Pell and Associated Pell Braid Sequences as GCDs of Sums of k Consecutive Pell, Balancing, and Related Numbers

aBa Mbirika^{1} and Janee Schrader Department of Mathematics University of Wisconsin-Eau Claire Eau Claire, WI 54702 USA

                             mbirika@uwec.edu
                           schradjm5500@uwec.edu

Jürgen Spilker Institute of Mathematics University of Freiburg 79085 Freiburg im Breisgau Germany

                        juergen.spilker@t-online.de

Abstract

We consider the greatest common divisor (GCD) of all sums of k consecutive terms of a sequence (S_{n})_{n≥0} where the terms S_{n} come from exactly one of following six wellknown sequences’ terms: Pell P_{n}, associated Pell Q_{n}, balancing B_{n}, Lucas-balancing C_{n}, cobalancing b_{n}, and Lucas-cobalancing c_{n} numbers. For each of the six GCDs, we provide closed forms dependent on k. Moreover, each of these closed forms can be realized as braid sequences of Pell and associated Pell numbers in an intriguing manner. We end with partial results on GCDs of sums of squared terms and open questions.

1 Introduction

In 2021, Guyer and Mbirika gave closed forms for the greatest common divisor (GCD) of all sums of k consecutive generalized Fibonacci numbers [?]. Further, in 2022, Mbirika and Spilker generalized those results to the setting of the GCD of all sums of k consecutive squares of generalized Fibonacci numbers [?]. In this current paper, we extend the results of 2021 to the following six well-known sequences: Pell (P_{n})_{n≥0}, associated Pell (Q_{n})_{n≥0}, balancing (B_{n})_{n≥0}, Lucas-balancing (C_{n})_{n≥0}, cobalancing (b_{n})_{n≥0}, and Lucas-cobalancing (c_{n})_{n≥0}. Moreover, these GCDs can be realized as braid sequences. A braid sequence arises when we intertwine two sequences. For example in Figure 1, we intertwine the sequence (P_{n})_{n≥1} in the top row with the associated Pell sequence (Q_{n})_{n≥1} in the bottom row.

1corresponding author

\begin{matrix}P_{1} & P_{2} & P_{3} & P_{4} & P_{5} & P_{6} & P_{7} & P_{8} & ⋯ \\ Q_{1} & Q_{2} & Q_{3} & Q_{4} & Q_{5} & Q_{6} & Q_{7} & Q_{8} & ⋯ \end{matrix}

Figure 1: The braiding of (P_{n})_{n≥1} and (Q_{n})_{n≥1}.

The red-dotted path is the sequence (gcd(B_{n}, b_{n}))_{n≥1} and the blue-solid path is the sequence (gcd(B_{n}, b_{n} + 1))_{n≥1}. Both braid sequences easily follow from the identities

B_{n} = P_{n}Q_{n}, b_{n} = \{ \begin{matrix}P_{n}Q_{n−1}, & if n is even; \\ P_{n−1}Q_{n}, & if n is odd, \end{matrix} b_{n} + 1 = \{ \begin{matrix}P_{n−1}Q_{n}, & if n is even; \\ P_{n}Q_{n−1}, & if n is odd, \end{matrix}

and the facts that gcd(P_{n−1}, P_{n}) = 1 and gcd(Q_{n−1}, Q_{n}) = 1.

Convention 1. When (S_{n})_{n≥0} is any of the six sequences (P_{n})_{n≥0}, (Q_{n})_{n≥0}, (B_{n})_{n≥0}, (C_{n})_{n≥0}, (b_{n})_{n≥0}, or (c_{n})_{n≥0}, we establish the notation �^{m}(k) to denote the GCD of all sums of k consecutive m^{th} powers of sequence terms, in the respective six settings:

�^{m}(k) = gcd {(∑_{i=1}^{k} P_{n+i}^{m})_{n≥0}} ,  �^{m}(k) = gcd {(∑_{i=1}^{k} Q_{n+i}^{m})_{n≥0}} ,

�^{m}(k) = gcd {(∑_{i=1}^{k} B_{n+i}^{m})_{n≥0}} ,  �^{m}(k) = gcd {(∑_{i=1}^{k} C_{n+i}^{m})_{n≥0}} ,

�^{m}(k) = gcd {(∑_{i=1}^{k} b_{n+i}^{m})_{n≥0}} ,  �^{m}(k) = gcd {(∑_{i=1}^{k} c_{n+i}^{m})_{n≥0}} .

Remark 2. When m = 1, we omit the superscript and simply write �(k).

In this paper we give closed forms for the braid sequence of GCD-values (�^{m}(k))_{k≥1} when m = 1 for each of the six sequences and give partial results for when m = 2. We chose these six sequences, in particular, since the GCD-values (�(k))_{k≥1} for each of the six sequences all involve Pell and associated Pell numbers in an intriguing manner. The breakdown of this paper is as follows. In Section 2, we provide definitions of the six sequences and some historical origins of the relatively newer sequences (B_{n})_{n≥0}, (C_{n})_{n≥0}, (b_{n})_{n≥0}, and (c_{n})_{n≥0}. In Section 3, we give preliminary identities used to prove our main results, which are in Sections 4 and 5. Finally, in Section 6, we address our progress towards the m = 2 setting and provide some open questions.

1.1 Motivation: a new proof of an old result

At the 20^{th} International Conference on Fibonacci Numbers and Their Applications in Sarajevo in 2022, Mbirika presented his and collaborator Guyer’s results on the GCD of all sums of k consecutive generalized Fibonacci numbers [?]. Conference participant Florian Luca communicated to Mbirika an observation that leads to a simple proof in the Fibonacci setting when k is even. In this setting, the Guyer-Mbirika result was the following:

F(k) = \{ \begin{matrix}F_{k/2}, & if k ≡ 0 (mod 4); \\ L_{k/2}, & if k ≡ 2 (mod 4), \end{matrix}

where F(k) denotes the GCD of all sums of k consecutive Fibonacci numbers. For ease of notation, set σ_{F} (k, n) to be the sum ∑_{i=1}^{k} F_{n+i}. Luca’s observations was the following:

σ_{F} (k, n) = F_{n+k+2} − F_{n+2} = \{ \begin{matrix}F_{k/2}L_{(k/2+2)+n}, & if k ≡ 0 (mod 4); \\ L_{k/2}F_{(k/2+2)+n}, & if k ≡ 2 (mod 4). \end{matrix}  (1)

From this identity, the Guyer-Mbirika result is easily proven. For example for k = 20, using Identity (1) and the fact that consecutive Lucas numbers are relatively prime, we have

F(20) = gcd(σ_{F} (20, 0), σ_{F} (20, 1), σ_{F} (20, 2), …) = gcd(F_{10}L_{12}, F_{10}L_{13}, F_{10}L_{14}, …)
= F_{10} · gcd(L_{12}, L_{13}, L_{14}, …)
= F_{10},

as expected. For Identity (1), Luca noted that the first equality is easily shown if we utilize the fact that ∑_{i=1}^{k} F_{i} = F_{k+2} − 1, and the second equality follows by the known result

F_{a} − F_{b} = \{ \begin{matrix}F_{\frac{a−b}{2}} L_{\frac{a+b}{2}} , & if a − b ≡ 0 (mod 4); \\ L_{\frac{a−b}{2}} F_{\frac{a+b}{2}} , & if a − b ≡ 2 (mod 4), \end{matrix}  (2)

if we set a := n + k + 2 and b := n + 2. Identity (2) follows directly from a 1963 result proven by Ruggles [?, p. 77]. In this current paper, we generalize Identity (2) into the Pell and associated Pell settings in Lemmas 12 and 13. Moreover, using the latter two lemmas we generalize Identity (1) to compute the sums σ_{S}(k, n) for five (of our six) different sequences (S_{n})_{n≥0} in one of our main results given in Theorem 14.

