On k-Fibonacci Brousseau Sums

Prabha Sivaraman Nair Department of Mathematics Baby John Memorial Government College, Chavara Kerala 691583 India

                prabhamaths@gmail.com

Rejikumar Karunakaran Department of Mathematics NSS College, Cherthala Kerala 688541 India

                  rkkmaths@gmail.com

Abstract

Using elementary methods, we provide formulas for evaluating the Brousseau sum ∑_{i=1}^{n} i^{p}F_{k,i} and the shifted Brousseau sum ∑_{i=1}^{n} i^{p}F_{k,m+i} for all integers m, p ≥ 0, where (F_{k,i})_{i≥0} is the k-Fibonacci sequence defined by the two-term linear recurrence F_{k,i} = kF_{k,i−1} + F_{k,i−2} for i ≥ 2 with initial values F_{k,0} = 0 and F_{k,1} = 1.

1 Introduction

In the second issue of Fibonacci Quarterly in 1963, Brousseau [4] proposed a problem to discover an expression for the Fibonacci sum of the form

∑_{i=1}^{n} i^{3}F_{i},

where F_{i} is the i^{th} Fibonacci number. In the following year, Erbacher and Fuchs [7] gave a solution for this problem in terms of F_{n+2} and F_{n+3}. Later, Ledin [12], Brousseau [5], and Zeitlin [16] developed various methods to determine expressions for the Brousseau sum

∑_{i=1}^{n} i^{p}F_{i},

where p is a non-negative integer. Ledin showed that the solution of Erbacher and Fuchs can be expressed in the form

∑_{i=1}^{n} i^{3}F_{i} = (n^{3} − 6n^{2} + 24n − 50)F_{n+1} + (n^{3} − 3n^{2} + 15n − 31)F_{n} + 50.  (1)

Recently, Ollerton [13], Shannon [14], Hendel [11], and Adegoke [1] derived expressions for such sums. Dresden [6] used a different technique to find the sum ∑_{i=1}^{n} i^{p}F_{i} using just the binomial coefficients. Motivated by this, we develop formulas for the sums

∑_{i=1}^{n} i^{p}F_{k,i}, and ∑_{i=1}^{n} i^{p}F_{k,m+i},

for integers m, p ≥ 0, where F_{k,i} is the i^{th} k-Fibonacci number. For a positive integer k, the k-Fibonacci sequence (see [8]) (F_{k,n})_{n≥0} is defined by the two-term linear recurrence

F_{k,n} = kF_{k,n−1} + F_{k,n−2},  (2)

with initial values F_{k,0} = 0 and F_{k,1} = 1. The numbers F_{k,n} are sometimes called the “metallic” or “metallonacci” numbers. The numbers F_{1,n} are the “regular” Fibonacci numbers and F_{2,n} are the Pell numbers P_{n}. The Pell numbers F_{2,n} are sometimes called the silver Fibonacci numbers, and the numbers F_{3,n} are sometimes called the bronze Fibonacci numbers. For k = 1, 2, 3, and 4, the numbers F_{k,n} are \underline{A000045}, \underline{A000129}, \underline{A006190}, and \underline{A001076}, respectively, in the OEIS [15]. The k-Fibonacci numbers can be extended to negative subscripts by

F_{k,n} = F_{k,n+2} − kF_{k,n+1}, for n < 0.

The Lucas sequence (L_{n})_{n≥0} and the Pell-Lucas sequence (Q_{n})_{n≥0} are defined, respectively, by

L_{0} = 2, L_{1} = 1, L_{n} = L_{n−1} + L_{n−2} for n ≥ 2,

and

Q_{0} = Q_{1} = 2, Q_{n} = 2Q_{n−1} + Q_{n−2} for n ≥ 2.

The numbers L_{n} and Q_{n} are \underline{A000032} and \underline{A002203}, respectively, in the OEIS. Falcón [9] proved that the 4-Fibonacci numbers F_{4,n} are just F_{3n}/F_{3}, the 11-Fibonacci numbers F_{11,n} are just F_{5n}/F_{5}, the 29-Fibonacci numbers F_{29,n} are just F_{7n}/F_{7}, and so on. In general, for all odd indexed Lucas numbers L_{m}, the L_{m}-Fibonacci numbers F_{L_{m},n} are just F_{mn}/F_{m}. The sequence 1, 4, 11, 29, 76, … is \underline{A002878} in the OEIS. Falcón [9] also proved that the 14-Fibonacci numbers F_{14,n} are just P_{3n}/P_{3}, the 82-Fibonacci numbers F_{82,n} are just P_{5n}/P_{5}, the 478-Fibonacci numbers F_{478,n} are just P_{7n}/P_{7}, and so on. In general, for all odd indexed Pell-Lucas numbers Q_{m}, the Q_{m}-Fibonacci numbers are just P_{mn}/P_{m}. The sequence 2, 14, 82, 478, 2786, … is \underline{A077444} in the OEIS.

Hendel proved that

2 · ∑_{i=1}^{n} i^{3}P_{i} = (n^{3} − 3n^{2} + 6n − 7)P_{n+1} + (n^{3} + 3n − 3)P_{n} + 7.  (3)

We can see the clear similarities between Eqs. (1) and (3). For further clarity, we rewrite Eqs. (1) and (3) as

∑_{i=1}^{n} i^{3}F_{i} = (n^{3} − 6n^{2} + 24n − 50)F_{n+1} + ((n + 1)^{3} − 6(n + 1)^{2} + 24(n + 1) − 50)F_{n} + 50, (4)

and

2 · ∑_{i=1}^{n} i^{3}P_{i} = (n^{3} − 3n^{2} + 6n − 7)P_{n+1} + ((n + 1)^{3} − 3(n + 1)^{2} + 6(n + 1) − 7)F_{n} + 7. (5)

The identity (4) appears in the OEIS at \underline{A259546}. If we use the 3-Fibonacci numbers as another example, then we would have

3·∑_{i=1}^{n} i^{3}F_{3,i} = (n^{3}−2n^{2}+\frac{8}{3}n−\frac{22}{9})F_{3,n+1}+((n+1)^{3}−2(n+1)^{2}+\frac{8}{3}(n+1)−\frac{22}{9})F_{3,n}+\frac{22}{9}. (6)

Eqs. (4), (5), and (6) naturally suggest the following interesting generalization about the Brousseau sums of the k-Fibonacci numbers:

k · ∑_{i=1}^{n} i^{p}F_{k,i} = (C_{k}^{(p)}(n))F_{k,n+1} + (C_{k}^{(p)}(n + 1))F_{k,n} − C_{k}^{(p)}(0),  (7)

where C_{k}^{(p)}(n) is a “coefficient polynomial” in n of degree p with rational coefficients. The main task is to find an expression for the polynomial C_{k}^{(p)}(n). We do this using some simple recursion formulas involving just binomial coefficients. We use a similar technique as that of Dresden. Throughout this paper, we assume \binom{0}{0} = 1.

