k n) (j k) 4^{j} x^{j} .

T_{n}^{r} = s = 0 \sum ⌊ n r /2 ⌋ k = 0 \sum ⌊ r /2 ⌋ i = 0 \sum ⌊ n (r - 2 k) /2 ⌋ (k r) (2 i n ( r - 2 k )) (s - nk i) 2^{1 + 2 t - r n} \frac{( - 1 ) ^{nk}}{1 + δ _{r /2, k}} x^{s} .

T_{n}^{r} = s = 0 \sum ⌊ n r /2 ⌋ k = 0 \sum ⌊ r /2 ⌋ i = 0 \sum ⌊ n (r - 2 k) /2 ⌋ (k r) (2 i n ( r - 2 k )) (s - nk i) 2^{1 + 2 t - r n} \frac{( - 1 ) ^{nk}}{1 + δ _{r /2, k}} x^{s} .

δ_{p, q} = {1, 0, p = q, p \neq = q .

δ_{p, q} = {1, 0, p = q, p \neq = q .

y_{j}^{(n)} = i = 0 \sum ⌊ j /2 ⌋ (i j - i) T_{n}^{j - 2 i} (- D_{n})^{i} .

y_{j}^{(n)} = i = 0 \sum ⌊ j /2 ⌋ (i j - i) T_{n}^{j - 2 i} (- D_{n})^{i} .

(y_{mn} x y_{mn - 1} y_{mn - 1} x y_{mn - 2})

(y_{mn} x y_{mn - 1} y_{mn - 1} x y_{mn - 2})

= (y_{m}^{(n)} - x y_{n - 2} y_{m - 1}^{(n)} x y_{n - 1} y_{m - 1}^{(n)} y_{n - 1} y_{m - 1}^{(n)} y_{m}^{(n)} - y_{n - 1} y_{m - 1}^{(n)}) .

y_{mn - 1} = y_{n - 1} y_{m - 1}^{(n)} .

y_{mn - 1} = y_{n - 1} y_{m - 1}^{(n)} .

s = 0 \sum ⌊(mn - 1) /2 ⌋ (s mn - 1 - s) x^{s} = s = 0 \sum ⌊(mn - 1) /2 ⌋ i, j, k, t \sum 2^{1 + 2 t - mn + n} \frac{( - 1 ) ^{nk + i (n + 1)}}{1 + δ _{(m - 1) /2, i + k}} (i m - 1 - i) (k m - 1 - 2 i) \times (2 j n ( m - 1 - 2 ( i + k ))) (t - n ( i + k ) j) (s - t n - 1 - s + t) x^{s} .

s = 0 \sum ⌊(mn - 1) /2 ⌋ (s mn - 1 - s) x^{s} = s = 0 \sum ⌊(mn - 1) /2 ⌋ i, j, k, t \sum 2^{1 + 2 t - mn + n} \frac{( - 1 ) ^{nk + i (n + 1)}}{1 + δ _{(m - 1) /2, i + k}} (i m - 1 - i) (k m - 1 - 2 i) \times (2 j n ( m - 1 - 2 ( i + k ))) (t - n ( i + k ) j) (s - t n - 1 - s + t) x^{s} .

\frac{1}{2 ^{n - 2 s - 1}} j = s \sum ⌊ n /2 ⌋ (2 j n) (s j) = \frac{n}{n - s} (s n - s) .

\frac{1}{2 ^{n - 2 s - 1}} j = s \sum ⌊ n /2 ⌋ (2 j n) (s j) = \frac{n}{n - s} (s n - s) .

f (mn) = f (m) f (n),

f (mn) = f (m) f (n),

f (m) f (n) = d ∣ (m, n) \sum f (\frac{mn}{d ^{2}}) f_{A} (d)

f (m) f (n) = d ∣ (m, n) \sum f (\frac{mn}{d ^{2}}) f_{A} (d)

f (p^{n + 1}) = f (p) f (p^{n}) - g (p) f (p^{n - 1}), n \geq 1,

f (p^{n + 1}) = f (p) f (p^{n}) - g (p) f (p^{n - 1}), n \geq 1,

f (p^{k}) = j = 0 \sum ⌊ k /2 ⌋ (- 1)^{j} (j k - j) f (p)^{k - 2 j} g (p)^{j} .

f (p^{k}) = j = 0 \sum ⌊ k /2 ⌋ (- 1)^{j} (j k - j) f (p)^{k - 2 j} g (p)^{j} .