2 Definitions of the six sequences and some remarks

We first recall the recursive definitions of the six sequences used in this paper, and then we follow with their well-known Binet forms.

Definition 3. The Pell sequence (P_{n})_{n≥0} and the associated Pell sequence (Q_{n})_{n≥0} are defined by the recurrence relations P_{n} = 2P_{n−1} + P_{n−2} and Q_{n} = 2Q_{n−1} + Q_{n−2}, respectively, with initial conditions P_{0} = 0, P_{1} = 1, Q_{0} = 1, and Q_{1} = 1. In the OEIS, these are sequences \underline{A000129} and \underline{A001333}, respectively [?].

Remark 4. In the literature, there is unfortunately some discrepancy on the precise definition of the Pell-Lucas sequence. Though many sources attribute the OEIS sequence \underline{A002203} as the “companion Pell sequence” (or equivalently, the Pell-Lucas sequence), we choose to follow Koshy [?] and many others in the literature who define the Pell-Lucas sequence as we have done in Definition 3 and call (Q_{n})_{n≥0} the “associated Pell sequence”. Remark 5. The associated Pell (respectively, Pell) sequence is the sequence of numerators (respectively, denominators) of the rational convergents to \sqrt{2}; that is, lim_{n→∞} \frac{Q_{n}}{P_{n}} = \sqrt{2}.

Before we give the recursive definition of the remaining four sequences, we first discuss how these four sequences were originally defined. In 1999, Behera and Panda [?] defined an integer n ∈ ℕ to be a balancing number if it is a solution to the Diophantine equation

1 + 2 + ⋯ + (n − 1) = (n + 1) + (n + 2) + ⋯ + (n + r),  (3)

where r is the balancer corresponding to n. The terms in the sequence of balancing numbers and their corresponding balancers are denoted B_{n} and R_{n}, respectively. For example, B_{2} = 6 and R_{2} = 2 since 1 + 2 + ⋯ + 5 = 7 + 8. Later in 2005, Panda and Ray [?] slightly modified Equation (3) to the Diophantine equation

1 + 2 + ⋯ + n = (n + 1) + (n + 2) + ⋯ + (n + r).  (4)

In this new setting, they called the value n a cobalancing number and the corresponding r a cobalancer. The terms in the sequence of cobalancing numbers and their corresponding cobalancers are denoted b_{n} and r_{n}, respectively. It turns out that every balancer is also a cobalancing number in the following sense: R_{n} = b_{n}. Moreover, every cobalancer is also a balancing number in the following sense: r_{n+1} = B_{n}. Hence, in this paper we consider the sequences (B_{n})_{n≥0} and (b_{n})_{n≥0} and neither (R_{n})_{n≥0} nor (r_{n})_{n≥0}.

Behera and Panda also showed that B_{n} is a balancing number if and only if 8B_{n}^{2} + 1 is a perfect square. So consider the sequence, denoted (C_{n})_{n≥0}, of positive roots of \sqrt{8B_{n}^{2} + 1} for each n ≥ 0. This sequence is called the Lucas-balancing sequence and is named so since the value C_{n} is associated to B_{n} in many manners similar to the relationship between L_{n} and F_{n} [?]. Lastly, it is known that b_{n} is a cobalancing number if and only if 8b_{n}^{2} + 8b_{n} + 1 is a perfect square. So consider the sequence, denoted (c_{n})_{n≥0}, of positive roots of \sqrt{8b_{n}^{2} + 8b_{n} + 1} for each n ≥ 1 and set c_{0} := −1. This sequence is called the Lucas-cobalancing sequence.

We now give the recursive definitions of the four sequences (B_{n})_{n≥0}, (C_{n})_{n≥0}, (b_{n})_{n≥0}, and (c_{n})_{n≥0}. Then in Table 1, we give the first eleven terms of each of the six sequences.

Definition 6. The balancing sequence (B_{n})_{n≥0} and the Lucas-balancing sequence (C_{n})_{n≥0} are defined by the recurrence relations B_{n} = 6B_{n−1} −B_{n−2} and C_{n} = 6C_{n−1} −C_{n−2}, respectively, with initial conditions B_{0} = 0, B_{1} = 1, C_{0} = 1, and C_{1} = 3. In the OEIS, these are sequences \underline{A001109} and \underline{A001541}, respectively [?].

Definition 7. The cobalancing sequence (b_{n})_{n≥0} and the Lucas-cobalancing sequence (c_{n})_{n≥0} are defined by the recurrence relations b_{n} = 6b_{n−1} − b_{n−2} + 2 and c_{n} = 6c_{n−1} − c_{n−2}, respectively, with initial conditions b_{0} = 0, b_{1} = 0, c_{0} = −1, and c_{1} = 1. In the OEIS, these are sequences \underline{A053141} and \underline{A002315}, respectively [?].

Finally, let γ = 1 + \sqrt{2} and δ = 1 − \sqrt{2}. Then we have the following well-known Binet forms for the sequence terms P_{n}, Q_{n}, B_{n}, C_{n}, b_{n}, and c_{n}, respectively:

{P_{n} = \frac{γ^{n}−δ^{n}}{2\sqrt{2}} , \atop Q_{n} = \frac{γ^{n}+δ^{n}}{2} ,}  {B_{n} = \frac{γ^{2n}−δ^{2n}}{4\sqrt{2}} , \atop C_{n} = \frac{γ^{2n}+δ^{2n}}{2} ,}  {b_{n} = \frac{γ^{2n−1}−δ^{2n−1}}{4\sqrt{2}} − \frac{1}{2}, \atop c_{n} = \frac{γ^{2n−1}+δ^{2n−1}}{2} .}

\begin{matrix}n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ P_{n} & 0 & 1 & 2 & 5 & 12 & 29 & 70 & 169 & 408 & 985 & 2378 \\ Q_{n} & 1 & 1 & 3 & 7 & 17 & 41 & 99 & 239 & 577 & 1393 & 3363 \\ B_{n} & 0 & 1 & 6 & 35 & 204 & 1189 & 6930 & 40391 & 235416 & 1372105 & 7997214 \\ C_{n} & 1 & 3 & 17 & 99 & 577 & 3363 & 19601 & 114243 & 665857 & 3880899 & 22619537 \\ b_{n} & 0 & 0 & 2 & 14 & 84 & 492 & 2870 & 16730 & 97512 & 568344 & 3312554 \\ c_{n} & −1 & 1 & 7 & 41 & 239 & 1393 & 8119 & 47321 & 275807 & 1607521 & 9369319 \end{matrix}

Table 1: The first 11 Pell P_{n}, associated Pell Q_{n}, balancing B_{n}, cobalancing b_{n}, Lucasbalancing C_{n}, and Lucas-cobalancing c_{n} numbers.

3 Some old and new identities

In this section, we provide the preliminary identities used to prove our main results in Sections 4 and 5. Some of these identities are well known, but most are new.

3.1 Sum identities, Cassini’s identities, and GCD identities

The following lemma follows from results in Koshy’s book [?], the Binet formulas given in Section 2, or Catarino et al. [?].

Lemma 8. For all k ≥ 1, the following sum identities hold for the six sequences Pell (P_{n})_{n≥0}, associated Pell (Q_{n})_{n≥0}, balancing (B_{n})_{n≥0}, Lucas-balancing (C_{n})_{n≥0},cobalancing (b_{n})_{n≥0}, and Lucas-cobalancing (c_{n})_{n≥0}:

\begin{aligned}∑_{i=1}^{k} P_{i} & = \frac{1}{2}(Q_{k+1} − 1), & (5) \\ ∑_{i=1}^{k} Q_{i} & = P_{k+1} − 1, & (6) \\ ∑_{i=1}^{k} B_{i} & = \frac{1}{4}(P_{2k+1} − 1), & (7) \end{aligned}  \begin{aligned}∑_{i=1}^{k} C_{i} & = \frac{1}{2}(Q_{2k+1} − 1), & (8) \\ ∑_{i=1}^{k} b_{i} & = \frac{1}{4}(b_{k+1} − b_{k} − 2k), & (9) \\ ∑_{i=1}^{k} c_{i} & = \frac{1}{2}(Q_{2k} − 1). & (10) \end{aligned}

Proof of Identity (5). This is well known (see [?, Identity (10.1)]).

