Further Extensions of Sury’s Identity

Gregory Dresden Washington & Lee University Lexington, VA 24450 USA

                   dresdeng@wlu.edu

Xiaoya Gao Ürümqi Bayi High School Ürümqi, Xinjiang China

                   lexie131@163.com

Abstract

The equation commonly known as Sury’s identity is a deceptively simple summation formula that connects the Lucas numbers, Fibonacci numbers, and powers of two. Many authors have given extensions and generalizations over the years; in this paper, we take a different approach that allows us to produce a good number of new summation formulas, all from elementary (but non-trivial) methods.

1 Introduction

The identity

2^{n+1}F_{n+1} = ∑_{i=0}^{n} 2^{i}L_{i}  (1)

which connects the Fibonacci numbers, the Lucas numbers, and the powers of 2, is often called Sury’s identity thanks to his deceptively simple proof [23] that takes the sum of two geometric series. Fifteen years before Sury’s proof, Benjamin and Quinn [5] proved the same identity by counting the tilings of a bracelet with squares and dominoes.

Of course, there are many variations on this identity. Two years after Sury’s proof, Martinjak [18] used a telescoping sum to prove that

(−1/2)^{n}F_{n+1} = ∑_{i=0}^{n}(−1/2)^{i}L_{i+1},  (2)

and if we multiply both sides by (−2)^{n} then this becomes the convolution identity

F_{n+1} = ∑_{i=0}^{n}(−2)^{n−i}L_{i+1}.  (3)

Marques [17] used a quick induction argument to produce a slightly more complicated version of equation (1) with 3^{i} instead of 2^{i}, giving us

3^{n+1}F_{n+1} = ∑_{i=0}^{n} 3^{i}(L_{i} + F_{i+1}),  (4)

and a clever rearrangement of sums by Edgar [12] produced the more general equation

t^{n+1}F_{n+1} = ∑_{i=0}^{n} t^{i}(L_{i} + (t − 2)F_{i+1}).  (5)

We note that Edgar’s equation (5) will easily give us Sury’s equation (1) when t = 2, and Marques’ equation (4) when t = 3. It is a rather pleasant challenge to show that it also gives Martinjak’s equation (3) when t = −1/2. For two different methods of proving equation (5), with colorful tilings and with generating functions, see Martinjack and Prodinger [19].

Dafnis [8] used a telescoping sum to give a variation on Edgar’s equation (5), which in our notation would be

(−1)^{n}F_{n+1} = ∑_{i=0}^{n}(−1)^{i}t^{n−i}(L_{i+1} + (t − 2)F_{i}).  (6)

Dafnis went on to prove a similar identity involving the tribonacci numbers.

Abd-Elhameed and Zeyada took Edgar’s equation (5) even further by taking up generalized Fibonacci and Lucas sequences, defined as

U_{n+2}^{a,b} = aU_{n+1}^{a,b} + bU_{n}^{a,b} with U_{0}^{a,b} = 0, U_{1}^{a,b} = 1,
V_{n+2}^{a,b} = aV_{n+1}^{a,b} + bV_{n}^{a,b} with V_{0}^{a,b} = 2, V_{1}^{a,b} = a.

(7)

Their version of Sury’s identity (for a, b, t all non-zero) is

t^{n+1}U_{n+1}^{a,b} = \frac{1}{a} ∑_{i=0}^{n} t^{i}(V_{i}^{a,b} + (at − 2)U_{i+1}^{a,b}),  (8)

and we can clearly see how this is a generalization of Edgar’s equation (5). Bhatnagar considered even more geneal sequences by replacing the constants a, b in equation (7) with sequences a_{n}, b_{n}. For other approaches, see recent papers by Adegoke and Frontczak [2], Chung, Yao, and Zhou [7], Dafnis, Philippou, and Livieris [9], De Micheli Vitturi [10], Kuhapatanakul and Thongsing [15], Kwong [16], Melham [20], and Philippou and Dafnis [21].

Despite all this, we have discovered a new way to generalize Sury’s identity. Every other approach has tried to replace the 2^{i} in Sury’s equation (1) with something like the 3^{i} in equation (4) and then the t^{i} as seen in equation (5), which leads to an adjustment in the other term in the summation, from the simple L_{i} in equation (1) to the slightly more complex L_{i}+F_{i+1} in equation (4) to the rather complicated L_{i}+(t−2)F_{i+1} in equation (5), culminating in the general expression V_{i}^{a,b} + (at − 2)U_{i+1}^{a,b} from Abd-Elhameed and Zeyada’s equation (8). By contrast, our approach allows us to select the sequence (the Fibonacci numbers, or the Pell numbers, or really any linear recurrence sequence of order 2) inside the summation, and also the initial position, and then produce the particular “t” to make the formula work out nicely.

This is best explained by a demonstration. Recall that Sury and Martinjak began their sums in equations (1) and (2) with L_{0} and L_{1} respectively, but with our approach we can begin with L_{2} or L_{3} or even L_{4}, giving us the new formulas

F_{n+1} = ∑_{i=0}^{n}(−2)^{n−i}L_{i+1} = \frac{1}{3} ∑_{i=0}^{n}(−1/3)^{n−i}L_{i+2}

= \frac{1}{4} ∑_{i=0}^{n}(−3/4)^{n−i}L_{i+3}

= \frac{1}{7} ∑_{i=0}^{n}(−4/7)^{n−i}L_{i+4}.

