1 Introduction and outline
In mathematics and applied science, Fibonacci and Lucas numbers are well known. They are defined by the same recurrence relations
but with different initial conditions
According to these recurrence relations, it is not hard to check that both definitions for Fibonacci and Lucas numbers can be extended to negative integer indices by
In 1965, Carlitz [Reference Carlitz2] discovered an elegant identity for circular sums:
There are three different proofs that can be found in [Reference Benjamin, Rouse and Howard1, Reference Chu5, Reference Mikic8]. Denote an m-tuple of integers by $\mathbf {k}=(k_{1},k_{2},\ldots ,k_{m})\in \mathbb {N}_{0}^{m}$ . We define two component-related sums by
Recently, the author [Reference Chu5] found the following alternating counterpart:
where $u_{n}$ and $v_{n}$ are two periodic sequences (see A010892 and A049347 recorded in [Reference Sloane9]) and related by $u_{2n+1}=v_{n}$ and given explicitly by
To reduce lengthy expressions, the following abbreviations will be used throughout the paper. For a real number x, the greatest integer not exceeding it will be denoted by $\lfloor {x}\rfloor $ . We shall use $\chi $ for the logical function with $\chi (\text {true})=1$ and $\chi (\text {false})=0$ . For three integers $i,j,m$ with $m>0$ , the notation $i\equiv _{m}j$ stands for ‘i is congruent to j modulo m’.
The aim of the present paper is to evaluate two further alternating circular sums of binomial coefficients by making use of the generating function approach:
To fulfil this task, the Lagrange expansion formula [Reference Lagrange7] will be crucial. We give it in the following form (see Comtet [Reference Comtet and Reidel6, Section 3.8] and Chu [Reference Chu3, Reference Chu4]). For a formal power series $\varphi (x)$ subject to the condition $\varphi (0)\not =0$ , the functional equation $y=x/\varphi (x)$ determines x as an implicit function of y. Then for another formal power series $F(x)$ in the variable x, the following expansions hold for both composite series:
where $[x^{k}]\phi (x)$ denotes the coefficient of $x^{k}$ in the formal power series $\phi (x)$ .□
2 The first alternating sum $\boldsymbol {\Phi _{m}(n)}$
Observe that $\Phi _{m}(n)$ defined in (1.2) can be rewritten as
where $\sum k_{2i-1}$ denotes the sum of the odd indexed components of $\mathbf {k}$ . Then we have the following theorem.
Theorem 2.1 (generating functions: $m,n\in \mathbb {N}$ ).
We have
Recall the Binet formulae:
When m is even with $m=2\lambda $ , we can rewrite the generating function and then decompose it into partial fractions:
By extracting the coefficient of $x^{n}$ across the above equation, we find the following counterpart of Carlitz’ formula (1.1).
Proposition 2.2 (explicit formula: $n,\lambda \in \mathbb {N}$ ).
We have
However, when m is odd with $m=2\lambda -1$ , there exists no such closed formula. By expanding the corresponding generating function into power series
and then extracting the coefficient of $y^{n}$ , we get the binomial sum expression.
Proposition 2.3 (explicit formula: $n,\lambda \in \mathbb {N}$ ).
The first five values are highlighted in the following corollary.
Corollary 2.4 (explicit formula: $n\in \mathbb {N}$ ).
We have
Proof of Theorem 2.1.
First, it is routine to verify the relations
They can be used to express the binomial sum with respect to $k_{1}$ in $\Phi _{m}(n)$ as
We proceed next with the binomial sum with respect to $k_{2}$ in $\Phi _{m}(n)$ :
Repeat this process for the binomial sum with respect to $k_{3}$ in $\Phi _{m}(n)$ :
and then the binomial sum with respect to $k_{4}$ in $\Phi _{m}(n)$ :
By means of the induction principle, we can show that for $1\le j\le m/2$ , the binomial sums with respect to $k_{2j-1}$ and $k_{2j-2}$ in $\Phi _{m}(n)$ are respectively given by
According to the parity of m, we can determine $\Phi _{m}(n)$ separately as follows. For even $m=2\lambda $ , we can evaluate
which simplifies to
By means of the Lagrange expansion formula, these expressions can be simplified further. In fact, specialise first in (1.4) by
Then we can deduce that
together with the generating function
After some routine simplification, we find the elegant formula
Analogously for odd $m=2\lambda -1$ , we can evaluate
which simplifies to
By means of the Lagrange expansion formula, these expressions can be simplified further. In fact, specialise now in (1.4) by
Then we can deduce that
together with the generating function below
After some routine simplification, we confirm another elegant formula,
Hence, both statements of Theorem 2.1 are proved.
3 The second alternating sum $\boldsymbol {\Psi _{m}(n)}$
Analogously, we can reformulate the circular sum $\Psi _{m}(n)$ as
where $\sum k_{2i}$ stands for the sum of the even indexed components of $\mathbf {k}$ . We have the following theorem.
Theorem 3.1 (generating functions).
Remark 3.2. Comparing the generating functions in Theorems 2.1 and 3.1, we can see that $\Phi _{m}(n)=\Psi _{m}(n)$ for all $n\in \mathbb {N}$ when m is even. This is not surprising because in this case, both sums $\Psi _{m}(n)$ and $\Phi _{m}(n)$ , defined respectively in (1.3) and (1.2), are equivalent if we reverse the order for the components of $\mathbf {k}$ . When m is odd, we have instead $\Psi _{m}(n)=\Phi _{m+6}(n)$ for all $n\in \mathbb {N}$ .
According to this remark, we have the following summation formulae, where $\Psi _{5}(n)$ , $\Psi _{7}(n)$ and $\Psi _{9}(n)$ are integer sequences A006190, A004254 and A041025, respectively, recorded in [Reference Sloane9]:
Proof of Theorem 3.1.
We only need to prove the case when m is odd with $m=2\lambda -1$ . According to the binomial relations
we can express the binomial sum with respect to $k_{1}$ in $\Psi _{m}(n)$ as
We proceed next with the binomial sum with respect to $k_{2}$ in $\Psi _{m}(n)$ :
Repeat this process for the binomial sums with respect to $k_{3}$ in $\Psi _{m}(n)$ :
and then with respect to $k_{4}$ in $\Psi _{m}(n)$ :
By means of the induction principle, we can show that for $1\le j\le m/2$ , the binomial sums with respect to $k_{2j-1}$ and $k_{2j-2}$ in $\Psi _{m}(n)$ are respectively given by
Since $m=2\lambda -1$ is odd, we can further evaluate
which simplifies to
Remark 3.3. When ${m=3}$ , the triple sum $\Psi _{3}(n)$ results in
By means of the Lagrange expansion formula, we can further simplify the generating function for $\Psi _{m}(n)$ . In fact, specialise in (1.4) by
Then we can deduce that
together with the generating function
After some routine simplification, we arrive at
which is also valid for ${m=3}$ . This completes the proof of Theorem 3.1.
4 Open problems
For Carlitz’ identity (1.1), there is a combinatorial proof by Benjamin and Rouse [Reference Benjamin, Rouse and Howard1] through the domino tiling. It would be interesting to construct a similar proof for the identity displayed in Proposition 2.2. Another intriguing problem is to find a combinatorial interpretation of the identity $\Psi _{m}(n)=\Phi _{m+6}(n)$ for all $n\in \mathbb {N}$ and odd $m\in \mathbb {N}$ .
Acknowledgement
The author expresses his sincere gratitude to an anonymous referee for the careful reading and valuable comments.