Proof of Identity (6). This is well known (see [?, Identity (10.2)]).

Proof of Identity (7). By the Binet formulas, we have B_{i} = \frac{γ^{2i}−δ^{2i}}{4\sqrt{2}} = \frac{1}{2} · \frac{γ^{2i}−δ^{2i}}{2\sqrt{2}} = \frac{1}{2}P_{2i}. It follows that ∑_{i=1}^{k} B_{i} = \frac{1}{2} ∑_{i=1}^{k} P_{2i} = \frac{1}{2} (\frac{1}{2}P_{2k+1} − \frac{1}{2}) = \frac{1}{4}(P_{2k+1} − 1), where the second equality holds by a well-known identity for the sum of the first k even-indexed Pell numbers (see [?, Identity (10.4)]).

Proof of Identity (8). By the Binet formulas, we have C_{i} = \frac{γ^{2i}−δ^{2i}}{2} = Q_{2i}. It follows that ∑_{i=1}^{k} C_{i} = ∑_{i=1}^{k} Q_{2i} = \frac{1}{2}(Q_{2k+1}−1), where the second equality holds by a well-known identity for the sum of the first k even-indexed associated Pell numbers (see [?, Identity (10.5)]).

Proof of Identity (9). This is well known (see [?, Proposition 3.6]).

Proof of Identity (10). By the Binet formulas, we have c_{i} = \frac{γ^{2i−1}−δ^{2i−1}}{2} = Q_{2i−1}. It follows that ∑_{i=1}^{k} c_{i} = ∑_{i=1}^{k} Q_{2i−1} = \frac{1}{2}(Q_{2k} − 1), where the second equality holds by a wellknown identity for the sum of the first k odd-indexed associated Pell numbers (see [?, Identity (10.6)]).

Cassini’s identity for the Fibonacci numbers has an analogue in both the Pell and associated Pell settings. We use the following Cassini’s identities in the next two lemmas to prove the closed forms of �(k) and �(k) given in Theorems 15 and 16, respectively.

Lemma 9 (Cassini’s identity for (P_{n})_{n≥0}). For all k ≥ 1, we have P_{k−1}P_{k+1} = P_{k}^{2} + (−1)^{k}.

Proof. See Horadam [?, Identity (30)].

Koshy mentions Cassini’s identity in the associated Pell setting [?, Identity (35)]; however, he provides no proof. As we could not find a proof in the literature of this identity, we provide our own proof using the Binet formula for Q_{n}.

Lemma 10 (Cassini’s identity for (Q_{n})_{n≥0}). For all k ≥ 1, we have

Q_{k−1}Q_{k+1} = Q_{k}^{2} + 2(−1)^{k−1}.

Proof. By the Binet formula for the associated Pell sequence, we have

Q_{k+1}Q_{k−1} − Q_{k}^{2} = (\frac{γ^{k+1} + δ^{k+1}}{2} ) (\frac{γ^{k−1} + δ^{k−1}}{2} ) − (\frac{γ^{k} + δ^{k}}{2} )^{2}

= \frac{γ^{2k} + γ^{k+1}δ^{k−1} + γ^{k−1}δ^{k+1} + δ^{2k}}{4} − \frac{γ^{2k} + 2(γδ)^{k} + δ^{2k}}{4}

= \frac{γ^{2}(γδ)^{k−1} + δ^{2}(γδ)^{k−1} − 2(γδ)^{k}}{4}

= \frac{(γ^{2} + δ^{2})(−1)^{k−1} − 2(−1)^{k}}{4}  (since γδ = −1)

= \frac{6(−1)^{k−1} + 2(−1)^{k−1}}{4}

= 2(−1)^{k−1},

where the fifth equality holds since 3 = Q_{2} = \frac{γ^{2}+δ^{2}}{2} implies γ^{2} + δ^{2} = 6.

Lemma 11. For all n ≥ 1, we have the following five identities:

gcd(P_{n}, P_{n+1}) = 1, (11)
gcd(Q_{n}, Q_{n+1}) = 1, (12)
gcd(P_{2n}, P_{2n+2}) = 2, (13)

gcd(P_{2n−1}, P_{2n+1}) = 1,  (14)_{gcd(Q_{n}, Q_{n+2}) = 1.  (15)}

Proof. Identity (11) follows from Lemma 9, while Identity (12) follows from Lemma 10 and the fact that associated Pell numbers are always odd. Identities (13) and (14) hold by Flórez et al. [?] in Proposition 2 part (2) if we set x := 1, and Identity (15) holds by Flórez et al. [?] in Proposition 2 part (1) if we set x := 1.

3.2 New identities used to prove our main results

Lemma 12. For all s, r ≥ 1 where s is even, the following identity holds:

P_{s+r} − P_{r} = \{ \begin{matrix}2P_{s/2}Q_{s/2+r}, & if s ≡ 0 (mod 4); \\ 2Q_{s/2}P_{s/2+r}, & if s ≡ 2 (mod 4). \end{matrix}

Proof. Let s, r ≥ 1 be given where s is even.

Case I. Suppose s ≡ 0 (mod 4). Then \frac{s}{2} is even and hence (γδ)^{s/2} = 1. Observe that

P_{s+r} − P_{r} = \frac{γ^{s+r} − δ^{s+r}}{2\sqrt{2}} − \frac{γ^{r} − δ^{r}}{2\sqrt{2}}

= \frac{1}{2\sqrt{2}} (γ^{s+r} − δ^{s+r} − (γδ)^{s/2}(γ^{r} − δ^{r}))  (since (γδ)^{s/2} = 1)

= \frac{1}{2\sqrt{2}} (γ^{s/2} − δ^{s/2}) (γ^{s/2+r} + δ^{s/2+r})

= 2 · \frac{γ^{s/2} − δ^{s/2}}{2\sqrt{2}} · \frac{γ^{s/2+r} + δ^{s/2+r}}{2}

= 2P_{s/2}Q_{s/2+r}.

Case II. Suppose s ≡ 2 (mod 4). Then \frac{s}{2} is odd and hence (γδ)^{s/2} = −1. Observe that

P_{s+r} − P_{r} = \frac{γ^{s+r} − δ^{s+r}}{2\sqrt{2}} − \frac{γ^{r} − δ^{r}}{2\sqrt{2}}
= \frac{1}{2\sqrt{2}} (γ^{s+r} − δ^{s+r} + (γδ)^{s/2}(γ^{r} − δ^{r}))  (since (γδ)^{s/2} = −1)

= \frac{1}{2\sqrt{2}} (γ^{s/2} + δ^{s/2}) (γ^{s/2+r} − δ^{s/2+r})

= 2 · \frac{γ^{s/2} + δ^{s/2}}{2} · \frac{γ^{s/2+r} − δ^{s/2+r}}{2\sqrt{2}}

= 2Q_{s/2}P_{s/2+r}.

Lemma 13. For all s, r ≥ 1 where s is even, the following identity holds:

Q_{s+r} − Q_{r} = \{ \begin{matrix}4P_{s/2}P_{s/2+r}, & if s ≡ 0 (mod 4); \\ 2Q_{s/2}Q_{s/2+r}, & if s ≡ 2 (mod 4). \end{matrix}

Proof. Let s, r ≥ 1 be given where s is even.