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Full text

Further Combinatorial Identities deriving from the $n$ -th power of a

$2\times 2$ matrix

J. Mc Laughlin

Mathematics Department

Trinity College

300 Summit Street, Hartford, CT 06106-3100

[email protected]

and

Nancy J. Wyshinski

Mathematics Department

Trinity College

300 Summit Street, Hartford, CT 06106-3100

[email protected]

(Date: March 4th, 2004)

Abstract.

In this paper we use a formula for the $n$ -th power of a $2\times 2$ matrix $A$ (in terms of the entries in $A$ ) to derive various combinatorial identities. Three examples of our results follow.

We show that if $m$ and $n$ are positive integers and $s\in\{0,1,2,\dots,$ $\lfloor(mn-1)/2\rfloor\}$ , then

[TABLE]

The generalized Fibonacci polynomial $f_{m}(x,s)$ can be expressed as

[TABLE]

We prove that the following functional equation holds:

[TABLE]

If an arithmetical function $f$ is multiplicative and for each prime $p$ there is a complex number $g(p)$ such that

[TABLE]

then $f$ is said to be specially multiplicative. We give another derivation of the following formula for a specially multiplicative function $f$ evaluated at a prime power:

[TABLE]

We also prove various other combinatorial identities.

1. Introduction

Throughout the paper, let $I$ denote the $2\times 2$ -identity matrix and $n$ an arbitrary positive integer. In [7], the first author proved the following theorem, which gives a formula for the $n$ -th power of a $2\times 2$ matrix in terms of its entries:

Theorem 1.

Let

[TABLE]

be an arbitrary $2\times 2$ matrix and let $T=a+d$ denote its trace and $D=ad-bc$ its determinant. Let

[TABLE]

Then, for $n\geq 1$ ,

[TABLE]

The proof used the fact that

[TABLE]

This theorem was then used to derive various binomial identities. As an example, we cite the following corollary.

Corollary 1.

Let $n$ be a positive integer and let $m$ be an integer with $0\leq m\leq 2n$ . Then for $-n\leq w\leq n$ ,

[TABLE]

In this present paper we use Theorem 1 to derive some further identities.

2. A Binomial Identity deriving from $(A^{m})^{n}=A^{mn}$

We use the trivial identity $(A^{m})^{n}=A^{mn}$ to prove the following theorem.

Theorem 2.

Let $m$ and $n$ be positive integers and let $s\in\{0,1,2,\dots,\lfloor(mn-1)/2\rfloor\}$ . Then

[TABLE]

where $i$ , $j$ , $k$ and $t$ run through integral values which keep all binomial entries in (2.1) non-negative, and

[TABLE]

Proof.

Let

[TABLE]

From Theorem 1 and (1.1) we have that

[TABLE]

with

[TABLE]

Let $T_{n}$ denote the trace of $A^{n}$ and $D_{n}$ the determinant of $A^{n}$ (so $D_{n}=(-x)^{n}$ ). From (2.2) we have that

[TABLE]

Thus the sequence $\{T_{n}\}$ satisfies the same recurrence relation as the sequence $\{y_{n}\}$ , namely

[TABLE]

This leads to the explicit formula

[TABLE]

After some straightforward but tedious calculations, we derive from the first of these equalities, for integral $r\geq 0$ , that

[TABLE]

As usual,

[TABLE]

For integral $j\geq 0$ define

[TABLE]

Then Theorem 1 and the trivial identity $A^{mn}=(A^{n})^{m}$ give that

[TABLE]

If we compare $(1,2)$ entries of the first and last matrices, we have that

[TABLE]

Upon combining (2.7), (2.6) and (2.3), we get that

[TABLE]

Here $i$ , $j$ , $k$ , and $t$ run through all sets of integers which keep all binomial entries non-negative. The result now follows upon comparing coefficients of like powers of $x$ . ∎

Upon comparing like powers of $x$ on each side of (2.4), using (2.3) and (2.5), we get the following.

Corollary 2.

Let $n$ be a positive integer. Then for each integer $s$ , $0\leq s\leq\lfloor n/2\rfloor$ ,

[TABLE]

This identity is also found in [1] (page 442) and [3] (formula 3.120).