2 k-Fibonacci numbers and powers

We begin with the following set of identities, which are similar to those with the “regular” Fibonacci numbers [6].

\begin{aligned}F_{k,n} & = n + ∑_{i=1}^{n} (ki − 2 · \binom{1}{1})F_{k,n−i}, \\ F_{k,n} & = n^{2} + ∑_{i=1}^{n} (ki^{2} − 2 · \binom{2}{1}i)F_{k,n−i}, \\ F_{k,n} & = n^{3} + ∑_{i=1}^{n} (ki^{3} − 2 · \binom{3}{1}i^{2} − 2 · \binom{3}{3})F_{k,n−i}, \\ F_{k,n} & = n^{4} + ∑_{i=1}^{n} (ki^{4} − 2 · \binom{4}{1}i^{3} − 2 · \binom{4}{3}i)F_{k,n−i}, \\ F_{k,n} & = n^{5} + ∑_{i=1}^{n} (ki^{5} − 2 · \binom{5}{1}i^{4} − 2 · \binom{5}{3}i^{2} − 2 · \binom{5}{5})F_{k,n−i}, \end{aligned}

and so on. Our first theorem involves the generalization of these formulas.

Theorem 1. For all integers n, p ≥ 1, we have

F_{k,n} = n^{p} + ∑_{i=1}^{n} (ki^{p} − 2 ∑_{j=1}^{⌈ \frac{p}{2} ⌉} \binom{p}{2j − 1}i^{p−2j+1})F_{k,n−i}.  (8)

Proof. We use induction on n. When n = 1, the left-hand side of Eq. (8) is F_{k,1} = 1 and the right-hand side is

1 + ∑_{i=1}^{1} (ki^{p} − 2 ∑_{j=1}^{⌈ \frac{p}{2} ⌉} \binom{p}{2j − 1}i^{p−2j+1})F_{k,1−i}

= 1 + (k − 2 ∑_{j=1}^{⌈ \frac{p}{2} ⌉} \binom{p}{2j − 1})F_{k,0}

= 1 + 0 = 1.

When n = 2, the left-hand side of Eq. (8) is F_{k,2} = k and the right-hand side is

2^{p} + ∑_{i=1}^{2} (ki^{p} − 2 ∑_{j=1}^{⌈ \frac{p}{2} ⌉} \binom{p}{2j − 1}i^{p−2j+1})F_{k,2−i}

= 2^{p} + (k − 2 ∑_{j=1}^{⌈ \frac{p}{2} ⌉} \binom{p}{2j − 1})F_{k,1} + (k2^{p} − 2 ∑_{j=1}^{⌈ \frac{p}{2} ⌉} \binom{p}{2j − 1}2^{p−2j+1})F_{k,0}

= 2^{p} + (k − 2 · 2^{p−1}) + 0

= k.

Thus, Eq. (8) holds for n = 1 and n = 2. Now fix n ≥ 2. Assume that Eq. (8) holds for n − 1 and n. Then

F_{k,n} = n^{p} + ∑_{i=1}^{n} (ki^{p} − 2 ∑_{j=1}^{⌈ \frac{p}{2} ⌉} \binom{p}{2j − 1}i^{p−2j+1})F_{k,n−i},  (9)

and

F_{k,n−1} = (n − 1)^{p} + ∑_{i=1}^{n−1} (ki^{p} − 2 ∑_{j=1}^{⌈ \frac{p}{2} ⌉} \binom{p}{2j − 1}i^{p−2j+1})F_{k,n−1−i}.  (10)

Since F_{k,n−i} = 0 for i = n, Eq. (9) can be put in the form

F_{k,n} = n^{p} + ∑_{i=1}^{n−1} (ki^{p} − 2 ∑_{j=1}^{⌈ \frac{p}{2} ⌉} \binom{p}{2j − 1}i^{p−2j+1})F_{k,n−i}.  (11)

Multiplying Eq. (11) by k and adding it to Eq. (10), we get

kF_{k,n} + F_{k,n−1} = kn^{p} + (n − 1)^{p} + ∑_{i=1}^{n−1} (ki^{p} − 2 ∑_{j=1}^{⌈ \frac{p}{2} ⌉} \binom{p}{2j − 1}i^{p−2j+1})(kF_{k,n−i} + F_{k,n−i−1}).

Using the k-Fibonacci recurrence (2), this becomes

F_{k,n+1} = kn^{p} + (n − 1)^{p} + ∑_{i=1}^{n−1} (ki^{p} − 2 ∑_{j=1}^{⌈ \frac{p}{2} ⌉} \binom{p}{2j − 1}i^{p−2j+1})F_{k,n−i+1}.  (12)

Using the binomial expansion, we have the identity

(n + 1)^{p} − (n − 1)^{p} = 2 ∑_{j=1}^{⌈ \frac{p}{2} ⌉} \binom{p}{2j − 1}n^{p−2j+1},

and hence

(n − 1)^{p} = (n + 1)^{p} − 2 ∑_{j=1}^{⌈ \frac{p}{2} ⌉} \binom{p}{2j − 1}n^{p−2j+1}.  (13)

Now using Eqs. (12) and (13), we obtain

F_{k,n+1} = kn^{p} + (n + 1)^{p} − 2 ∑_{j=1}^{⌈ \frac{p}{2} ⌉} \binom{p}{2j − 1}n^{p−2j+1}

+ ∑_{i=1}^{n−1} (ki^{p} − 2 ∑_{j=1}^{⌈ \frac{p}{2} ⌉} \binom{p}{2j − 1}i^{p−2j+1})F_{k,n+1−i}

= (n + 1)^{p} + (kn^{p} − 2 ∑_{j=1}^{⌈ \frac{p}{2} ⌉} \binom{p}{2j − 1}n^{p−2j+1})

+ ∑_{i=1}^{n−1} (ki^{p} − 2 ∑_{j=1}^{⌈ \frac{p}{2} ⌉} \binom{p}{2j − 1}i^{p−2j+1})F_{k,n+1−i}.