Case I. Suppose s ≡ 0 (mod 4). Then \frac{s}{2} is even and hence (γδ)^{s/2} = 1. Observe that

Q_{s+r} − Q_{r} = \frac{γ^{s+r} + δ^{s+r}}{2} − \frac{γ^{r} + δ^{r}}{2}
= \frac{1}{2} (γ^{s+r} + δ^{s+r} − (γδ)^{s/2}(γ^{r} + δ^{r}))  (since (γδ)^{s/2} = 1)

= \frac{1}{2} (γ^{s/2} − δ^{s/2}) (γ^{s/2+r} − δ^{s/2+r})

= 4 · \frac{γ^{s/2} − δ^{s/2}}{2\sqrt{2}} · \frac{γ^{s/2+r} − δ^{s/2+r}}{2\sqrt{2}}

= 4P_{s/2}P_{s/2+r}.

Case II. Suppose s ≡ 2 (mod 4). Then \frac{s}{2} is odd and hence (γδ)^{s/2} = −1. Observe that

Q_{s+r} − Q_{r} = \frac{γ^{s+r} + δ^{s+r}}{2} − \frac{γ^{r} + δ^{r}}{2}
= \frac{1}{2} (γ^{s+r} + δ^{s+r} + (γδ)^{s/2}(γ^{r} + δ^{r}))  (since (γδ)^{s/2} = −1)

= \frac{1}{2} (γ^{s/2} + δ^{s/2}) (γ^{s/2+r} + δ^{s/2+r})

= 2 · \frac{γ^{s/2} + δ^{s/2}}{2} · \frac{γ^{s/2+r} + δ^{s/2+r}}{2}

= 2Q_{s/2}Q_{s/2+r}.

Using the latter Lemmas 12 and 13, we are now ready to prove our main sum identities in the following theorem, which we use to prove our main results in Section 4.

Theorem 14. For all k ≥ 1, set σ_{S}(k, n) := ∑_{i=1}^{k} S_{n+i} where (S_{n})_{n≥0} is any sequence. Then the following identities hold for the five sequences Pell (P_{n})_{n≥0}, associated Pell (Q_{n})_{n≥0}, balancing (B_{n})_{n≥0}, Lucas-balancing (C_{n})_{n≥0}, and Lucas-cobalancing (c_{n})_{n≥0}:

σ_{P} (k, n) = \frac{1}{2}(Q_{n+k+1} − Q_{n+1}) = \{ \begin{matrix}2P_{k/2}P_{k/2+n+1}, & if k ≡ 0 (mod 4); \\ Q_{k/2}Q_{k/2+n+1}, & if k ≡ 2 (mod 4). \end{matrix}  (16)

σ_{Q}(k, n) = P_{n+k+1} − P_{n+1} = \{ \begin{matrix}2P_{k/2}Q_{k/2+n+1}, & if k ≡ 0 (mod 4); \\ 2Q_{k/2}P_{k/2+n+1}, & if k ≡ 2 (mod 4). \end{matrix}  (17)

σ_{B}(k, n) = \frac{1}{4} (P_{2k+2n+1} − P_{2n+1}) = \{ \begin{matrix}\frac{1}{2}P_{k}Q_{k+2n+1}, & if k is even; \\ \frac{1}{2}Q_{k}P_{k+2n+1}, & if k is odd. \end{matrix}  (18)

σ_{C}(k, n) = \frac{1}{2} (Q_{2k+2n+1} − Q_{2n+1}) = \{ \begin{matrix}2P_{k}P_{k+2n+1}, & if k is even; \\ Q_{k}Q_{k+2n+1}, & if k is odd. \end{matrix}  (19)

σ_{c}(k, n) = \frac{1}{2} (Q_{2k+2n} − Q_{2n}) = \{ \begin{matrix}2P_{k}P_{k+2n}, & if k is even; \\ Q_{k}Q_{k+2n}, & if k is odd. \end{matrix}  (20)

Proof of Identity (16). Let k ≥ 2 be given where k is even. Observe that

σ_{P} (k, n) := ∑_{i=1}^{k} P_{n+i} = ∑_{i=1}^{k+n} P_{i} − ∑_{i=1}^{n} P_{i}

= \frac{1}{2}(Q_{k+n+1} − Q_{n+1}),

where the last equality holds by Identity (5) of Lemma 8. By Lemma 13, if we set s := k and r := n + 1, then we have

σ_{P} (k, n) = \{ \begin{matrix}\frac{1}{2} (4P_{k/2}P_{k/2+n+1}) = 2P_{k/2}P_{k/2+n+1}, & if k ≡ 0 (mod 4); \\ \frac{1}{2} (2Q_{k/2}Q_{k/2+n+1}) = Q_{k/2}Q_{k/2+n+1}, & if k ≡ 2 (mod 4). \end{matrix}

Proof of Identity (17). Let k ≥ 2 be given where k is even. Observe that

σ_{Q}(k, n) := ∑_{i=1}^{k} Q_{n+i} = ∑_{i=1}^{k+n} Q_{i} − ∑_{i=1}^{n} Q_{i}
= P_{k+n+1} − P_{n+1},

where the last equality holds by Identity (6) of Lemma 8. By Lemma 12, if we set s := k and r := n + 1, then we have

σ_{Q}(k, n) = \{ \begin{matrix}2P_{k/2}Q_{k/2+n+1}, & if k ≡ 0 (mod 4); \\ 2Q_{k/2}P_{k/2+n+1}, & if k ≡ 2 (mod 4). \end{matrix}

Proof of Identity (18). Let k ≥ 1 be given. Observe that

σ_{B}(k, n) := ∑_{i=1}^{k} B_{n+i} = ∑_{i=1}^{k+n} B_{i} − ∑_{i=1}^{n} B_{i}

= \frac{1}{4} (P_{2k+2n+1} − P_{2n+1}) ,

where the last equality holds by Identity (7) of Lemma 8. By Lemma 12, if we set s := 2k and r := 2n + 1, then we have

σ_{B}(k, n) = \{ \begin{matrix}\frac{1}{4} (2P_{2k/2}Q_{2k/2+2n+1}) = \frac{1}{2}P_{k}Q_{k+2n+1}, & if k is even; \\ \frac{1}{4} (2Q_{2k/2}P_{2k/2+2n+1}) = \frac{1}{2}Q_{k}P_{k+2n+1}, & if k is odd, \end{matrix}

since s ≡ 0 (mod 4) if and only if k is even, and s ≡ 2 (mod 4) if and only if k is odd.

Proof of Identity (19). Let k ≥ 1 be given. Observe that

σ_{C}(k, n) := ∑_{i=1}^{k} C_{n+i} = ∑_{i=1}^{k+n} C_{i} − ∑_{i=1}^{n} C_{i}

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

where the last equality holds by Identity (8) of Lemma 8. By Lemma 13, if we set s := 2k and r := 2n + 1, then we have

σ_{C}(k, n) = \{ \begin{matrix}\frac{1}{2} (4P_{2k/2}P_{2k/2+2n+1}) = 2P_{k}P_{k+2n+1}, & if k is even; \\ \frac{1}{2} (2Q_{2k/2}Q_{2k/2+2n+1}) = Q_{k}Q_{k+2n+1}, & if k is odd, \end{matrix}

since s ≡ 0 (mod 4) if and only if k is even, and s ≡ 2 (mod 4) if and only if k is odd.

Proof of Identity (20). Let k ≥ 1 be given. Observe that

σ_{c}(k, n) := ∑_{i=1}^{k} c_{n+i} = ∑_{i=1}^{k+n} c_{i} − ∑_{i=1}^{n} c_{i}

= \frac{1}{2} (Q_{2k+2n} − Q_{2n}) ,

where the last equality holds by Identity (10) of Lemma 8. By Lemma 13, if we set s := 2k and r := 2n, then we have

σ_{c}(k, n) = \{ \begin{matrix}\frac{1}{2} (4P_{2k/2}P_{2k/2+2n}) = 2P_{k}P_{k+2n}, & if k is even; \\ \frac{1}{2} (2Q_{2k/2}Q_{2k/2+2n}) = Q_{k}Q_{k+2n}, & if k is odd, \end{matrix}

since s ≡ 0 (mod 4) if and only if k is even, and s ≡ 2 (mod 4) if and only if k is odd.