However, our personal favorites are these generalizations of Sury’s formula (and again, these seem to be new),

\begin{aligned}∑_{i=0}^{n}(−2)^{i}L_{3i} & = 2 − (−2)^{n+1}F_{3n}, \\ 4 ∑_{i=0}^{n} 9^{i}L_{6i} & = 9^{n+1}F_{6n} + 8, \\ 17 ∑_{i=0}^{n}(−38)^{i}L_{9i} & = 34 − (−38)^{n+1}F_{9n}. \end{aligned}  (9)

These are similar (in spirit but not in fact) to the formulas we can derive from Theorem 2.1 in Adegoke and Frontczak’s recent paper [2], which give us (after some manipulation)

∑_{i=0}^{n}(1/2)^{i}L_{3i} = F_{3(n+1)}/2^{n},

4 ∑_{i=0}^{n}(1/9)^{i}L_{6i} = F_{6(n+1)}/9^{n},

17 ∑_{i=0}^{n}(1/38)^{i}L_{9i} = F_{9(n+1)}/38^{n}.

We can also produce this next collection of formulas with the Fibonacci and Lucas numbers (which, again, we believe are new):

\begin{aligned}F_{2n+2} & = 1 · ∑_{i=0}^{n} 1^{n−i}F_{2i+1} = 1 · ∑_{i=0}^{n}(−1)^{n−i}L_{2i+1}, \\ F_{3n+3} & = 2 · ∑_{i=0}^{n} 1^{n−i}F_{3i+1} = 2 · ∑_{i=0}^{n}(−3)^{n−i}L_{3i+1}, \\ F_{4n+4} & = 3 · ∑_{i=0}^{n} 2^{n−i}F_{4i+1} = 3 · ∑_{i=0}^{n}(−4)^{n−i}L_{4i+1}. \end{aligned}  (10)

Moving on to the Pell numbers (which are found at \underline{A000129} in the On-Line Encyclopedia of Integer Sequences (OEIS) [22]), we have

\begin{aligned}P_{n+1} & = \frac{1}{2} ∑_{i=0}^{n}(−1/2)^{n−i}P_{i+2} \\ & = \frac{1}{5} ∑_{i=0}^{n}(−2/5)^{n−i}P_{i+3} \\ & = \frac{1}{12} ∑_{i=0}^{n}(−5/12)^{n−i}P_{i+4}. \end{aligned}  (11)

We compare these to equation (23) in Abd-Elhameed and Zeyada’s recent paper [1], which gives us

(1/3)^{n+1}Q_{n+1} = 1 − \frac{2}{3} ∑_{i=0}^{n}(1/3)^{i}P_{i−1},  (12)

where Q_{n} represents the nth Pell-Lucas number \underline{A001333} as defined later in equation (66).

Finally, if we think of the Pell numbers (with initial values P_{0} = 0 and P_{1} = 1 and recurrence P_{n} = 2P_{n−1}+P_{n−2}) as the “silver Fibonacci numbers”, then the “bronze Fibonacci numbers” would be the sequence \underline{A006190} with same initial values but with recurrence B_{n} = 3B_{n−1} + B_{n−2}. Here are some interesting (and mostly new) formulas for these bronze Fibonacci numbers, which remind us of the similar identities in equation (10):

\begin{aligned}B_{2n+2} & = 3 · ∑_{i=0}^{n} 1^{n−i}B_{2i+1}, \\ B_{3n+3} & = 10 · ∑_{i=0}^{n} 3^{n−i}B_{3i+1}, \\ B_{4n+4} & = 33 · ∑_{i=0}^{n} 10^{n−i}B_{4i+1}. \end{aligned}  (13)

There are countless variations on these formulas. Again, what sets our approach apart from all the previous work is that our weighted sums can start at an arbitrary position in our sequence, and our weighted sums avoid the complicated summands as seen in equations (5) and (8). Our proofs rely mostly on induction, along with generalized versions of Cassini’s and d’Ocagne’s identities.

2 Main results

We begin with a theorem associated with a very particular type of sequence.

Theorem 1. Let (A_{n})_{n≥0} be a second-order linear recurrence sequence with initial term A_{0} = 1 and with recurrence

A_{n} = c_{1}A_{n−1} + c_{2}A_{n−2}.

If we set t = (c_{1} − A_{1}) then so long as t ≠ 0, we have

A_{n+2} − A_{1}A_{n+1} = (A_{2} − A_{1}^{2}) ∑_{i=0}^{n} t^{n−i}A_{i}.  (14)

Proof. We proceed by induction. For the base case n = 0, we find that equation (14) simplifies to

A_{2} − A_{1}A_{1} = (A_{2} − A_{1}^{2}) · A_{0},

and since A_{0} = 1 we are done.

Next, we assume the identity holds for some fixed n, and we try to prove it for n + 1. If we let R represent the right-hand side of equation (14) with n + 1 in place of n, we have

R = (A_{2} − A_{1}^{2}) ∑_{i=0}^{n+1} t^{n+1−i}A_{i},

and if we separate out that last term in the summation and then factor out t then we have

R = (A_{2} − A_{1}^{2}) (A_{n+1} + t ∑_{i=0}^{n} t^{n−i}A_{i}) .

We multiply through to obtain

R = (A_{2} − A_{1}^{2})A_{n+1} + t · (A_{2} − A_{1}^{2}) ∑_{i=0}^{n} t^{n−i}A_{i},

and we use our induction hypothesis on that last term to give us

R = (A_{2} − A_{1}^{2})A_{n+1} + t · (A_{n+2} − A_{1}A_{n+1}).

Recalling that t = c_{1} − A_{1}, the above expression becomes

R = (A_{2} − A_{1}^{2})A_{n+1} + (c_{1} − A_{1}) · (A_{n+2} − A_{1}A_{n+1}).

We see that the A_{1}^{2}A_{n+1} terms cancel, and so we now have

R = A_{2}A_{n+1} + (c_{1} − A_{1}) · A_{n+2} − c_{1}A_{1}A_{n+1}.  (15)

When we replace that leading A_{2} term with

A_{2} = c_{1}A_{1} + c_{2}A_{0} = c_{1}A_{1} + c_{2}1 = c_{1}A_{1} + c_{2},

then equation (15) becomes

R = (c_{1}A_{1} + c_{2})A_{n+1} + (c_{1} − A_{1}) · A_{n+2} − c_{1}A_{1}A_{n+1}.