Since F_{k,1} = 1, we rewrite this as

F_{k,n+1} = (n + 1)^{p} + ∑_{i=1}^{n} (ki^{p} − 2 ∑_{j=1}^{⌈ \frac{p}{2} ⌉} \binom{p}{2j − 1}i^{p−2j+1})F_{k,n+1−i}.

Since F_{k,n+1−i} = 0 when i = n + 1, we may simply add the corresponding term to the summand on the right-hand side of the above equation to get

F_{k,n+1} = (n + 1)^{p} + ∑_{i=1}^{n+1} (ki^{p} − 2 ∑_{j=1}^{⌈ \frac{p}{2} ⌉} \binom{p}{2j − 1}i^{p−2j+1})F_{k,n+1−i},

and this completes the induction step. This concludes the proof.

3 Convolutions

Applying Theorem 1 along with the k-Fibonacci sum [8, Proposition 8]

∑_{i=0}^{n} F_{k,n} = \frac{1}{k}(F_{k,n+1} + F_{k,n} − 1),

we can recursively find the following convolution identities:

k · ∑_{i=0}^{n} F_{k,n−i} = F_{k,n+1} + F_{k,n} − 1,

k · ∑_{i=0}^{n} iF_{k,n−i} = \frac{1}{k}(2F_{k,n+1} + (−k + 2)F_{k,n} − (kn + 2))

= (\frac{2}{k})F_{k,n+1} + (\frac{−k + 2}{k})F_{k,n} − \frac{kn + 2}{k},

k · ∑_{i=0}^{n} i^{2}F_{k,n−i} = \frac{1}{k^{2}} (8F_{k,n+1} + (k^{2} − 4k + 8)F_{k,n} − (k^{2}n^{2} + 4kn + 8))

= (\frac{8}{k^{2}} )F_{k,n+1} + (\frac{k^{2} − 4k + 8}{k^{2}})F_{k,n} − \frac{k^{2}n^{2} + 4kn + 8}{k^{2}},

k · ∑_{i=0}^{n} i^{3}F_{k,n−i} = \frac{1}{k^{3}} ((2k^{2} + 48)F_{k,n+1} + (−k^{3} + 8k^{2} − 24k + 48)F_{k,n}

− (k^{3}n^{3} + 6k^{2}n^{2} + 24kn + 2k^{2} + 48)),

= (\frac{2k^{2} + 48}{k^{3}})F_{k,n+1} + (\frac{−k^{3} + 8k^{2} − 24k + 48}{k^{3}})F_{k,n}

− \frac{k^{3}n^{3} + 6k^{2}n^{2} + 24kn + 2k^{2} + 48}{k^{3}},

and so on. Each sum on the left-hand side of the above set of equations is a convolution of the powers of i and the k-Fibonacci numbers. We define

T_{k,n}^{(p)} = \begin{cases}∑_{i=0}^{n} F_{k,n−i}, & if p = 0; \\ ∑_{i=0}^{n} i^{p}F_{k,n−i}, & if p ≥ 1. \end{cases}  (14)

A pattern is evident in the above set of equations. Note that each equation is of the form

k · T_{k,n}^{(p)} = Φ_{k}^{(p)}(0)F_{k,n+1} + Φ_{k}^{(p)}(−1)F_{k,n} − Φ_{k}^{(p)}(n),  (15)

where Φ_{k}^{(p)}(n) is a polynomial in n of degree p. To find an explicit formula for Φ_{k}^{(p)}(n), we must define the sequence (A_{k}^{(p)})_{p≥0} as follows:

Definition 2. The sequence (A_{k}^{(p)})_{p≥0} of numbers is defined by the recurrence

A_{k}^{(p)} = \begin{cases}1, & if p = 0; \\ \frac{2}{k} ∑_{j=1}^{⌈ \frac{p}{2} ⌉} \binom{p}{2j−1}A_{k}^{(p−2j+1)}, & if p ≥ 1. \end{cases}  (16)

The recurrence A_{k}^{(p)} generates the sequence

1, \frac{2}{k}, \frac{8}{k^{2}} , \frac{2k^{2} + 48}{k^{3}}, \frac{32k^{2} + 384}{k^{4}}, ….

Note that these numbers are the coefficients of F_{k,n+1} in the above set of equations. The numbers A_{k}^{(p)} for p = 1, 2, 3, and 4 are given below:

(i) A_{1}^{(p)} : 1, 2, 8, 50, 416, 4322, 53888, . . .. This sequence is \underline{A000557} in the OEIS [15].

(ii) A_{2}^{(p)} : 1, 1, 2, 7, 32, 181, 1232, . . .. This sequence is \underline{A006154} in the OEIS [15].

(iii) A_{3}^{(p)} : 1, \frac{2}{3}, \frac{8}{9}, \frac{22}{9}, \frac{224}{27}, \frac{2774}{81}, \frac{13952}{81}, . . ..

(iv) A_{4}^{(p)} : 1, \frac{1}{2}, \frac{1}{2}, \frac{5}{4}, \frac{7}{2}, \frac{47}{4}, \frac{197}{4}, . . ..

From the last equation in the set of convolution identities given above, we recognize that

\begin{aligned}Φ_{k}^{(3)}(n) & = n^{3} + \frac{6}{k}n^{2} + \frac{24}{k^{2}}n + \frac{2k^{2} + 48}{k^{3}} \\ & = 1 · \binom{3}{0}n^{3} + \frac{2}{k} · \binom{3}{1}n^{2} + \frac{8}{k^{2}} · \binom{3}{2}n + \frac{2k^{2} + 48}{k^{3}} · \binom{3}{3}, \end{aligned}

where 1, 2/k, 8/k^{2}, (2k^{2} + 48)/k^{3} are the first four terms of the sequence (A_{k}^{(p)})_{p≥0}. With all this in mind, we make the following definition:

Definition 3. For all integers p ≥ 0, we define the polynomial Φ_{k}^{(p)}(n) as

Φ_{k}^{(p)}(n) = \begin{cases}A_{k}^{(p)}, & if n = 0; \\ ∑_{r=0}^{p} A_{k}^{(r)}\binom{p}{r}n^{p−r}, & if n ≠ 0. \end{cases}  (17)

Rewriting Eq. (15) in terms of A_{k}^{(p)}, we have the following theorem:

Theorem 4. If T_{k,n}^{(p)} is as defined in Eq. (14), then for all integers p ≥ 0, we have

k · T_{k,n}^{(p)} = A_{k}^{(p)}F_{k,n+1} + ∑_{r=0}^{p} A_{k}^{(r)}\binom{p}{r}((−1)^{p−r}F_{k,n} − n^{p−r}).  (18)

Proof. We use induction on p. When p = 0, the left-hand side of Eq. (18) is

k · T_{k,n}^{(0)} = k ∑_{i=0}^{n} F_{k,n−i} = F_{k,n+1} + F_{k,n} − 1,

and the right-hand side is

A_{k}^{(0)}F_{k,n+1} + A_{k}^{(0)}(F_{k,n} − 1) = \frac{1}{k}(F_{k,n+1} + F_{k,n} − 1).