4 Main results for �(k), �(k), �(k), �(k), and �(k)

These are the braids for braid sequences (�(k))_{k≥1} in solid blue and (�(k))_{k≥1} in dotted red. In Theorems 15 and 16, respectively, we give the proofs of these two braids.

\begin{matrix}1 & 2Q_{1} & 1 & 2P_{2} & 1 & 2Q_{3} & 1 & 2P_{4} & ⋯ \\ 1 & Q_{1} & 1 & 2P_{2} & 1 & Q_{3} & 1 & 2P_{4} & ⋯ \end{matrix}

\begin{matrix}k & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\ �(k) & 1 & 1 & 1 & 4 & 1 & 7 & 1 & 24 & 1 & 41 & 1 & 140 & 1 & 239 \\ �(k) & 1 & 2 & 1 & 4 & 1 & 14 & 1 & 24 & 1 & 82 & 1 & 140 & 1 & 478 \end{matrix}

Table 2: The first 14 terms of the sequences (�(k))_{k≥1} and (�(k))_{k≥1}.

Theorem 15. For all k ≥ 1, the GCD of all sums of k consecutive Pell numbers is

�(k) = \{ \begin{matrix}2P_{k/2}, & if k ≡ 0 (mod 4); \\ Q_{k/2}, & if k ≡ 2 (mod 4); \\ 1, & if k ≡ 1, 3 (mod 4). \end{matrix}

Proof. Let k ≥ 1 be given. Recall σ_{P} (k, n) = ∑_{i=1}^{k} P_{n+i}. So by definition of �(k), we have

�(k) = gcd (σ_{P} (k, 0), σ_{P} (k, 1), σ_{P} (k, 2), …) .  (21)

Case I. Suppose k ≡ 0 (mod 4). By Identity (16) of Theorem 14, it follows that

�(k) = gcd(2P_{k/2}P_{k/2+1}, 2P_{k/2}P_{k/2+2}, 2P_{k/2}P_{k/2+3}, …)
= 2P_{k/2} · gcd(P_{k/2+1}, P_{k/2+2}, P_{k/2+3}, …)
= 2P_{k/2},

where the last equality holds by Identity (11) of Lemma 11.

Case II. Suppose k ≡ 2 (mod 4). By Identity (16) of Theorem 14, it follows that

�(k) = gcd(Q_{k/2}Q_{k/2+1}, Q_{k/2}Q_{k/2+2}, Q_{k/2}Q_{k/2+3}, …)
= Q_{k/2} · gcd(Q_{k/2+1}, Q_{k/2+2}, Q_{k/2+3}, …)
= Q_{k/2},

where the last equality holds by Identity (12) of Lemma 11.

Case III. Suppose k is odd. Assume by way of contradiction that p^{j} divides �(k) for some prime p with j ≥ 1. By Equation (21), it follows that p^{j} divides σ_{P} (k, n) for all n ≥ 0. Hence p^{j} divides both σ_{P} (k, 1) − σ_{P} (k, 0) and σ_{P} (k, 2) − σ_{P} (k, 1). Observe that

σ_{P} (k, 1) − σ_{P} (k, 0) = (P_{2} + P_{3} + ⋯ + P_{k+1}) − (P_{1} + P_{2} + ⋯ + P_{k}) = P_{k+1} − 1, and
σ_{P} (k, 2) − σ_{P} (k, 1) = (P_{3} + P_{4} + ⋯ + P_{k+2}) − (P_{2} + P_{3} + ⋯ + P_{k+1}) = P_{k+2} − 2.

Hence p^{j} divides both P_{k+1} − 1 and P_{k+2} − 2. Since P_{k} + 2P_{k+1} = P_{k+2}, we have

P_{k} = P_{k+2} − 2P_{k+1}
= P_{k+2} − 2 − 2P_{k+1} + 2
= (P_{k+2} − 2) − 2(P_{k+1} − 1),

and therefore p^{j} divides P_{k}. Thus P_{k}P_{k+2} ≡ 0 (mod p^{j}). Moreover, since p^{j} divides P_{k+1} − 1, we have p^{j} divides P_{k+1}^{2} − 1, and thus P_{k+1}^{2} + 1 ≡ 2 (mod p^{j}). By the Pell Cassini Identity, Lemma 9, we have P_{k}P_{k+2} = P_{k+1}^{2} + 1 since k is odd. It follows that 2 ≡ 0 (mod p^{j}) and thus p^{j} divides 2, forcing p = 2 and j = 1. Since p^{j} divides P_{k+1} − 1 and we know p^{j} = 2, this implies P_{k+1} − 1 is even. However, k being odd implies P_{k+1} is even and hence P_{k+1} − 1 is also odd, which yields a contradiction. Thus there exists no prime that divides �(k) when k is odd, and hence �(k) = 1 for all odd k.

Theorem 16. For all k ≥ 1, the GCD of all sums of k consecutive associated Pell numbers is

�(k) = \{ \begin{matrix}2P_{k/2}, & if k ≡ 0 (mod 4); \\ 2Q_{k/2}, & if k ≡ 2 (mod 4); \\ 1, & if k ≡ 1, 3 (mod 4). \end{matrix}

Proof. Let k ≥ 1 be given. Recall σ_{Q}(k, n) = ∑_{i=1}^{k} Q_{n+i}. So by definition of �(k), we have

�(k) = gcd (σ_{Q}(k, 0), σ_{Q}(k, 1), σ_{Q}(k, 2), …) .  (22)

Case I. Suppose k ≡ 0 (mod 4). By Identity (17) of Theorem 14, it follows that

�(k) = gcd(2P_{k/2}Q_{k/2+1}, 2P_{k/2}Q_{k/2+2}, 2P_{k/2}Q_{k/2+3}, …)
= 2P_{k/2} · gcd(Q_{k/2+1}, Q_{k/2+2}, Q_{k/2+3}, …)
= 2P_{k/2},

where the last equality holds by Identity (12) of Lemma 11.

Case II. Suppose k ≡ 2 (mod 4). By Identity (17) of Theorem 14, it follows that

�(k) = gcd(2Q_{k/2}P_{k/2+1}, 2Q_{k/2}P_{k/2+2}, 2Q_{k/2}P_{k/2+3}, …)
= 2Q_{k/2} · gcd(P_{k/2+1}, P_{k/2+2}, P_{k/2+3}, …)
= 2Q_{k/2},

where the last equality holds by Identity (11) of Lemma 11.

Case III. Suppose k is odd. Assume by way of contradiction that p^{j} divides �(k) for some prime p with j ≥ 1. By Equation (22), it follows that p^{j} divides σ_{Q}(k, n) for all n ≥ 0. Hence p^{j} divides both σ_{Q}(k, 1) − σ_{Q}(k, 0) and σ_{Q}(k, 2) − σ_{Q}(k, 1). Observe that

σ_{Q}(k, 1) − σ_{Q}(k, 0) = (Q_{2} + Q_{3} + ⋯ + Q_{k+1}) − (Q_{1} + Q_{2} + ⋯ + Q_{k}) = Q_{k+1} − 1, and
σ_{Q}(k, 2) − σ_{Q}(k, 1) = (Q_{3} + Q_{4} + ⋯ + Q_{k+2}) − (Q_{2} + Q_{3} + ⋯ + Q_{k+1}) = Q_{k+2} − 3.