Further simplification reduces this to

R = c_{2}A_{n+1} + (c_{1} − A_{1}) · A_{n+2}.

Since c_{1}A_{n+2} + c_{2}A_{n+1} is equal to A_{n+3}, then we have

R = A_{n+3} − A_{1}A_{n+2},

and this is exactly equal to the left-hand side of equation (14) with n + 1 in place of n, thus completing our induction step.

With all that taken care of, we are now ready for our next theorem, which covers a much more general collection of sequences.

Theorem 2. For X_{n} a second-order recurrence sequence with initial values X_{0} and X_{1} and with recurrence

X_{n} = c_{1}X_{n−1} + c_{2}X_{n−2},  (16)

then for any fixed integer k such that X_{k} and X_{k−1} are both non-zero, we have

X_{0}X_{n+2} − X_{1}X_{n+1} = \frac{X_{0}X_{2} − X_{1}^{2}}{X_{k}} ∑_{i=0}^{n} (−c_{2}X_{k−1}/X_{k})^{n−i} X_{i+k}.  (17)

We note that we allow k to be negative. Given that we have fixed initial values X_{0} and X_{1} for our sequence, then we can always define X_{−1} and X_{−2} and so on by the identity

X_{n−2} = \frac{1}{c_{2}} (X_{n} − c_{1}X_{n−1}) ,

which follows from the definition of our sequence in equation (16). Our requirement of a second-order recurrence sequence tells us that c_{2} ≠ 0, and thus X_{−1} and X_{−2} (and so on) are well-defined.

Proof of Theorem 2. For k fixed, we begin by setting

A_{n} = \frac{X_{n+k}}{X_{k}} .  (18)

This gives us the new sequence (A_{n})_{n≥0} with initial term A_{0} = 1 but with the same recurrence as the sequence (X_{n})_{n≥0} in the statement of the theorem. Next, for convenience we will set

t = c_{1} − A_{1},

which means that

t = c_{1} − X_{k+1}/X_{k} = (c_{1}X_{k} − X_{k+1})/X_{k} = −c_{2}X_{k−1}/X_{k},

and since X_{k−1} is also assumed to be non-zero then t ≠ 0. This means we can now apply Theorem 1 to the sequence (A_{n})_{n≥0}, giving us

A_{n+2} − A_{1}A_{n+1} = (A_{2} − A_{1}^{2}) ∑_{i=0}^{n} t^{n−i}A_{i},  (19)

and it remains to show that we can transform equation (19) into equation (17) in the statement of the theorem.

The first step, of course, is to use equation (18) to re-write equation (19) as

\frac{X_{n+k+2}}{X_{k}} − \frac{X_{k+1}}{X_{k}} \frac{X_{n+k+1}}{X_{k}} = (\frac{X_{k+2}}{X_{k}} − \frac{X_{k+1}^{2}}{X_{k}^{2}} ) ∑_{i=0}^{n} t^{n−i} \frac{X_{k+i}}{X_{k}} ,  (20)

and if we multiply through by X_{k}^{2} we arrive at

X_{n+k+2}X_{k} − X_{k+1}X_{n+k+1} = \frac{X_{k+2}X_{k} − X_{k+1}^{2}}{X_{k}} ∑_{i=0}^{n} t^{n−i}X_{k+i}.  (21)

This is almost, but not quite, a match for equation (17). On the left of equation (21) we will show that

X_{n+k+2}X_{k} − X_{k+1}X_{n+k+1} = (−c_{2})^{k}(X_{n+2}X_{0} − X_{n+1}X_{1}),  (22)

and on the right of equation equation (21) we will show that

X_{k+2}X_{k} − X_{k+1}^{2} = (−c_{2})^{k}(X_{2}X_{0} − X_{1}^{2}).  (23)

After canceling (−c_{2})^{k} from both sides, we will obtain a perfect match for equation (17).

So, it remains to establish the identities (22) and (23) for all integer values of k. Equation (22) is just a generalization of d’Ocagne’s identity, and it follows immediately from taking the determinant of both sides of the matrix equation

(\begin{matrix}X_{n+k+2} & X_{k+1} \\ X_{n+k+1} & X_{k} \end{matrix}) = (\begin{matrix}c_{1} & c_{2} \\ 1 & 0 \end{matrix})^{k} · (\begin{matrix}X_{n+2} & X_{1} \\ X_{n+1} & X_{0} \end{matrix}) ,

which holds for k positive, negative, or zero.

As for equation (23), this is a generalization of Cassini’s identity, and we will prove it by induction (twice), once for k > 0 and again for k < 0. First, take k to be positive. We note that equation (23) is certainly true for k = 0, and if we assume it to be true for some fixed k, then at k + 1 we can write

X_{k+3}X_{k+1} − X_{k+2}^{2} = (c_{1}X_{k+2} + c_{2}X_{k+1}) X_{k+1} − X_{k+2}^{2},

and we re-arrange the right-hand side to get

X_{k+3}X_{k+1} − X_{k+2}^{2} = c_{2}X_{k+1}^{2} + X_{k+2} (c_{1}X_{k+1} − X_{k+2}) .

Since X_{k+2} = c_{1}X_{k+1} + c_{2}X_{k},then the above equation becomes

X_{k+3}X_{k+1} − X_{k+2}^{2} = c_{2}X_{k+1}^{2} − c_{2}X_{k+2}X_{k} = (−c_{2}) (X_{k+2}X_{k} − X_{k+1}^{2}) ,

and now our induction hypothesis comes into play for that last expression on the right to give us

X_{k+3}X_{k+1} − X_{k+2}^{2} = (−c_{2})(−c_{2})^{k}(X_{2}X_{0} − X_{1}^{2}) = (−c_{2})^{k+1}(X_{2}X_{0} − X_{1}^{2}),

thus establishing equation (23) for k positive.