Thus, Eq. (18) holds for p = 0. Now, fix p ≥ 1. Assume that Eq. (18) holds for all non-negative integers less than p. First, we rewrite Eq. (8) in Theorem 1 as

F_{k,n} = n^{p} + k ∑_{i=1}^{n} i^{p}F_{k,n−i} − 2 ∑_{i=1}^{n} ∑_{j=1}^{⌈ \frac{p}{2} ⌉} \binom{p}{2j − 1}i^{p−2j+1}F_{k,n−i}.

By switching the order of summation in the double summand, we get

F_{k,n} = n^{p} + k ∑_{i=1}^{n} i^{p}F_{k,n−i} − 2 ∑_{j=1}^{⌈ \frac{p}{2} ⌉} \binom{p}{2j − 1} ∑_{i=1}^{n} i^{p−2j+1}F_{k,n−i}.

Since i^{p}F_{k,n−i} = 0 for i = 0, we can start the first summand at i = 0 instead of at i = 1, and thus we obtain

F_{k,n} = n^{p} + k ∑_{i=0}^{n} i^{p}F_{k,n−i} − 2 ∑_{j=1}^{⌈ \frac{p}{2} ⌉} \binom{p}{2j − 1} ∑_{i=1}^{n} i^{p−2j+1}F_{k,n−i}.  (19)

If p is even, then Eq. (19) can be put in the form

F_{k,n} = n^{p} + k ∑_{i=0}^{n} i^{p}F_{k,n−i} − 2 ∑_{j=1}^{⌈ \frac{p}{2} ⌉} \binom{p}{2j − 1} ∑_{i=0}^{n} i^{p−2j+1}F_{k,n−i},  (20)

because the term corresponding to i = 0 in the last summand is 0^{p−2j+1}F_{k,n} = 0 for all j = 1, 2, …, p/2. On the other hand, if p is odd, then Eq. (19) can be written as

F_{k,n} = n^{p} + k ∑_{i=0}^{n} i^{p}F_{k,n−i} − 2 ∑_{j=1}^{⌈ \frac{p}{2} ⌉} \binom{p}{2j − 1} ∑_{i=0}^{n} i^{p−2j+1}F_{k,n−i} + 2F_{k,n},

because the term corresponding to i = 0 in the last summation is 0^{p−2j+1}F_{k,n} = 0 for j ≠ (p + 1)/2 and F_{k,n} for j = (p + 1)/2. Therefore, if p is odd, we have

−F_{k,n} = n^{p} + k ∑_{i=0}^{n} i^{p}F_{k,n−i} − 2 ∑_{j=1}^{⌈ \frac{p}{2} ⌉} \binom{p}{2j − 1} ∑_{i=0}^{n} i^{p−2j+1}F_{k,n−i}.  (21)

Thus, from Eqs. (20) and (21), we conclude that

(−1)^{p}F_{k,n} = n^{p} + k · T_{k,n}^{(p)} − 2 ∑_{j=1}^{⌈ \frac{p}{2} ⌉} \binom{p}{2j − 1}T_{k,n}^{(p−2j+1)},

k · T_{k,n}^{(p)} = (−1)^{p}F_{k,n} − n^{p} + \frac{2}{k} ∑_{j=1}^{⌈ \frac{p}{2} ⌉} \binom{p}{2j − 1}k · T_{k,n}^{(p−2j+1)}.  (22)

Now, by induction hypothesis, for 1 ≤ j ≤ ⌈\frac{p}{2}⌉, we have

k · T_{k,n}^{(p−2j+1)} =A_{k}^{(p−2j+1)}F_{k,n+1} + ∑_{r=0}^{p−2j+1} A_{k}^{(r)}\binom{p − 2j + 1}{r}((−1)^{p−2j+1−r}F_{k,n} − n^{p−2j+1−r}).

Substituting this in Eq. (22), we get

\begin{aligned}k · T_{k,n}^{(p)} & = (−1)^{p}F_{k,n} − n^{p} + \frac{2}{k} ∑_{j=1}^{⌈ \frac{p}{2} ⌉} \binom{p}{2j − 1}A_{k}^{(p−2j+1)}F_{k,n+1} \\ & + \frac{2}{k} ∑_{j=1}^{⌈ \frac{p}{2} ⌉} \binom{p}{2j − 1} ∑_{r=0}^{p−2j+1} A_{k}^{(r)}\binom{p − 2j + 1}{r}((−1)^{p−2j+1−r}F_{k,n} − n^{p−2j+1−r}). \end{aligned}

Using the Definition 2, this can be written as

\begin{aligned}k · T_{k,n}^{(p)} & = (−1)^{p}F_{k,n} − n^{p} + A_{k}^{(p)}F_{k,n+1} \\ & + \frac{2}{k} ∑_{j=1}^{⌈ \frac{p}{2} ⌉} ∑_{r=0}^{p−2j+1} A_{k}^{(r)}\binom{p}{2j − 1}\binom{p − 2j + 1}{r}((−1)^{p−2j+1−r}F_{k,n} − n^{p−2j+1−r}). \end{aligned}

If we execute the change of variable r′ = 2j + r − 1, this becomes

\begin{aligned}k · T_{k,n}^{(p)} & = (−1)^{p}F_{k,n} − n^{p} + A_{k}^{(p)}F_{k,n+1} \\ & + \frac{2}{k} ∑_{j=1}^{⌈ \frac{p}{2} ⌉} ∑_{r′=2j−1}^{p} A_{k}^{(r′−2j+1)}\binom{p}{2j − 1}\binom{p − 2j + 1}{r′ − 2j + 1}((−1)^{p−r′}F_{k,n} − n^{p−r′}). \end{aligned}