Hence p^{j} divides both Q_{k+1} − 1 and Q_{k+2} − 3. Since Q_{k} + 2Q_{k+1} = Q_{k+2}, we have

Q_{k} = Q_{k+2} − 2Q_{k+1}

= Q_{k+2} − 3 − 2Q_{k+1} + 3
= (Q_{k+2} − 3) − 2(Q_{k+1} − 1) + 1,

and therefore p^{j} divides Q_{k} − 1, and thus Q_{k} ≡ 1 (mod p^{j}). Also, since p^{j} divides Q_{k+2} − 3, we have Q_{k+2} ≡ 3 (mod p^{j}). Therefore, Q_{k}Q_{k+2} ≡ 3 (mod p^{j}). Moreover, since p^{j} divides Q_{k+1}−1, we have p^{j} divides Q_{k+1}^{2}−1, and thus Q_{k+1}^{2}−2 ≡ −1 (mod p^{j}). By the associated Pell Cassini Identity, Lemma 10, we have Q_{k}Q_{k+2} = Q_{k+1}^{2} − 2 since k is odd. It follows that 3 ≡ −1 (mod p^{j}), and hence p^{j} divides 4, so p = 2 is forced. Since p^{j} divides Q_{k+2} − 3 and we know p = 2, this implies Q_{k+2} − 3 is even. However, for any k we have Q_{k+2} being odd and hence Q_{k+2} − 3 is also odd, which yields a contradiction. Thus there exists no prime that divides �(k) when k is odd, and hence �(k) = 1 for all odd k.

These are the braids for braid sequences (�(k))_{k≥1} in solid blue and (�(k))_{k≥1} in dotted red. In Theorems 17 and 18, respectively, we give the proofs of these two braids.

\begin{matrix}Q_{1} & 2P_{2} & Q_{3} & 2P_{4} & Q_{5} & 2P_{6} & Q_{7} & 2P_{8} & ⋯ \\ Q_{1} & \frac{1}{2}P_{2} & Q_{3} & \frac{1}{2}P_{4} & Q_{5} & \frac{1}{2}P_{6} & Q_{7} & \frac{1}{2}P_{8} & ⋯ \end{matrix}

\begin{matrix}k & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 \\ �(k) & 1 & 1 & 7 & 6 & 41 & 35 & 239 & 204 & 1393 & 1189 & 8119 & 6930 & 47321 \\ �(k) & 1 & 4 & 7 & 24 & 41 & 140 & 239 & 816 & 1393 & 4756 & 8119 & 27720 & 47321 \end{matrix}

Table 3: The first 13 terms of the sequences (�(k))_{k≥1} and (�(k))_{k≥1}.

Theorem 17. For all k ≥ 1, the GCD of all sums of k consecutive balancing numbers is

�(k) = \{ \begin{matrix}\frac{1}{2}P_{k}, & if k is even; \\ Q_{k}, & if k is odd. \end{matrix}

Proof. Let k ≥ 1 be given. Recall σ_{B}(k, n) = ∑_{i=1}^{k} B_{n+i}. So by definition of �(k), we have

�(k) = gcd (σ_{B}(k, 0), σ_{B}(k, 1), σ_{B}(k, 2), …) .  (23)

Case I. Suppose k is even. By Identity (18) of Theorem 14, it follows that

�(k) = gcd (\frac{1}{2}P_{k}Q_{k+1}, \frac{1}{2}P_{k}Q_{k+3}, \frac{1}{2}P_{k}Q_{k+5}, …)

= \frac{1}{2}P_{k} · gcd (Q_{k+1}, Q_{k+3}, Q_{k+5}, …)

= \frac{1}{2}P_{k},

where the last equality holds by Identity (15) of Theorem 11. Case II. Suppose k is odd. By Identity (18) of Theorem 14, it follows that

�(k) = gcd (\frac{1}{2}Q_{k}P_{k+1}, \frac{1}{2}Q_{k}P_{k+3}, \frac{1}{2}Q_{k}P_{k+5}, …)

= \frac{1}{2}Q_{k} · gcd (P_{k+1}, P_{k+3}, P_{k+5}, …)

= Q_{k},

where the last equality holds by Identity (13) of Theorem 11. Theorem 18. For all k ≥ 1, the GCD of all sums of k consecutive Lucas-balancing numbers is

�(k) = \{ \begin{matrix}2P_{k}, & if k is even; \\ Q_{k}, & if k is odd. \end{matrix}

Proof. Let k ≥ 1 be given. Recall σ_{C}(k, n) = ∑_{i=1}^{k} C_{n+i}. So by definition of �(k), we have

�(k) = gcd (σ_{C}(k, 0), σ_{C}(k, 1), σ_{C}(k, 2), …) .  (24)

Case I. Suppose k is even. By Identity (19) of Theorem 14, it follows that

�(k) = gcd (2P_{k}P_{k+1}, 2P_{k}P_{k+3}, 2P_{k}P_{k+5}, …)
= 2P_{k} · gcd(P_{k+1}, P_{k+3}, P_{k+5}, …)
= 2P_{k},

where the last equality holds by Identity (14) of Theorem 11. Case II. Suppose k is odd. By Identity (19) of Theorem 14, it follows that

�(k) = gcd (Q_{k}Q_{k+1}, Q_{k}Q_{k+3}, Q_{k}Q_{k+5}, …)
= Q_{k} · gcd(Q_{k+1}, Q_{k+3}, Q_{k+5}, …)
= Q_{k},

where the last equality holds by Identity (15) of Theorem 11.

These are the braids for braid sequences (�(k))_{k≥1} in solid blue and (�(k))_{k≥1} in dotted red. In Theorem 19, we give the proof of the (�(k))_{k≥1} braid. However, we leave the proof of the (�(k))_{k≥1} braid to Section 5.

gcd(P_{1}, 1)4P_{2}  gcd(P_{3}, 3)4P_{4}  gcd(P_{5}, 5)4P_{6}  gcd(P_{7}, 7)4P_{8}  ⋯
Q_{1}  2 gcd(Q_{2}, 2) Q_{3}  2 gcd(Q_{4}, 4) Q_{5}  2 gcd(Q_{6}, 6) Q_{7}  2 gcd(Q_{8}, 8) ⋯

\begin{matrix}k & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 \\ �(k) & 2 & 2 & 2 & 4 & 2 & 2 & 2 & 8 & 2 & 2 & 2 & 12 & 2 \\ �(k) & 1 & 8 & 7 & 48 & 41 & 280 & 239 & 1632 & 1393 & 9512 & 8119 & 55440 & 47321 \end{matrix}

Table 4: The first 13 terms of the sequences (�(k))_{k≥1} and (�(k))_{k≥1}.

Theorem 19. For all k ≥ 1, the GCD of all sums of k consecutive Lucas-cobalancing numbers is

�(k) = \{ \begin{matrix}4P_{k}, & if k is even; \\ Q_{k}, & if k is odd. \end{matrix}

Proof. Let k ≥ 1 be given. Recall σ_{c}(k, n) = ∑_{i=1}^{k} c_{n+i}. So by definition of �(k), we have

�(k) = gcd (σ_{c}(k, 0), σ_{c}(k, 1), σ_{c}(k, 2), …) .  (25)

Case I. Suppose k is even. By Identity (20) of Theorem 14, it follows that

�(k) = gcd (2P_{k}P_{k}, 2P_{k}P_{k+2}, 2P_{k}P_{k+4}, …)
= 2P_{k} · gcd(P_{k}, P_{k+2}, P_{k+4}, …)
= 4P_{k},

where the last equality holds by Identity (13) of Theorem 11.

Case II. Suppose k is odd. By Identity (20) of Theorem 14, it follows that

�(k) = gcd (Q_{k}Q_{k}, Q_{k}Q_{k+2}, Q_{k}Q_{k+4}, …)
= Q_{k} · gcd(Q_{k}, Q_{k+2}, Q_{k+4}, …)
= Q_{k},

where the last equality holds by Identity (15) of Theorem 11.