Finally, take k to be negative. We again note that equation (23) is certainly true for k = 0, and if we assume it to be true for some fixed k, then at k − 1 we can write

X_{k+1}X_{k−1} − X_{k}^{2} = X_{k+1} (\frac{1}{c_{2}} X_{k+1} − \frac{c_{1}}{c_{2}} X_{k}) − X_{k}^{2},

and we re-arrange the right-hand side to get

X_{k+1}X_{k−1} − X_{k}^{2} = \frac{1}{c_{2}} X_{k+1}^{2} − \frac{X_{k}}{c_{2}} (c_{1}X_{k+1} + c_{2}X_{k}).

Since X_{k+2} = c_{1}X_{k+1} + c_{2}X_{k},then the above equation becomes

X_{k+1}X_{k−1} − X_{k}^{2} = \frac{1}{c_{2}} X_{k+1}^{2} − \frac{X_{k}}{c_{2}} (X_{k+2}) = (−1/c_{2}) (X_{k+2}X_{k} − X_{k+1}^{2}) ,

and now our induction hypothesis comes into play for that last expression on the right to give us

X_{k+1}X_{k−1} − X_{k}^{2} = (−1/c_{2})(−c_{2})^{k}(X_{2}X_{0} − X_{1}^{2}) = (−c_{2})^{k−1}(X_{2}X_{0} − X_{1}^{2}),

thus establishing equation (23) for k negative.

3 Examples

We now show how to use Theorems 1 and 2 to produce many of the identities from the Introduction.

Here is an overview of our approach: we begin with a particular sequence (say, the Lucas numbers or the Pell numbers), and then we try out different variations and offsets so that we can apply one of our two theorems. Once we have obtained a generic summation identity from our theorem, we can then attempt various simplifications in order to arrive at an aesthetically-pleasing formula. For example, we might start with the Lucas numbers L_{n}, but then decide to apply our theorems with either L_{n−1} (giving us Sury’s identity in section 3.1 below) or L_{n} (as seen in section 3.2, with various offsets) or L_{jn} (as done in Theorem 4) or even L_{jn+k}/L_{k} (which we use to give us equation (54) in Theorem 5).

3.1 Sury’s identity

To prove that

2^{n+1}F_{n+1} = ∑_{i=0}^{n} 2^{i}L_{i},

we will use Theorem 2 with X_{n} = L_{n−1}, which gives us X_{0} = −1, X_{1} = 2, and c_{1} = c_{2} = 1. Taking k = 1 in Theorem 2, we get

−L_{n+1} − 2L_{n} = \frac{−1 − 2^{2}}{2} ∑_{i=0}^{n} (1/2)^{n−i} L_{i}.  (24)

After cancelling the negatives, and re-writing L_{n+1} + 2L_{n} = L_{n+2} + L_{n} = 5F_{n+1}, we have

5F_{n+1} = \frac{5}{2} ∑_{i=0}^{n} (1/2)^{n−i} L_{i}.  (25)

Finally, we cancel the 5’s and we multiply both sides by 2^{n+1}, to give us our desired formula,

2^{n+1}F_{n+1} = ∑_{i=0}^{n} 2^{i}L_{i}.

3.2 Martinjak’s convolution and its generalizations

From equation (3) in the introduction we have

F_{n+1} = ∑_{i=0}^{n}(−2)^{n−i}L_{i+1},

which is part of the family of identities

F_{n+1} = ∑_{i=0}^{n}(−2)^{n−i}L_{i+1} = \frac{1}{3} ∑_{i=0}^{n}(−1/3)^{n−i}L_{i+2}

= \frac{1}{4} ∑_{i=0}^{n}(−3/4)^{n−i}L_{i+3}

= \frac{1}{7} ∑_{i=0}^{n}(−4/7)^{n−i}L_{i+4}

from the introduction. The general form is

F_{n+1} = \frac{1}{L_{k}} ∑_{i=0}^{n}(−L_{k−1}/L_{k})^{n−i}L_{i+k},  (26)

and to produce all of these from Theorem 2 we take X_{n} = L_{n} so that we have X_{0} = 2, X_{1} = 1, and c_{1} = c_{2} = 1. As a result, equation (17) in Theorem 2 becomes

2L_{n+2} − L_{n+1} = \frac{2 · 3 − 1^{2}}{L_{k}} ∑_{i=0}^{n} (−L_{k−1}/L_{k})^{n−i} L_{i+k}.  (27)

Since 2L_{n+2} − L_{n+1} = 5F_{n+1}, we easily produce equation (26).

3.3 Fibonacci and Lucas j-step numbers

We define the “j-step numbers” of a sequence to be the new sequence made up of every jth term in the original sequence. For example, when we talk about the 3-step Fibonacci numbers then we could be referring to (F_{3n})_{n≥0} or perhaps (F_{3n+1})_{n≥0} or even (F_{3n+2})_{n≥0}, depending on our choice of offset. We assume throughout this section that j ≥ 1, and we note that if j = 1 then we get the “regular” Fibonacci (or Lucas) numbers without any skipping.

3.3.1 Fibonacci numbers with zero offset

For our first case, we will consider the j-step Fibonacci numbers such as F_{3n} for j = 3 or F_{4n} for j = 4, with no offset. Here is our result.