Now, by switching the order of summation, we obtain

\begin{aligned}k · T_{k,n}^{(p)} & = (−1)^{p}F_{k,n} − n^{p} + A_{k}^{(p)}F_{k,n+1} \\ & + \frac{2}{k} ∑_{r′=1}^{p} ∑_{j=1}^{⌈ \frac{r′}{2} ⌉} A_{k}^{(r′−2j+1)}\binom{p}{2j − 1}\binom{p − 2j + 1}{r′ − 2j + 1}((−1)^{p−r′}F_{k,n} − n^{p−r′}). \end{aligned}

Using the well-known binomial identity \binom{n}{r} = \binom{n}{n−r}, we rewrite this as

\begin{aligned}k · T_{k,n}^{(p)} & = (−1)^{p}F_{k,n} − n^{p} + A_{k}^{(p)}F_{k,n+1} \\ & + \frac{2}{k} ∑_{r′=1}^{p} ∑_{j=1}^{⌈ \frac{r′}{2} ⌉} A_{k}^{(r′−2j+1)}\binom{p}{p − 2j + 1}\binom{p − 2j + 1}{p − r′}((−1)^{p−r′}F_{k,n} − n^{p−r′}). \end{aligned} (23)

Next, using the binomial identity [2, Identity 134, p. 67] \binom{p}{q}\binom{q}{r} = \binom{p}{r}\binom{p−r}{p−q}, we have

\binom{p}{p − 2j − 1}\binom{p − 2j + 1}{p − r′} = \binom{p}{p − r′}\binom{r′}{2j − 1} = \binom{p}{r′}\binom{r′}{2j − 1}.

When we substitute this in Eq. (23), we obtain

k · T_{k,n}^{(p)} = (−1)^{p}F_{k,n} − n^{p} + A_{k}^{(p)}F_{k,n+1}

+ \frac{2}{k} ∑_{r′=1}^{p} ∑_{j=1}^{⌈ \frac{r′}{2} ⌉} A_{k}^{(r′−2j+1)}\binom{p}{r′}\binom{r′}{2j − 1}((−1)^{p−r′}F_{n,k} − n^{p−r′})

= (−1)^{p}F_{k,n} − n^{p} + A_{k}^{(p)}F_{k,n+1}

+ ∑_{r′=1}^{p} (\frac{2}{k} ∑_{j=1}^{⌈ \frac{r′}{2} ⌉} \binom{r′}{2j − 1}A_{k}^{(r′−2j+1)})\binom{p}{r′}((−1)^{p−r′}F_{n,k} − n^{p−r′}).

Now, using Definition 2, this becomes

k · T_{k,n}^{(p)} = A_{k}^{(p)}F_{k,n+1} + ((−1)^{p}F_{k,n} − n^{p}) + ∑_{r′=1}^{p} A_{k}^{(r′)}\binom{p}{r′}((−1)^{p−r′}F_{n,k} − n^{p−r′})).

Since A_{k}^{(r′)} = 1 at r′ = 0, we conclude that

k · T_{k,n}^{(p)} = A_{k}^{(p)}F_{k,n+1} + ∑_{r′=0}^{p} A_{k}^{(r′)}\binom{p}{r′}((−1)^{p−r′}F_{n,k} − n^{p−r′}).

Hence, by induction, Eq. (18) holds for all p ≥ 0. This completes the proof.

4 The Brousseau sums

Let us begin this section with finding the identities about the Brousseau sums ∑_{i=0}^{n} i^{p}F_{k,i} for p = 1, 2, 3, . . .. Consider the case p = 1.

k · ∑_{i=0}^{n} iF_{k,i} = k · ∑_{i=0}^{n}(n − i)F_{k,n−i}

= nk · ∑_{i=0}^{n} F_{k,n−i} − k · ∑_{i=0}^{n} iF_{k,n−i}

= n(F_{k,n+1} + F_{k,n} − 1) − (\frac{2}{k}F_{k,n+1} + \frac{−k + 2}{k}F_{k,n} − \frac{kn + 2}{k})

= \frac{1}{k}((kn − 2)F_{k,n+1} + (kn + k − 2)F_{k,n} + 2).

Proceeding like this, we get the following set of identities:

\begin{aligned}k · ∑_{i=0}^{n} F_{k,i} & = F_{k,n+1} + F_{k,n} − 1, \\ k · ∑_{i=0}^{n} iF_{k,i} & = \frac{1}{k}((kn − 2)F_{k,n+1} + (kn + k − 2)F_{k,n} + 2) \\ & = \frac{1}{k}((kn − 2)F_{k,n+1} + (k(n + 1) − 2)F_{k,n} + 2), \\ k · ∑_{i=0}^{n} i^{2}F_{k,i} & = \frac{1}{k^{2}} ((k^{2}n^{2} − 4kn + 8)F_{k,n+1} + (k^{2}n^{2} + 2k(k − 2)n + k^{2} − 4k + 8)F_{k,n} − 8) \\ & = \frac{1}{k^{2}} ((k^{2}n^{2} − 4kn + 8)F_{k,n+1} + (k^{2}(n + 1)^{2} − 4k(n + 1) + 8)F_{k,n} − 8), \\ k · ∑_{i=0}^{n} i^{3}F_{k,i} & = \frac{1}{k^{3}} ((k^{3}n^{3} − 6k^{2}n^{2} + 24kn − 2k^{2} − 48)F_{k,n+1} \\ & + (k^{3}n^{3} + 3k^{2}(k − 2)n^{2} + 3k(k^{2} − 4k + 8)n + k^{3} − 8k^{2} + 24k − 48)F_{k,n} \\ & + 2k^{2} + 48) \\ & = \frac{1}{k^{3}} ((k^{3}n^{3} − 6k^{2}n^{2} + 24kn − 2k^{2} − 48)F_{k,n+1} \\ & + (k^{3}(n + 1)^{3} − 6k^{2}(n + 1)^{2} + 24k(n + 1) − 2k^{2} − 48)F_{k,n} + 2k^{2} + 48), \end{aligned} ; (24)

and so on. As we expected, these equations also follow a pattern. If we define the sums S_{k,n}^{(p)} as