5 Main results for �(k)

For the five sequences Pell (P_{n})_{n≥0}, associated Pell (Q_{n})_{n≥0}, balancing (B_{n})_{n≥0}, Lucasbalancing (C_{n})_{n≥0}, and Lucas-cobalancing (c_{n})_{n≥0}, the closed forms of the GCD of all sums of k consecutive terms involved braids of Pell and associate Pell numbers. However, in the setting of the cobalancing numbers, something much different occurs, namely we have the following closed form for the GCD of all sums of k consecutive cobalancing numbers:

�(k) = \{ \begin{matrix}gcd(P_{k}, k), & if k is even; \\ 2 gcd(Q_{k}, k), & if k is odd. \end{matrix}  (26)

To prove this, we first derive an intermediary form of �(k) in Theorem 23 of Subsection 5.1. Then we prove our main result, Identity (26), in Theorem 28 of Subsection 5.2.

5.1 An intermediary result for �(k)

In the proof of Theorem 23, we use the following easily-derived GCD result (see [?, Lemma 3.1]).

Lemma 20. Let (a_{i})_{i≥0} be a sequence of integers. Then the following identity holds:

gcd(a_{0}, a_{1}, a_{2}, a_{3} …) = gcd(a_{0}, a_{1} − a_{0}, a_{2} − a_{1}, a_{3} − a_{2}, …).

Theorem 21. For all k ≥ 1, set σ_{b}(k, n) := ∑_{i=1}^{k} b_{n+i}. Then the following identity holds:

σ_{b}(k, n) = \frac{1}{2}(B_{k+n} − B_{n} − k).  (27)

Proof. Let k ≥ 1 be given. Observe that

σ_{b}(k, n) := ∑_{i=1}^{k} b_{n+i} = ∑_{i=1}^{k+n} b_{i} − ∑_{i=1}^{n} b_{i}

= \frac{1}{4} (b_{k+n+1} − b_{k+n} − 2(k + n)) − \frac{1}{4} (b_{n+1} − b_{n} − 2n)

= \frac{1}{4} (2B_{k+n} − 2k − 2n) − \frac{1}{4} (2B_{n} − 2n)

= \frac{1}{4} (2B_{k+n} − 2B_{n} − 2k)

= \frac{1}{2} (B_{k+n} − B_{n} − k) ,

where the second equality holds by Identity (9) of Lemma 8, and the third equality holds from the identity 2B_{r} = b_{r+1} − b_{r} for all r ≥ 1 by Panda and Ray [?, Corollary 4.2].

Lemma 22. For all i, k ≥ 1, the following identity holds:

\frac{1}{2}(P_{2k+2i} − P_{2i}) − \frac{1}{2}(P_{2k+2i−2} − P_{2i−2}) = \{ \begin{matrix}2P_{k}Q_{k+2i−1}, & if k is even; \\ 2Q_{k}P_{k+2i−1}, & if k is odd. \end{matrix}

Proof. For ease of notation, set t_{i} := \frac{1}{2}(P_{2k+2i} − P_{2i}) − \frac{1}{2}(P_{2k+2i−2} − P_{2i−2}). Then we have

t_{i} = \frac{1}{2}((P_{2k+2i} − P_{2k+2i−2}) − (P_{2i} − P_{2i−2}))
= \frac{1}{2}(2P_{2k+2i−1} − 2P_{2i−1})  (by the Pell recurrence)

= P_{2k+2i−1} − P_{2i−1}

= \{ \begin{matrix}2P_{k}Q_{k+2i}, & if k is even; \\ 2Q_{k}P_{k+2i}, & if k is odd, \end{matrix}

where the last equality holds by Lemma 12, if we set s := 2k and r := 2i − 1 in the third equality. Then observe that s ≡ 0 (mod 4) if and only if k is even, and s ≡ 2 (mod 4) if and only if k is odd.

Theorem 23. For all k ≥ 1, the GCD of all sums of k consecutive cobalancing numbers is

�(k) = \{ \begin{matrix}gcd (\frac{1}{2}(B_{k} − k), P_{k}) , & if k is even; \\ gcd (\frac{1}{2}(B_{k} − k), 2Q_{k}) , & if k is odd. \end{matrix}

Proof. Let k ≥ 1 be given. Recall σ_{b}(k, n) = ∑_{i=1}^{k} b_{n+i}. So by definition of �(k), we have

�(k) = gcd (σ_{b}(k, 0), σ_{b}(k, 1), σ_{b}(k, 2), …)

= gcd (\frac{1}{2}(B_{k} − k), \frac{1}{2}(B_{k+1} − B_{1} − k), \frac{1}{2}(B_{k+2} − B_{2} − k), …) ,

where the second equality holds by Theorem 21. For ease of notation, set r_{i} := B_{k+i} −B_{i} −k and s_{i} := B_{k+i} − B_{i} − (B_{k+i−1} − B_{i−1}) and t_{i} := \frac{1}{2}(P_{2k+2i} − P_{2i}) − \frac{1}{2}(P_{2k+2i−2} − P_{2i−2}). It is clear that s_{i} = r_{i} − r_{i−1} for all i ≥ 1. Moreover by the Binet formulas, B_{i} = \frac{P_{2i}}{2} holds, so we have s_{i} = t_{i} for all i ≥ 1. Observe that

�(k) = \frac{1}{2} gcd (r_{0}, r_{1}, r_{2}, …)
= \frac{1}{2} gcd (B_{k} − k, s_{1}, s_{2}, …)  (by Lemma 20)

= \frac{1}{2} gcd (B_{k} − k, gcd(s_{1}, s_{2}, …))

= \frac{1}{2} gcd (B_{k} − k, gcd(t_{1}, t_{2}, …)) .

Case I. Suppose k is even. Then t_{i} = 2P_{k}Q_{k+2i−1} for all i ≥ 1 by Lemma 22. It follows that

�(k) = \frac{1}{2} gcd (B_{k} − k, gcd(2P_{k}Q_{k+1}, 2P_{k}Q_{k+3}, 2P_{k}Q_{k+5}, …)

= \frac{1}{2} gcd (B_{k} − k, 2P_{k} · gcd(Q_{k+1}, Q_{k+3}, Q_{k+5}, …))

= \frac{1}{2} gcd (B_{k} − k, 2P_{k})

= gcd (\frac{1}{2}(B_{k} − k), P_{k}) ,

where the third equality holds by Identity (15) of Theorem 11.

Case II. Suppose k is odd. Then t_{i} = 2Q_{k}P_{k+2i−1} for all i ≥ 1 by Lemma 22. It follows that

�(k) = \frac{1}{2} gcd (B_{k} − k, gcd(2Q_{k}P_{k+1}, 2Q_{k}P_{k+3}, 2Q_{k}P_{k+5}, …))

= \frac{1}{2} gcd (B_{k} − k, 2Q_{k} · gcd(P_{k+1}, P_{k+3}, P_{k+5}, …))

= \frac{1}{2} gcd (B_{k} − k, 4Q_{k})

= gcd (\frac{1}{2}(B_{k} − k), 2Q_{k}) ,

where the third equality holds by Identity (13) of Theorem 11.

5.2 Our main result for �(k) involving gcd(P_{k}, k) and gcd(Q_{k}, k)

To prove the results in this subsection, we use the p-adic valuation function and some of its well-known properties in Lemma 25 whose proofs we omit. In this subsection, we sometimes apply Lemma 25 without reference.

Definition 24. For each n ≥ 1 and p a prime, the p-adic valuation of n, denoted ν_{p}(n), is the smallest nonnegative integer k such that p^{k} divides n.