Theorem 3. For j ≥ 1 an integer, the j-step Fibonacci numbers satisfy

L_{j}^{n+1}F_{j(n−1)} + (−1)^{j(n+1)}F_{2j} = ∑_{i=0}^{n}(−1)^{j(n+i)}L_{j}^{i}F_{ji}.  (28)

Proof. We begin with the well-known formula [13, p. 112]

F_{a+b} = L_{b}F_{a} + (−1)^{b+1}F_{a−b},  (29)

and if we replace a with j(n − 1) and b with j, we get

F_{jn} = L_{j}F_{j(n−1)} + (−1)^{j+1}F_{j(n−2)}.  (30)

For j ≥ 1 fixed, we will use Theorem 2 with X_{n} = F_{jn}, and since Theorem 2 requires both X_{k} and X_{k−1} to be non-zero, we will set k = 2. Thanks to equation (30), we have c_{1} = L_{j} and c_{2} = (−1)^{j+1}. Since X_{0} = 0 and X_{1} = F_{j}, then the expression on the left-hand side of equation (17) of Theorem 2 is

X_{0}X_{n+2} − X_{1}X_{n+1} = 0 − F_{j}F_{j(n+1)},  (31)

and the expression in front of the sum on the right-hand side of equation (17) is

\frac{X_{0}X_{2} − X_{1}^{2}}{X_{2}} = \frac{0 − F_{j}^{2}}{F_{2j}} = \frac{−F_{j}}{L_{j}} ,  (32)

where we have used the well-known identity F_{2j} = F_{j}L_{j}. Likewise, the expression inside the sum in equation (17) is

(−c_{2}X_{1}/X_{2})^{n−i}X_{i+2} = ((−1)^{j}F_{j}/F_{2j})^{n−i}F_{j(i+2)} = ((−1)^{j}/L_{j})^{n−i}F_{j(i+2)}.  (33)

When we substitute equations (31), (32), and (33) into equation (17), and cancel the −F_{j} term from each side, we get

F_{j(n+1)} = \frac{1}{L_{j}} ∑_{i=0}^{n}((−1)^{j}/L_{j})^{n−i}F_{j(i+2)}.

This equation actually looks a bit nicer if we multiply both sides by (L_{j})^{n+1} and simplify, giving us

L_{j}^{n+1}F_{j(n+1)} = ∑_{i=0}^{n}(−1)^{j(n+i)}L_{j}^{i}F_{j(i+2)}.  (34)

Further simplification is needed. First, we replace i with i − 2 in the right-hand summation, giving us

L_{j}^{n+1}F_{j(n+1)} = ∑_{i=2}^{n+2}(−1)^{j(n+i)}L_{j}^{i−2}F_{ji}.  (35)

Then, we multiply both sides by L_{j}^{2} to give us

L_{j}^{n+3}F_{j(n+1)} = ∑_{i=2}^{n+2}(−1)^{j(n+i)}L_{j}^{i}F_{ji}.  (36)

Next, we add in the “missing term” for i = 1 to both sides, giving us

L_{j}^{n+3}F_{j(n+1)} + (−1)^{j(n+1)}L_{j}F_{j} = ∑_{i=1}^{n+2}(−1)^{j(n+i)}L_{j}^{i}F_{ji}.  (37)

Finally, we replace n with n − 2, we replace L_{j}F_{j} with F_{2j}, and we start the summation at i = 0 instead of at i = 1 to give us

L_{j}^{n+1}F_{j(n−1)} + (−1)^{j(n+1)}F_{2j} = ∑_{i=0}^{n}(−1)^{j(n+i)}L_{j}^{i}F_{ji},  (38)

which is equation (28) as desired.

Here are the results for small values of j:

(at j = 2): 3^{n+1}F_{2(n−1)} + 3 = ∑_{i=1}^{n} 3^{i}F_{2i},

(at j = 3): 4^{n+1}F_{3(n−1)} + 8(−1)^{n+1} = (−1)^{n} ∑_{i=1}^{n}(−4)^{i}F_{3i},

(at j = 4): 7^{n+1}F_{4(n−1)} + 21 = ∑_{i=1}^{n} 7^{i}F_{4i},

and it is interesting to compare them to this similar collection of formulas from Dresden and Tulskikh [11, p. 2],

3 − F_{2(n+2)}/3^{n} = ∑_{i=0}^{n}(1/3)^{i}F_{2i},

F_{3(n+2)}/4^{n} − 8 = ∑_{i=0}^{n}(1/4)^{i}F_{3i},

21 − F_{4(n+2)}/7^{n} = ∑_{i=0}^{n}(1/7)^{i}F_{4i}.

3.3.2 Lucas numbers with zero offset

For the j-step Lucas numbers such as L_{2n} for j = 2 or L_{3n} for j = 3, we will again use Theorem 2 but this time with X_{n} = L_{jn} and with k = 1. Here is our result.

Theorem 4. For j ≥ 1 an integer, the j-step Lucas numbers satisfy

2(L_{j}/2)^{n+1}(F_{jn}/F_{j}) + 2(−1)^{jn} = ∑_{i=0}^{n}(−1)^{j(n+i)}(L_{j}/2)^{i}L_{ji},  (39)

Proof. To begin, we will need the formula

L_{a+b} = L_{b}L_{a} + (−1)^{b+1}L_{a−b},  (40)

which follows from adding together two copies of equation (29), the first with a + 1 in place of a, and the second with a − 1 in place of a. If we replace a with j(n − 1) and b with j in equation (40), we get

L_{jn} = L_{j}L_{j(n−1)} + (−1)^{j+1}L_{j(n−2)}.  (41)

We set X_{n} = L_{jn}, which means that X_{0} = 2 and X_{1} = L_{j}. Thanks to equation (41) we have c_{1} = L_{j} and c_{2} = (−1)^{j+1}.

We now turn our attention to the expression on the left-hand side of equation (17) of Theorem 2, which is

X_{0}X_{n+2} − X_{1}X_{n+1} = 2L_{j(n+2)} − L_{j}L_{j(n+1)}.  (42)

From equation (41) with n + 2 in place of n we have

L_{j(n+2)} = L_{j}L_{j(n+1)} + (−1)^{j+1}L_{jn},

and from Exercise 55 in [14, p. 118] we have

L_{j(n+2)} = 5F_{j}F_{j(n+1)} − (−1)^{j+1}L_{jn},

and so if we add these together we will get

2L_{j(n+2)} = L_{j}L_{j(n+1)} + 5F_{j}F_{j(n+1)}.