S_{k,n}^{(p)} = \begin{cases}∑_{i=0}^{n} F_{k,i}, & if p = 0; \\ ∑_{i=0}^{n} i^{p}F_{k,i}, & if p ≥ 1, \end{cases}  (25)

then each equation is of the form

k · S_{k,n}^{(p)} = C_{k}^{(p)}(n)F_{k,n+1} + C_{k}^{(p)}(n + 1)F_{k,n} − C_{k}^{(p)}(0),

where C_{k}^{(p)}(n) is a polynomial in n of degree p. Let us try to investigate the rule of formation of the coefficients of this polynomial, C_{k}^{(p)}(n). From the last equation in Eq. (24), we identify that

C_{k}^{(3)}(n) = \frac{1}{k^{3}} (k^{3}n^{3} − 6k^{2}n^{2} + 24kn − 2k^{2} − 48)

= n^{3} − (\frac{6}{k})n^{2} + (\frac{24}{k^{2}})n − \frac{2k^{2} + 48}{k^{3}}

= 1 · \binom{3}{0}n^{3} − \frac{2}{k} · \binom{3}{1}n^{2} + \frac{8}{k^{2}} · \binom{3}{2}n − \frac{2k^{2} + 48}{k^{3}} · \binom{3}{3},

where the numbers 1, 2/k, 8/k^{2}, (2k^{2} + 48)/k^{3} are the first four terms of the sequence (A_{k}^{(p)})_{p≥0}. With this in mind, we define the coefficient polynomial, C_{k}^{(p)}(n), in n of degree p as follows: Definition 5. For all integers p ≥ 0, we define

C_{k}^{(p)}(n) = \begin{cases}(−1)^{p}A(p), & if n = 0; \\ ∑_{r=0}^{p}(−1)^{r}A_{k}^{(r)}\binom{p}{r}n^{p−r}, & if n ≠ 0. \end{cases}  (26)

It should be noted that, for k > 2, the polynomial C_{k}^{(p)}(n) generally doesn’t have integer coefficients. From Eqs. (17) and (26), it is clear that Φ_{k}^{(p)}(n) = (−1)^{p}C_{k}^{(p)}(−n). Consequently, we may rewrite Eq. (15) as

k · T_{k,n}^{(p)} = (−1)^{p}(C_{k}^{(p)}(0)F_{k,n+1} + C_{k}^{(p)}(1)F_{k,n} − C_{k}^{(p)}(−n)).

The central result (7) about the Brousseau sums of the k-Fibonacci numbers can now be established. Theorem 6. If S_{k,n}^{(p)} is defined as in Eq. (25), then for all p ≥ 0, we have

k · S_{k,n}^{(p)} = C_{k}^{(p)}(n)F_{k,n+1} + C_{k}^{(p)}(n + 1)F_{k,n} − C_{k}^{(p)}(0),  (27)

where C_{k}^{(p)}(n) is the “coefficient polynomial” as defined in Eq. (26). Proof. If p = 0, then the left-hand side of Eq. (27) is

k · S_{k,n}^{(p)} = k · ∑_{i=0}^{n} F_{k,i} = F_{k,n+1} + F_{k,n} − 1,

and the right-hand side is

\begin{aligned}C_{k}^{(0)}(n)F_{k,n+1} + C_{k}^{(0)}(n + 1)F_{k,n} − C_{k}^{(0)}(0) & = A_{k}^{(0)}F_{k,n+1} + A_{k}^{(0)}F_{k,n} − 1 \\ & = F_{k,n+1} + F_{k,n} − 1. \end{aligned}

Thus, Eq. (27) holds for p = 0. Now fix p ≥ 1. Then, using the binomial expansion, we have

S_{k,n}^{(p)} = ∑_{i=0}^{n} i^{p}F_{k,i}

= ∑_{i=0}^{n}(n − i)^{p}F_{k,n−i}

= ∑_{i=0}^{n} ( ∑_{r=0}^{p} \binom{p}{r}n^{p−r}(−i)^{r})F_{k,n−i}

= ∑_{r=0}^{p}(−1)^{r}\binom{p}{r}n^{p−r}( ∑_{i=0}^{n} i^{r}F_{k,n−i})

= ∑_{r=0}^{p}(−1)^{r}\binom{p}{r}n^{p−r}T_{k,n}^{(r)}.

Thus,

k · S_{k,n}^{(p)} = ∑_{r=0}^{p}(−1)^{r}\binom{p}{r}n^{p−r}(k · T_{k,n}^{(r)}).  (28)

Now, using Theorem 4, we have

k · T_{k,n}^{(r)} = A_{k}^{(r)}F_{k,n+1} + ∑_{j=0}^{r} A_{k}^{(j)}\binom{r}{j}((−1)^{r−j}F_{k,n} − n^{r−j}).

When we substitute this in Equation (28), we obtain

k · S_{k,n}^{(p)} = ∑_{r=0}^{p}(−1)^{r}\binom{p}{r}n^{p−r}(A_{k}^{(r)}F_{k,n+1} + ∑_{j=0}^{r} A_{k}^{(j)}\binom{r}{j}((−1)^{r−j}F_{k,n} − n^{r−j}))

= ( ∑_{r=0}^{p}(−1)^{r}A_{k}^{(r)}\binom{p}{r}n^{p−r})F_{k,n+1}

+ ∑_{r=0}^{p} ∑_{j=0}^{r}(−1)^{r}A_{k}^{(j)}\binom{p}{r}\binom{r}{j}((−1)^{r−j}F_{k,n} − n^{r−j})n^{p−r}

= C_{k}^{(p)}(n)F_{k,n+1} + ∑_{r=0}^{p} ∑_{j=0}^{r}(−1)^{r}A_{k}^{(j)}\binom{p}{r}\binom{r}{j}((−1)^{r−j}F_{k,n} − n^{r−j})n^{p−r}.