3. A Proof of an Identity for Specially Multiplicative Functions

An arithmetical function $f$ is said to be multiplicative if $f(1)=1$ and

[TABLE]

whenever $(m,n)=1$ . If (3.1) holds for all $m$ and $n$ , then $f$ is said to be completely multiplicative. A multiplicative function $f$ is said to be specially multiplicative if there is a completely multiplicative function $f_{A}$ such that

[TABLE]

for all $m$ and $n$ . An alternative characterization of specially multiplicative functions is given below (see [5], for example):

If $f$ is multiplicative and for each prime $p$ there is a complex number $g(p)$ such that

[TABLE]

then $f$ is specially multiplicative. (In this case, $f_{A}(p)=g(p)$ , for all primes $p$ ).

We give an alternative proof of the following known result (also see [5], for example).

Proposition 1.

Let $f$ and $g$ be as at (3.2). Then for $k\geq 0$ and all primes $p$ ,

[TABLE]

Proof.

Clearly we can assume $k\geq 3$ . Equation 3.2 implies that

[TABLE]

The result now follows immediately from Theorem 1, upon comparing $(1,1)$ entries on each side. ∎

Remark: The Ramanujan $\tau$ function is specially multiplicative with $g(p)$ $=$ $p^{11}$ . We note in passing that the $\tau$ Conjecture for $p$ prime, namely that $|\tau(p)|<2p^{11/2}$ , is equivalent to the conjecture that $\lim_{k\to\infty}\tau(p^{k})/\tau(p^{k-1})$ does not exist. This follows from (3.3), the correspondence between matrices and continued fractions and Worpitzky’s Theorem for continued fractions.

4. A Recurrence Formula for the Generalized Fibonacci Polynomials

The Fibonacci polynomials $\{f_{m}(x,s)\}_{m=0}^{\infty}$ are defined by $f_{0}(x,s)=0$ , $f_{1}(x,s)=1$ and $f_{n+1}(x,s)=xf_{n}(x,s)+sf_{n-1}(x,s)$ , for $n\geq 1$ . They are given explicitly by the formula

[TABLE]

It is clear from Theorem 1 that the $f_{n}(x,s)$ satisfy

[TABLE]

We can now use the trivial identity $A^{m\,n}=(A^{m})^{n}$ applied to the matrix $\left(\begin{matrix}x&1\\ s&0\end{matrix}\right)$ , together with Theorem 1 applied to the $(1,2)$ -entries on each side to get the following functional equation for the Fibonacci polynomials.

Corollary 3.

Let $f_{i}(x,s)$ denote the $i$ -th Fibonacci polynomial and let $m$ and $n$ be positive integers. Then

[TABLE]

5. A Polynomial Identity of Bhatwadekar and Roy

In [8] Sury gave a proof of the following polynomial identity, which he attributes to Bhatwadekar and Roy [2]:

Corollary 4.

For every positive integer $n$ and all $x$ ,

[TABLE]

Proof.

Clearly we can assume $n\geq 2$ . One easily checks by induction that, for $n\geq 2$ ,

[TABLE]

The result is now immediate from Theorem 1. ∎

6. Other Elementary Identities

If we replace $n$ by $n+1$ in Equation 2.2 and take the determinant of the first and last matrices, we get

[TABLE]

Upon comparing coefficients of $x^{s}$ , for $0\leq s\leq n-1$ on each side, we get the following identity.

Corollary 5.

Let $n$ be a positive integer. If $s$ is an integer, $0\leq s\leq n-1$ , then

[TABLE]

Once again we start with the matrix $A=\left(\begin{smallmatrix}1&1\\ x&0\end{smallmatrix}\right)$ and then consider the identity $A^{mn}=\left(A^{m}\right)^{n}=\left(A^{n}\right)^{m}$ for small values of $m$ .

Corollary 6.

Let $n$ be a positive integer and let $s$ be an integer, $0\leq s\leq n-1$ . Then

[TABLE]

Proof.

With $A$ as defined above, we have

[TABLE]

If we compare the $(1,2)$ entries of $A^{2n}$ and $(A^{2})^{n}$ , using Theorem 1, we get that

[TABLE]

The equality of the first and third terms in (6.2) follows on comparing powers of $x$ . On the other hand, Theorem 1 also gives that

[TABLE]

where $y_{k}$ is as at (2.3). It is easy to show that

[TABLE]

If we compare the $(1,2)$ entries of $A^{2n}$ and $(A^{n})^{2}$ using (6.3) and Theorem 1, then

[TABLE]

The equality of the second and third terms in (6.2) now follows. ∎

A similar consideration of $A^{3n}$ and $(A^{3})^{n}$ gives the following identity.