When we substitute this into the right-hand side of equation (42) we get

X_{0}X_{n+2} − X_{1}X_{n+1} = 2L_{j(n+2)} − L_{j}L_{j(n+1)} = 5F_{j}F_{j(n+1)}.  (43)

Likewise, the expression in front of the sum on the right-hand side of equation (17) is

\frac{X_{0}X_{2} − X_{1}^{2}}{X_{1}} = \frac{2L_{2j} − L_{j}^{2}}{L_{j}} = \frac{5F_{j}^{2}}{L_{j}} ,  (44)

which follows from equation (43) with n = 0.

Now, since we are taking k = 1 in this section, then the expression inside the sum in equation (17) is

(−c_{2}X_{0}/X_{1})^{n−i}X_{i+1} = ((−1)^{j} · 2/L_{j})^{n−i}L_{j(i+1)}.  (45)

When we substitute equations (43), (44), and (45) into equation (17), and cancel the 5F_{j} term from each side, we get

F_{j(n+1)} = \frac{F_{j}}{L_{j}} ∑_{i=0}^{n}(2(−1)^{j}/L_{j})^{n−i}L_{j(i+1)}.

This equation actually looks a bit nicer if we multiply both sides by (L_{j}/2)^{n+1} and simplify, giving us

(L_{j}/2)^{n+1}F_{j(n+1)} = \frac{F_{j}}{2} ∑_{i=0}^{n}(−1)^{j(n+i)}(L_{j}/2)^{i}L_{j(i+1)}.  (46)

We will need to apply some further simplifications. First, we multiply both sides by 2/F_{j} and then replace i with i − 1 in the right-hand summation, giving us

2(L_{j}/2)^{n+1}(F_{j(n+1)}/F_{j}) = ∑_{i=1}^{n+1}(−1)^{j(n+i−1)}(L_{j}/2)^{i−1}L_{ji}.  (47)

We multiply both sides by (L_{j}/2) to produce

2(L_{j}/2)^{n+2}(F_{j(n+1)}/F_{j}) = ∑_{i=1}^{n+1}(−1)^{j(n+i−1)}(L_{j}/2)^{i}L_{ji}.  (48)

Next, we add in the “missing term” for i = 0 to both sides, giving us

2(L_{j}/2)^{n+2}(F_{j(n+1)}/F_{j}) + 2(−1)^{j(n+1)} = ∑_{i=0}^{n+1}(−1)^{j(n+i−1)}(L_{j}/2)^{i}L_{ji}.  (49)

Finally, we replace n with n − 1 to get

2(L_{j}/2)^{n+1}(F_{jn}/F_{j}) + 2(−1)^{jn} = ∑_{i=0}^{n}(−1)^{j(n+i)}(L_{j}/2)^{i}L_{ji},  (50)

and this is equation (39), as desired.

We note that equation (39) looks rather pleasant when L_{j} is even; this happens when j is a multiple of 3. For our first such case, we substitute j = 3 into equation (39) and multiply both sides by (−1)^{n} and simplify to get

2 − (−2)^{n+1}F_{3n} = ∑_{i=0}^{n}(−2)^{i}L_{3i}.  (51)

For the second, we use j = 6 and multiply both sides by 4 and simplify to get

9^{n+1}F_{6n} + 8 = 4 ∑_{i=0}^{n} 9^{i}L_{6i}.  (52)

These are the first two identities from equation (9) in the introduction, and they are quite nice and quite unexpected. (A similar procedure for j = 9 or j = 12 will give us additional equations.)

3.3.3 Non-zero offsets

The formulas with offsets −1, +1, and +2, when written down together, seem remarkably similar:

(offset −1): F_{j(n+1)} = F_{j} ∑_{i=0}^{n}(F_{j+1})^{n−i}F_{ji−1} for j > 1,

(offset +1): F_{j(n+1)} = F_{j} ∑_{i=0}^{n}(F_{j−1})^{n−i}F_{ji+1} for j > 1,

(offset +2): F_{j(n+1)} = F_{j} ∑_{i=0}^{n}(−F_{j−2})^{n−i}F_{ji+2} for j > 2.

Here are two additional offset formulas, and there does indeed seem to be a pattern:

(offset +3): F_{j(n+1)} = \frac{F_{j}}{2} ∑_{i=0}^{n}(F_{j−3}/2)^{n−i}F_{ji+3} for j > 3,

(offset +4): F_{j(n+1)} = \frac{F_{j}}{3} ∑_{i=0}^{n}(−F_{j−4}/3)^{n−i}F_{ji+4} for j > 4.

Here is the general form:

Theorem 5. For j > |k| integers, the j-step Fibonacci numbers with offset k satisfy

F_{j(n+1)} = \frac{F_{j}}{F_{k}} ∑_{i=0}^{n}((−1)^{k+1}F_{j−k}/F_{k})^{n−i}F_{ji+k}  for j > |k| > 0,  (53)

and for the Lucas numbers we have

F_{j(n+1)} = \frac{F_{j}}{L_{k}} ∑_{i=0}^{n}((−1)^{k}L_{j−k}/L_{k})^{n−i}L_{ji+k}  for j > |k|.  (54)

We note that if we set k = 0 in equation (54) then we obtain Theorem 2.1 from Adegoke and Frontczak’s recent paper [2], and if we set k = 1 in equations (53) and (54) then we obtain the collection of identities in equation (10).

Proof. To establish equation (53), we will use Theorem 1 with A_{n} = F_{jn+k}/F_{k}. Thus, we have A_{0} = 1, A_{1} = F_{j+k}/F_{k}, and since equation (29) tells us that

F_{jn+k} = L_{j}F_{j(n−1)+k} + (−1)^{j+1}F_{j(n−2)+k},

then we have

A_{n} = L_{j}A_{n−1} + (−1)^{j+1}A_{n−2}.

Thus, for the conditions in the statement of Theorem 1 we have c_{1} = L_{j} and so t = c_{1} −A_{1} = L_{j} − F_{j+k}/F_{k}, which gives us

t = L_{j} − F_{j+k}/F_{k} = \frac{L_{j}F_{k} − F_{j+k}}{F_{k}} ,

and thanks again to equation (29) which tells us that F_{k+j} = L_{j}F_{k} + (−1)^{j+1}F_{k−j}, this gives us

t = (−1)^{j}F_{k−j}/F_{k}.