By switching the order of summation, this becomes

k · S_{k,n}^{(p)} = C_{k}^{(p)}(n)F_{k,n+1} + ∑_{j=0}^{p} ∑_{r=j}^{p}(−1)^{j}A_{k}^{(j)}\binom{p}{r}\binom{r}{j}n^{p−r}F_{k,n}

− ∑_{j=0}^{p} ∑_{r=j}^{p}(−1)^{r}A_{k}^{(j)}\binom{p}{r}\binom{r}{j}n^{p−j}

= C_{k}^{(p)}(n)F_{k,n+1} + ∑_{j=0}^{p}(−1)^{j}A_{k}^{(j)}( ∑_{r=j}^{p} \binom{p}{r}\binom{r}{j}n^{p−r})F_{k,n}

− ∑_{j=0}^{p} A_{k}^{(j)}n^{p−j}( ∑_{r=j}^{p} \binom{p}{r}\binom{r}{j}(−1)^{r})

= C_{k}^{(p)}(n)F_{k,n+1} + ( ∑_{j=0}^{p}(−1)^{j}A_{k}^{(j)}\binom{p}{j}(n + 1)^{p−j})F_{k,n} − (−1)^{p}A_{k}^{(p)},

where the last equality follows from the binomial identities

∑_{r=j}^{p} \binom{p}{r}\binom{r}{j}n^{p−r} = \binom{p}{j}(n + 1)^{p−j},

and

∑_{r=j}^{p} \binom{p}{r}\binom{r}{j}(−1)^{r} = \begin{cases}0, & if j ≠ p; \\ (−1)^{p}, & if j = p, \end{cases}

from Gould’s collection [10, Identities 3.118, 3.119, p. 36]. Thus, we conclude that k · S_{k,n}^{(p)} = C_{k}^{(p)}(n)F_{k,n+1} + C_{k}^{(p)}(n + 1)F_{k,n} − C_{k}^{(p)}(0).

Example 7. Setting p = 2 in Eq. (27) we get

k · ∑_{i=1}^{n} i^{2}F_{k,i} = (n^{2} − \frac{4}{k}n + \frac{8}{k^{2}} )F_{k,n+1} + ((n + 1)^{2} − \frac{4}{k}(n + 1) + \frac{8}{k^{2}} )F_{k,n} − \frac{8}{k^{2}} .

In particular, when k = 11 this becomes

11 · ∑_{i=1}^{n} i^{2}F_{11,i} = (n^{2} − \frac{4}{11}n + \frac{8}{121})F_{11,n+1} + ((n + 1)^{2} − \frac{4}{11}(n + 1) + \frac{8}{121})F_{11,n} − \frac{8}{121}.

Since F_{11,i} = F_{5i}/F_{5} = F_{5i}/5 (see [9]), we obtain

11 · ∑_{i=1}^{n} i^{2}F_{5i} = (n^{2} − \frac{4}{11}n + \frac{8}{121})F_{5n+5} + ((n + 1)^{2} − \frac{4}{11}(n + 1) + \frac{8}{121})F_{5n} − \frac{5 · 8}{121},

which gives the identity about the Brousseau sums of the sequence (F_{5i})_{i≥1}. Example 8. Setting p = 2 and k = 14 in Eq. (27) we get

14 · ∑_{i=1}^{n} i^{2}F_{14,i} = (n^{2} − \frac{2}{7}n + \frac{2}{49})F_{14,n+1} + ((n + 1)^{2} − \frac{2}{7}(n + 1) + \frac{2}{49})F_{14,n} − \frac{2}{49}.

Since F_{14,i} = P_{3i}/P_{3} = P_{3i}/5 (see [9]), we obtain

14 · ∑_{i=1}^{n} i^{2}P_{3i} = (n^{2} − \frac{2}{7}n + \frac{2}{49})P_{3n+3} + ((n + 1)^{2} − \frac{2}{7}(n + 1) + \frac{2}{49})P_{3n} − \frac{5 · 2}{49},

which gives the identity about the Brousseau sums of the sequence (P_{3i})_{i≥1}.

Examples 7 and 8 suggest two interesting identities Eqs. (29) and (31). Eq. (29) is about the Brousseau sums of every m^{th} Fibonacci number, and Eq. (31) is that of every m^{th} Pell number, when m is odd.

Corollary 9. Let m ≥ 1 be an odd integer. Then for all integers p ≥ 0, the following identity holds:

L_{m} · ∑_{i=1}^{n} i^{p}F_{mi} = C_{L_{m}}^{(p)}(n)F_{m(n+1)} + C_{L_{m}}^{(p)}(n + 1)F_{mn} − C_{L_{m}}^{(p)}(0)F_{m},  (29)

where L_{m} is the m^{th} Lucas number. Proof. Setting k = L_{m} in Eq. (27) yields

L_{m} · ∑_{i=1}^{n} i^{p}F_{L_{m},i} = C_{L_{m}}^{(p)}(n)F_{L_{m},n+1} + C_{L_{m}}^{(p)}(n + 1)F_{L_{m},n} − C_{L_{m}}^{(p)}(0).  (30)

If m is odd, then we have (see [9])

F_{L_{m},i} = \frac{F_{mi}}{F_{m}} .

Applying this in Eq. (30) and multiplying through by F_{m}, we get Eq. (29). Corollary 10. Let m ≥ 1 be an odd number. Then for all integers p ≥ 0, the following identity holds:

Q_{m} · ∑_{i=1}^{n} i^{p}P_{mi} = C_{Q_{m}}^{(p)}(n)P_{m(n+1)} + C_{Q_{m}}^{(p)}(n + 1)P_{mn} − C_{Q_{m}}^{(p)}(0)P_{m},  (31)

where Q_{m} is the m^{th} Pell-Lucas number. Proof. The proof is similar to the proof of Corollary (9) by using the fact that (see [9])

F_{Q_{m},i} = \frac{P_{mi}}{P_{m}} ,

when m is odd.

5 Shifted Brousseau sums

In this section, we find the formula for the shifted Brousseau sums

∑_{i=1}^{n} i^{p}F_{k,m+i},

for all integers m, p ≥ 0. For example, if we take p = 1, then

k · ∑_{i=1}^{n} iF_{k,m+i} = k · ∑_{i=m+1}^{m+n}(i − m)F_{k,i}

= k · ∑_{i=m+1}^{m+n} iF_{k,i} − mk · ∑_{i=m+1}^{m+n} F_{k,i}

= k( ∑_{i=0}^{m+n} iF_{k,i} − ∑_{i=0}^{m} iF_{k,i}) − mk( ∑_{i=0}^{m+n} F_{k,i} − ∑_{i=0}^{m} F_{k,i}).