Corollary 7.

Let $n$ be a positive integer and $s$ an integer such that $0\leq s\leq\lfloor(3n-1)/2\rfloor$ . Then

[TABLE]

Proof.

Since

[TABLE]

comparing the $(1,2)$ entries of $A^{3n}$ and $(A^{3})^{n}$ , using Theorem 1, gives

[TABLE]

The results follows, after a little simplification, upon comparing coefficients of like powers of $x$ on each side of (6.5). ∎

More generally, one can use the identity $A^{m+n}=A^{m}A^{n}$ together with Theorem 1 to compare the $(1,1)$ entries on each side to get (again using the notation from (2.3)) that

[TABLE]

Upon collecting like powers of $x$ and equating coefficients on each side, we get the following identity.

Corollary 8.

Let $m$ and $n$ be a positive integer and $s$ an integer such that $0\leq s\leq\lfloor(m+n)/2\rfloor$ . Then

[TABLE]

7. Concluding Remarks

Some other interesting consequences follow readily from Theorem 1. We consider two more.

If we let $A=\left(\begin{smallmatrix}x&0\\ 0&y\end{smallmatrix}\right)$ , then Waring’s formula

[TABLE]

can be derived easily by considering the trace of $A^{n}$ .

If we set $A=\left(\begin{smallmatrix}x&1\\ 1&0\end{smallmatrix}\right)$ , then Theorem 1 and the correspondence between continued fractions and matrices give that, for $x>0$ ,

[TABLE]

Bibliography9

The reference list from the paper itself. Each links out to its DOI / PubMed record.

1[1] Berndt, Bruce C.; Evans, Ronald J.; Williams, Kenneth S. Gauss and Jacobi sums . Canadian Mathematical Society Series of Monographs and Advanced Texts. A Wiley-Interscience Publication. John Wiley & Sons, Inc., New York, 1998. xii+583 pp.
2[2] S. M. Bhatwadekar, A. Roy, Some results on embedding of a line in 3 3 3 -space. J. Algebra 142 (1991), no. 1, 101–109.
3[3] Gould, Henry W. Combinatorial identities. A standardized set of tables listing 500 binomial coefficient summations. Henry W. Gould, Morgantown, W.Va., 1972. viii+106 pp.
4[4] H. W. Gould, A history of the Fibonacci Q 𝑄 Q -matrix and a higher-dimensional problem. Fibonacci Quart. 19 (1981), no. 3, 250–257.
5[5] P. Haukkanen, Some characterizations of specially multiplicative functions. Int. J. Math. Math. Sci. 2003 , no. 37, 2335–2344.
6[6] R. C. Johnson, Matrix methods for Fibonacci and related sequences. http:// maths.dur.ac.uk/ dma 0rcj/PED/fib.pdf, (August 2003).
7[7] James Mc Laughlin, Combinatorial Identities Deriving from the n 𝑛 n -th Power of a 2 × 2 2 2 2\times 2 Matrix. – submitted
8[8] B. Sury, A curious polynomial identity. Nieuw Arch. Wisk. (4) 11 (1993), no. 2, 93–96.

TL;DR

Contribution

Findings

Abstract

Peer Reviews

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Further Combinatorial Identities deriving from the nnn-th power of a

Abstract.

1. Introduction

Theorem 1**.**

Corollary 1**.**

2. A Binomial Identity deriving from (Am)n=Amn(A^{m})^{n}=A^{mn}(Am)n=Amn

Theorem 2**.**

Proof.

Corollary 2**.**

3. A Proof of an Identity for Specially Multiplicative Functions

Proposition 1**.**

Proof.

4. A Recurrence Formula for the Generalized Fibonacci Polynomials

Corollary 3**.**

5. A Polynomial Identity of Bhatwadekar and Roy

Corollary 4**.**

Proof.

6. Other Elementary Identities

Corollary 5**.**

Corollary 6**.**

Proof.

Corollary 7**.**

Proof.

Corollary 8**.**

7. Concluding Remarks

Further Combinatorial Identities deriving from the $n$ -th power of a

Theorem 1.

Corollary 1.

2. A Binomial Identity deriving from $(A^{m})^{n}=A^{mn}$

Theorem 2.

Corollary 2.

Proposition 1.

Corollary 3.

Corollary 4.

Corollary 5.

Corollary 6.

Corollary 7.

Corollary 8.