Since k − j < 0, then we use the well-known fact that F_{−n} = (−1)^{n+1}F_{n}, giving us

t = (−1)^{j}(−1)^{k−j+1}F_{j−k}/F_{k} = (−1)^{k+1}F_{j−k}/F_{k}.  (55)

We are almost ready to use equation (14) from Theorem 1, but we still need to simplify the two expressions

A_{n+2} − A_{1}A_{n+1} and A_{2} − A_{1}^{2}

that we find in equation (14). From our definition of A_{n} as F_{jn+k}/F_{k}, these two expressions are

\frac{F_{j(n+2)+k}F_{k} − F_{j(n+1)+k}F_{j+k}}{F_{k}^{2}}  and  \frac{F_{2j+k}F_{k} − F_{j+k}^{2}}{F_{k}^{2}}  (56)

Here we will need to call upon a generalization of Catalan’s Identity [13, p. 108] which tells us that

F_{a+c}F_{b−c} − F_{a}F_{b} = (−1)^{b+c+1}F_{a+c−b}F_{c}.  (57)

For the first expression in equation (56), we use equation (57) with a = j(n+1)+k, b = j +k and c = j to give us

A_{n+2} − A_{1}A_{n+1} = \frac{F_{j(n+2)+k}F_{k} − F_{j(n+1)+k}F_{j+k}}{F_{k}^{2}} = \frac{(−1)^{k+1}F_{j(n+1)}F_{j}}{F_{k}^{2}} ,  (58)

For the second expression in equation (56), we simply take n = 0 in equation (58) to give us

A_{2} − A_{1}^{2} = \frac{F_{2j+k}F_{k} − F_{j+k}^{2}}{F_{k}^{2}} = \frac{(−1)^{k+1}F_{j}^{2}}{F_{k}^{2}} .  (59)

We can now use Theorem 1 with A_{n} = F_{jn+k}/F_{k}, and using our expressions in equations (55), (58), and (59) in equation (14) we get

\frac{(−1)^{k+1}F_{j(n+1)}F_{j}}{F_{k}^{2}} = \frac{(−1)^{k+1}F_{j}^{2}}{F_{k}^{2}} ∑_{i=0}^{n} ((−1)^{k+1}F_{j−k}/F_{k})^{n−i} \frac{F_{ji+k}}{F_{k}} .

With a bit of cancellation, this simplifies nicely to give us

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

which is equation (53) as desired.

To establish equation (54), we follow most of the same steps, but there are a few interesting new wrinkles. Proceeding as before, we will use Theorem 1 with A_{n} = L_{jn+k}/L_{k}. Thus, we have A_{0} = 1, A_{1} = L_{j+k}/L_{k}, and since equation (41) tells us that

L_{jn} = L_{j}L_{j(n−1)} + (−1)^{j+1}L_{j(n−2)},

then we have

A_{n} = L_{j}A_{n−1} + (−1)^{j+1}A_{n−2}.

Thus, for the conditions in the statement of Theorem 1 we have c_{1} = L_{j} and so t = c_{1} −A_{1} = L_{j} − L_{j+k}/L_{k}, which gives us

t = L_{j} − L_{j+k}/L_{k} = \frac{L_{j}L_{k} − L_{j+k}}{L_{k}} ,

and thanks to equation (40) which tells us that L_{k+j} = L_{j}L_{k} + (−1)^{j+1}L_{k−j}, this gives us

t = (−1)^{j}L_{k−j}/L_{k}.

Since k − j < 0, then we use the well-known fact that L_{−n} = (−1)^{n}L_{n}, giving us

t = (−1)^{j}(−1)^{k−j}L_{j−k}/L_{k} = (−1)^{k}L_{j−k}/L_{k}.  (60)

We are almost ready to use equation (14) from Theorem 1, but we still need to simplify the two expressions

A_{n+2} − A_{1}A_{n+1} and A_{2} − A_{1}^{2}

that we find in equation (14). From our definition of A_{n} as L_{jn+k}/L_{k}, these two expressions are

\frac{L_{j(n+2)+k}L_{k} − L_{j(n+1)+k}L_{j+k}}{L_{k}^{2}}  and  \frac{L_{2j+k}L_{k} − L_{j+k}^{2}}{L_{k}^{2}}  (61)

Here we will need to produce a variant of Catalan’s Identity but with Lucas numbers instead of Fibonacci numbers. If we take two versions of equation (57) with a + 1 and a − 1 in place of a, we have

F_{a+1+c}F_{b−c} − F_{a+1}F_{b} = (−1)^{b+c+1}F_{a+1+c−b}F_{c},
F_{a−1+c}F_{b−c} − F_{a−1}F_{b} = (−1)^{b+c+1}F_{a−1+c−b}F_{c},

and if we add these together we find that

L_{a+c}F_{b−c} − L_{a}F_{b} = (−1)^{b+c+1}L_{a+c−b}F_{c}.  (62)

We now repeat the process, but this time with equation (62) and this time with b + 1 and b − 1 in place of b, giving us

L_{a+c}F_{b+1−c} − L_{a}F_{b+1} = (−1)^{b+c}L_{a+c−b−1}F_{c},
L_{a+c}F_{b−1−c} − L_{a}F_{b−1} = (−1)^{b+c}L_{a+c−b+1}F_{c},

and if we add these together we find that

L_{a+c}L_{b−c} − L_{a}L_{b} = 5(−1)^{b+c}F_{a+c−b}F_{c}.  (63)

This is not quite as beautiful as the Fibonacci version in equation (57), but it will certainly meet our needs.