Now, using the first two identities in Eq. (24), we have

k · ∑_{i=1}^{n} iF_{k,m+i} = \frac{1}{k}((k(m + n) − 2)F_{k,m+n+1} + (k(m + n + 1) − 2)F_{k,m+n} − (km − 2)F_{k,m+1}

− (k(m + 1) − 2)F_{k,m}) − m(F_{k,m+n+1} + F_{k,m+n} − F_{k,m+1} − F_{k,m})

= \frac{1}{k}((kn − 2)F_{k,m+n+1} + (k(n + 1) − 2)F_{k,m+n} + 2F_{k,m+1} + (−k + 2)F_{k,m}).

We generalize this identity for all integers p ≥ 0 in the next theorem. Theorem 11. For all integers m, p ≥ 0, we have

k · ∑_{i=1}^{n} i^{p}F_{k,m+i} = C_{k}^{(p)}(n)F_{k,m+n+1} + C_{k}^{(p)}(n + 1)F_{k,m+n} − C_{k}^{(p)}(0)F_{k,m+1} − C_{k}^{(p)}(1)F_{k,m}. (32)

Proof. Since F_{k,0} = 0 and F_{k,1} = 1, the case m = 0 follows from Theorem 6. Now fix m ≥ 1. Then, using the binomial expansion, we have

∑_{i=1}^{n} i^{p}F_{k,m+i} = ∑_{i=m+1}^{m+n}(i − m)^{p}F_{k,i}

= ∑_{i=m+1}^{m+n} ∑_{j=0}^{p} \binom{p}{j}i^{p−j}(−m)^{j}F_{k,i}

= ∑_{j=0}^{p} \binom{p}{j}(−m)^{j} ∑_{i=m+1}^{m+n} i^{p−j}F_{k,i}

= ∑_{j=0}^{p} \binom{p}{j}(−m)^{j}(S_{k,m+n}^{(p−j)} − S_{k,m}^{(p−j)}).

Thus,

k · ∑_{i=1}^{n} i^{p}F_{k,m+i} = ∑_{j=0}^{p} \binom{p}{j}(−m)^{j}(k · S_{k,m+n}^{(p−j)} − k · S_{k,m}^{(p−j)}).

Now, applying Theorem 6, this becomes

\begin{aligned}k · ∑_{i=1}^{n} i^{p}F_{k,m+i} & = ∑_{j=0}^{p} \binom{p}{j}(−m)^{j}(C_{k}^{(p−j)}(m + n)F_{k,m+n+1} + C_{k}^{(p−j)}(m + n + 1)F_{k,m+n} \\ & − C_{k}^{(p−j)}(m)F_{k,m+1} − C_{k}^{(p−j)}(m + 1)F_{k,m}) \end{aligned} (33)

Consider

∑_{j=0}^{p} \binom{p}{j}(−m)^{j}C_{k}^{(p−j)}(m + n) = ∑_{j=0}^{p} \binom{p}{j}(−m)^{j} ∑_{r=0}^{p−j}(−1)^{r}A_{k}^{(r)}\binom{p − j}{r}(m + n)^{p−j−r}.

By switching the order of summation, this becomes

∑_{j=0}^{p} \binom{p}{j}(−m)^{j}C_{k}^{(p−j)}(m + n) = ∑_{r=0}^{p} ∑_{j=0}^{p−r}(−1)^{r}A_{k}^{(r)}\binom{p}{j}\binom{p − j}{r}(m + n)^{p−j−r}(−m)^{j}. (34)

Next, we use the binomial identity [2, Identity 134, p. 67] to get

\binom{p}{j}\binom{p − j}{r} = \binom{p}{p − j}\binom{p − j}{r} = \binom{p}{r}\binom{p − r}{j}.

Substituting this in Eq. (34), we obtain

\begin{aligned}∑_{j=0}^{p} \binom{p}{j}(−m)^{j}C_{k}^{(p−j)}(m + n) & = ∑_{r=0}^{p}(−1)^{r}A_{k}^{(r)}\binom{p}{r}( ∑_{j=0}^{p−r} \binom{p − r}{j}(m + n)^{p−r−j}(−m)^{j}) \\ & = ∑_{r=0}^{p}(−1)^{r}A_{k}^{(r)}\binom{p}{r}n^{p−r} \\ & = C_{k}^{(p)}(n). \end{aligned} (35)

Similarly, we can show that

∑_{j=0}^{p} \binom{p}{j}(−m)^{j}C_{k}^{(p−j)}(m + n + 1) = C_{k}^{(p)}(n + 1),  (36)
∑_{j=0}^{p} \binom{p}{j}(−m)^{j}C_{k}^{(p−j)}(m) = C_{k}^{(p)}(0),  (37)

and

∑_{j=0}^{p} \binom{p}{j}(−m)^{j}C_{k}^{(p−j)}(m + 1) = C_{k}^{(p)}(1).  (38)

Thus, Eq. (32) follows by substituting Eq. (35) through Eq. (38) in Eq. (33).

6 Conclusion

While all the results presented above assume that k is a positive integer, there is no reason not to extend them to nonzero real numbers as well. The only drawback is that the numbers F_{k,n} are not necessarily integers. The k-Fibonacci numbers are just the Fibonacci polynomials F_{n}(x) (see [3]) calculated at x = k. Hence, we strongly believe that all the above results are still valid if we allow non-integer values of k. For example, we can have the identity

\sqrt{2} · ∑_{i=1}^{n} i^{2}F_{i}(\sqrt{2}) = (n^{2} − 2\sqrt{2}n + 4)F_{n+1}(\sqrt{2}) + ((n + 1)^{2} − 2\sqrt{2}(n + 1) + 4)F_{n}(\sqrt{2}) − 4.

7 Acknowledgment

The authors would like to thank the anonymous referee for valuable comments and suggestions for improving the quality of the original version of this manuscript.

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[10] H. W. Gould, Combinatorial Identities, Revised edition published by the author, 1972.

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2020 Mathematics Subject Classification: Primary 11B39; Secondary 11B37, 11B65, 11B83. Keywords: Brousseau sum, Fibonacci number, Fibonacci polynomial, k−Fibonacci number, binomial coefficient, metallic number.

(Concerned with sequences \underline{A000032}, \underline{A000045}, \underline{A000129}, \underline{A000557}, \underline{A001076}, \underline{A002203}, \underline{A002878}, \underline{A006154}, \underline{A006190} \underline{A077444}, and \underline{A259546}.)

Received October 11 2023; revised versions received February 23 2024; July 17 2024. Published in Journal of Integer Sequences, July 17 2024.

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