Lemma 25. For all a, b ∈ ℤ and p a prime, the following identities hold:

ν_{p}(gcd(a, b)) = min (ν_{p}(a), ν_{p}(b)) ,  (28)
ν_{p}(a · b) = ν_{p}(a) + ν_{p}(b).  (29)

Lemma 26. For all k ≥ 1, we have ν_{2}(P_{k}) = ν_{2}(k).

Proof. This follows from a more general result by Sanna (see [?, Theorem 1.5]).

Lemma 27. For all k ≥ 2, we have ν_{2}(P_{k}Q_{k} − k) ≥ 2.

Proof. We first claim that if ℓ ≥ 4 is even, then ν_{2}(P_{ℓ} − ℓ) ≥ 3 holds. Observe that

P_{ℓ} = ∑_{i=1}^{ℓ/2} \binom{ℓ}{2i − 1}2^{i−1} ≡ ℓ + \binom{ℓ}{3}2 + \binom{ℓ}{5}2^{2} (mod 8),  (30)

where the first equality holds by [?, Identity (9.10)]. For ℓ ≥ 4, we have \binom{ℓ}{3} = \frac{ℓ·(ℓ−1)·(ℓ−2)}{2·3} and thus ν_{2}(\binom{ℓ}{3}) ≥ 2, so ν_{2}(\binom{ℓ}{3}2)= ν_{2}(\binom{ℓ}{3})+ν_{2}(2) ≥ 2+1 = 3 holds. Moreover for ℓ ≥ 6, we have \binom{ℓ}{5} = \frac{ℓ·(ℓ−1)·(ℓ−2)·(ℓ−3)·(ℓ−4)}{2^{3}·15} and thus ν_{2}(\binom{ℓ}{5}) ≥ 1, so ν_{2}(\binom{ℓ}{5}2^{2}) = ν_{2}(\binom{ℓ}{5}) + ν_{2}(2^{2}) ≥ 1 + 2 = 3 holds. Hence 8 divides both \binom{ℓ}{3}2 and \binom{ℓ}{5}2^{2}, and so Identity (30) implies P_{ℓ}−ℓ is also divisible by 8, and thus ν_{2}(P_{ℓ} − ℓ) ≥ 3, as desired. Now let k ≥ 2 be given. Observe that

ν_{2}(P_{k}Q_{k} − k) = ν_{2}(2P_{k}Q_{k} − 2k) − 1 = ν_{2}(P_{2k} − 2k) − 1 ≥ 3 − 1 = 2,

where the second equality holds since P_{2k} = 2P_{k}Q_{k}.

Theorem 28. For all k ≥ 1, the GCD of all sums of k consecutive cobalancing numbers is

�(k) = \{ \begin{matrix}gcd(P_{k}, k), & if k is even; \\ 2 gcd(Q_{k}, k), & if k is odd. \end{matrix}

Proof. By Theorem 23, it suffices to show that gcd (\frac{1}{2}(B_{k} − k), P_{k}) = gcd(P_{k}, k) when k is even, and that gcd (\frac{1}{2}(B_{k} − k), 2Q_{k}) = 2 gcd(Q_{k}, k) when k is odd.

Case I. Suppose k is even. We claim that ν_{p} (gcd(\frac{1}{2}(B_{k} − k), P_{k})) = ν_{p}(gcd(P_{k}, k)) for all primes p. Suppose p ≠ 2 and j ≥ 1. Then the following sequence of biconditionals holds:

p^{j} divides gcd (\frac{1}{2}(B_{k} − k), P_{k}) ⟺ p^{j} divides P_{k} and B_{k} − k
⟺ p^{j} divides P_{k} and k
⟺ p^{j} divides gcd(P_{k}, k),

where the first biconditional holds since p is an odd prime, and the second one holds since B_{k} = P_{k}Q_{k}. Thus ν_{p} (gcd(\frac{1}{2}(B_{k} − k), P_{k})) = ν_{p}(gcd(P_{k}, k)).

On the other hand, suppose p = 2. Since Q_{k} is odd, k is even, and ν_{2}(P_{k}) = ν_{2}(k) by Lemma 26, we have ν_{2}(P_{k}Q_{k} − k) > ν_{2}(P_{k}). It follows that ν_{2}(P_{k}Q_{k} − k) ≥ ν_{2}(P_{k}) + 1, and so ν_{2}(P_{k}Q_{k} − k) − 1 ≥ ν_{2}(P_{k}). Since B_{k} = P_{k}Q_{k}, then ν_{2}(B_{k} − k) = ν_{2}(P_{k}Q_{k} − k), and thus ν_{2} (\frac{1}{2}(B_{k} − k)) = ν_{2}(P_{k}Q_{k} − k) − 1 ≥ ν_{2}(P_{k}). By Identity (28) of Lemma 25, we have

ν_{2} (gcd (\frac{1}{2}(B_{k} − k), P_{k})) = min (ν_{2} (\frac{1}{2}(B_{k} − k)) , ν_{2}(P_{k}))

= ν_{2}(P_{k})
= min (ν_{2}(P_{k}), ν_{2}(k))
= ν_{2}(gcd(P_{k}, k)).

Case II. Suppose k is odd. We claim that ν_{p} (gcd(\frac{1}{2}(B_{k} − k), 2Q_{k})) = ν_{p}(2 gcd(Q_{k}, k)) for all primes p. Suppose p ≠ 2 and j ≥ 1. Then the following sequence of biconditionals holds:

p^{j} divides gcd (\frac{1}{2}(B_{k} − k), 2Q_{k}) ⟺ p^{j} divides Q_{k} and B_{k} − k
⟺ p^{j} divides Q_{k} and k
⟺ p^{j} divides gcd(Q_{k}, k)
⟺ p^{j} divides 2 gcd(Q_{k}, k),

where the first and fourth biconditionals hold since p is an odd prime, and the second one holds since B_{k} = P_{k}Q_{k}. Thus ν_{p} (gcd(\frac{1}{2}(B_{k} − k), 2Q_{k})) = ν_{p}(2 gcd(Q_{k}, k)).

On the other hand, suppose p = 2. Since Q_{k} and k are odd, we have gcd(Q_{k}, k) is odd and hence ν_{2}(2 gcd(Q_{k}, k)) = 1. Since B_{k} = P_{k}Q_{k}, Lemma 27 implies that ν_{2}(B_{k} − k) ≥ 2, and so ν_{2} (\frac{1}{2}(B_{k} − k)) ≥ 1. By Identity (28) of Lemma 25, we have

ν_{2} (gcd (\frac{1}{2}(B_{k} − k), 2Q_{k})) = min (ν_{2} (\frac{1}{2}(B_{k} − k)) , ν_{2}(2Q_{k})) = 1,

where the last equality holds since ν_{2} (\frac{1}{2}(B_{k} − k)) ≥ 1, and Q_{k} being odd implies ν_{2}(2Q_{k}) = 1. It follows that ν_{2} (gcd(\frac{1}{2}(B_{k} − k), 2Q_{k})) = 1 = ν_{2}(2 gcd(Q_{k}, k)).

6 Further directions and open questions

6.1 GCD of sums of k consecutive squares

Recall in Convention 1, we use the notation �^{m}(k) to denote the GCD of all sums of k consecutive m^{th} powers of sequence terms. Based on evidence collected from the software Mathematica, we have the following closed forms for the GCD of all sums of k consecutive squares of Pell and associated Pell numbers. We present this proof in a future paper.

Partial Result 29. For all k ≥ 1, the GCDs of all sums of k consecutive squares of Pell and associated Pell numbers, respectively, are

�^{2}(k) = \{ \begin{matrix}\frac{1}{2}P_{k}, & if k is even; \\ 1, & if k is odd, \end{matrix} and �^{2}(k) = \{ \begin{matrix}P_{k}, & if k is even; \\ 1, & if k is odd. \end{matrix}