If we turn now to the first expression in equation (61), we use equation (63) with a = j(n + 1) + k, b = j + k and c = j to give us

A_{n+2} − A_{1}A_{n+1} = \frac{L_{j(n+2)+k}L_{k} − L_{j(n+1)+k}L_{j+k}}{L_{k}^{2}} = \frac{5(−1)^{k}F_{j(n+1)}F_{j}}{L_{k}^{2}} ,  (64)

For the second expression in equation (61), we simply take n = 0 in equation (64) to give us

A_{2} − A_{1}^{2} = \frac{L_{2j+k}L_{k} − L_{j+k}^{2}}{L_{k}^{2}} = \frac{5(−1)^{k}F_{j}^{2}}{L_{k}^{2}} .  (65)

We can now use Theorem 1 with A_{n} = L_{jn+k}/L_{k}, and using our expressions in equations (60), (64), and (65) in equation (14) we get

\frac{5(−1)^{k}F_{j(n+1)}F_{j}}{L_{k}^{2}} = \frac{5(−1)^{k}F_{j}^{2}}{L_{k}^{2}} ∑_{i=0}^{n} ((−1)^{k}L_{j−k}/L_{k})^{n−i} \frac{L_{ji+k}}{L_{k}} .

With a bit of cancellation, this simplifies nicely to give us

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

which is equation (54) as desired.

3.4 Pell and Pell-Lucas numbers

We recall [14, p. 23] that the Pell and Pell-Lucas numbers P_{n} and Q_{n} are defined as

P_{n} = 2P_{n−1} + P_{n−2} with P_{0} = 0, P_{1} = 1,
Q_{n} = 2Q_{n−1} + Q_{n−2} with Q_{0} = 1, Q_{1} = 1.

(66)

These are the sequences \underline{A000129} and \underline{A001333}, respectively. If we apply Theorem 2 with X_{n} = P_{n} and with k ≠ 0, 1, then since P_{0}P_{n+2} − P_{1}P_{n+1} = −P_{n+1} and P_{0}P_{2} − P_{1}^{2} = −1, then equation (17) gives us

P_{n+1} = \frac{1}{P_{k}} ∑_{i=0}^{n}(−P_{k−1}/P_{k})^{n−i}P_{i+k}.

If we take k = 2, 3, and 4, we will produce the identities in equation (11). However, it is interesting to note that with k = −1 we produce the equation

P_{n+1} = ∑_{i=0}^{n} 2^{n−i}P_{i−1},  (67)

which is quite different from equation (12). However, it is equivalent to the almost-identical formula in [11, page 3],

P_{n+2} = 2^{n+1} + ∑_{i=0}^{n} 2^{n−i}P_{i}.

To see this, we multiply both sides of equation (67) by 2, we replace i − 1 with i, we pull out the nth term, we replace 2P_{n+1} with P_{n+2} − P_{n}, and we absorb P_{n} into the sum.

3.5 The bronze Fibonacci numbers

As mentioned in the introduction, we define the “bronze Fibonacci numbers” as B_{0} = 0, B_{1} = 1, and

B_{n} = 3B_{n−1} + B_{n−2}.

This gives us the sequence \underline{A006190}. To prove the first identity in equation (13), we will use Theorem 1 with A_{n} = B_{2n+1}, and so A_{0} = B_{1} = 1 and A_{1} = B_{3} = 10. Starting with

B_{n} = 3B_{n−1} + B_{n−2},

then it is easy to show that

B_{2n+1} = 11B_{2n−1} − B_{2n−3}

which means that

A_{n} = 11A_{n−1} − A_{n−2}.

So, using Theorem 1 with t = c_{1} − A_{1} = 11 − 10 = 1, we have

A_{n+2} − 10A_{n+1} = (A_{2} − A_{1}^{2}) ∑_{i=0}^{n} 1^{n−1}A_{i}.  (68)

Now, A_{n+2} − 10A_{n+1} simplifies as follows:

A_{n+2} − 10A_{n+1} = (11A_{n+1} − A_{n}) − 10A_{n+1}
= A_{n+1} − A_{n}
= B_{2n+3} − B_{2n+1}
= 3B_{2n+2}.

Furthermore, A_{2} − A_{1}^{2} is equal to B_{5} − B_{3}^{2}, and since B_{3} = 10 and B_{5} = 109, then putting everything together we see that equation (68) becomes

3B_{2n+2} = 9 ∑_{i=0}^{n} 1^{n−1}B_{2i+1},  (69)

and this simplifies nicely to give us the first identity in equation (13). The other two identities have similar proofs.

4 Conclusion

As we mentioned in the introduction, there are countless other identities that we can prove: we simply select a sequence, choose a starting value, and then apply our theorem. To take a random example, if we let a_{n} be the sequence with initial values a_{0} = 0 and a_{1} = 1 and with recurrence a_{n} = 4a_{n−1} + 3a_{n−2}, then we can use Theorem 2 to quickly produce the following identities for this sequence (which is sequence \underline{A015530} in the OEIS):

a_{n+1} = \frac{1}{4} ∑_{i=0}^{n}(−3/4)^{n−i}a_{i+2} = \frac{1}{19} ∑_{i=0}^{n}(−12/19)^{n−i}a_{i+3}.

We invite the reader to try their hand at developing new such identities. It would also be reasonable (and quite interesting) to try adapting the methods of this paper to the Fibonacci and Lucas polynomials, as seen in Adegoke and Frontczak’s recent article [2]. The possibilities are unlimited.

5 Acknowledgments

We are grateful to Pioneer Academics for arranging the summer research project which led to this paper.

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2020 Mathematics Subject Classification: Primary 11B39; Secondary 11B37. Keywords: Sury’s identity, Fibonacci number, Lucas number, convolution, weighted sum.

(Concerned with sequences \underline{A000032}, \underline{A000045}, \underline{A000129}, \underline{A001333}, \underline{A006190}, and \underline{A015530}.)

Received December 14 2025; revised version received May 9 2026. Published in Journal of Integer Sequences, May 15 2026.

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