Sums of Squares and Products of Jacobsthal Numbers

Zvonko Čerin Department of Mathematics University of Zagreb Bijenička 30 Zagreb 10 000 Croatia

                         cerin@math.hr

Abstract We consider sums of squares of odd and even terms of the Jacobsthal sequence and sums of their products. We also study the analogous alternating sums. These sums are related to products of appropriate Jacobsthal numbers and several integer sequences. The formulas that we discover show that a certain translation property for these sums holds, so that in practice, only sums of initial values and the information where the summation begins are necessary.

1 Introduction

The Jacobsthal and Jacobsthal-Lucas sequences J_{n} and j_{n} are defined by the recurrence relations

J_{0} = 0, J_{1} = 1, J_{n} = J_{n−1} + 2 J_{n−2} for n ⩾ 2,

and

j_{0} = 2, j_{1} = 1, j_{n} = j_{n−1} + 2 j_{n−2} for n ⩾ 2.

In Sections 2–4 we consider sums of squares of odd and even terms of the Jacobsthal sequence and sums of their products. These sums have nice representations as products of appropriate Jacobsthal and Jacobsthal-Lucas numbers.

The numbers J_{k} appear as the integer sequence A001045 from [9] while the numbers j_{k} is A014551. The properties of these numbers are summarized in [7]. For the convenience of the reader we shall now explicitly define these sequences.

The first eleven terms of the sequence J_{k} are 0, 1, 1, 3, 5, 11, 21, 43, 85, 171 and 341. It is given by the formula J_{n} = \frac{2^{n}−(−1)^{n}}{3} .

The first eleven terms of the sequence j_{k} are 2, 1, 5, 7, 17, 31, 65, 127, 257, 511 and 1025. It is given by the formula j_{n} = 2^{n} + (−1)^{n}.

In the last three sections we look into the alternating sums of squares of odd and even terms of the Jacobsthal sequence and the alternating sums of products of two consecutive Jacobsthal numbers. These sums also have nice representations as products of appropriate Jacobsthal and Jacobsthal-Lucas numbers.

These formulas for ordinary sums and for alternating sums have been discovered with the help of a PC computer and all algebraic identities needed for the verification of our theorems can be easily checked in either Derive, Mathematica or Maple V. Running times of all these calculations are in the range of a few seconds.

Similar results for Fibonacci, Lucas, Pell, and Pell-Lucas numbers have recently been discovered by G. M. Gianella and the author in papers [1, 2, 3, 4, 5, 6]. They improved some results in [8].

2 Jacobsthal even squares

The following lemma is needed to accomplish the inductive step in the proof of the first part of our first theorem.

For any n = 0, 1, 2, … let β_{n} = \frac{2^{4n+2}−1}{3} = J_{4n+2}, γ_{n} = \frac{2^{4n+2}+1}{5} = \frac{j_{4n+2}}{5} and τ_{n} = \frac{2^{4n}−1}{15} = \frac{J_{4n}}{5} .

Lemma 1. For every m ⩾ 0 and k ⩾ 0 the following equality holds

(β_{n+1} γ_{n+1} − β_{n} γ_{n}) J_{2k}^{2} + 8 (β_{n+1} τ_{n+1} − β_{n} τ_{n}) J_{2k}

= J_{2k+4n+4}^{2} + J_{2k+4n+2}^{2} − J_{4n+4}^{2} − J_{4n+2}^{2}. (2.1)

Proof. (P1). Let A = 2^{2k}, B = 2^{8n} and C = 2^{4n}. The difference of the left hand side and the right hand side of the relation (2.1) is equal to

\frac{544}{9} ((−1)^{1+2 k} + 1) AB + \frac{40}{9} ((−1)^{4 n+2 k} + (−1)^{1+4 n}) (A + 1)C

+ \frac{272}{9} ((−1)^{4 k} + 2 (−1)^{1+2 k} + 1) B + \frac{2}{9} ((−1)^{8 n} + (−1)^{1+8 n+4 k}) .

It is obvious that all coefficients in the above expression are zero and the proof is complete.

There are essentially three types of proofs in this paper that we indicate as P1, P2 and P3.

In P1 and P2 we use the substitutions A = 2^{2k}, B = 2^{8n}, C = 2^{4n} and M = 2^{k}, P = (−1)^{k}, respectively. We prove that the difference of the left hand side and the right hand side of the relation from the statement is equal to zero with algebraic simplification to an expression that is obviously vanishing. This part is easily done with the help of a computer (for example in Maple V) so that we shall only indicate the final expression.

In P3 we argue by induction on n taking care first of the initial value n = 0 and then showing that it holds for n = r + 1 under the assumption that it is true for n = r.

The following lemma is needed to accomplish the inductive step in the proof of the second part of our first theorem.

For any n = 0, 1, 2, … let η_{n} = 2^{4n+4} + 1 = j_{4n+4}.

Lemma 2. For every m ⩾ 0 and k ⩾ 0 the following equality holds

(τ_{n+2} η_{n+1} − τ_{n+1} η_{n}) J_{2k}^{2} + 8 (β_{n+1} τ_{n+2} − β_{n} τ_{n+1}) J_{2k}
= J_{2k+4n+6}^{2} + J_{2k+4n+4}^{2} − J_{4n+6}^{2} − J_{4n+4}^{2}. (2.2)

Proof. (P1).

\frac{8704}{9} ((−1)^{1+2 k} + 1) AB + \frac{160}{9} ((−1)^{4 n+2 k} + (−1)^{1+4 n}) (A + 1)C

+ \frac{4352}{9} ((−1)^{4 k} + 2 (−1)^{1+2 k} + 1) B + \frac{2}{9} ((−1)^{8 n} + (−1)^{1+8 n+4 k}) .

Theorem 1. For every m ⩾ 0 and k ⩾ 0 the following equalities hold

∑_{i=0}^{m} J_{2k+2i}^{2} = ∑_{i=0}^{m} J_{2i}^{2} + β_{n} J_{2 k} [γ_{n} J_{2k} + 8 τ_{n}] ,  (2.3)

if m = 2n and n = 0, 1, 2, … and

∑_{i=0}^{m} J_{2k+2i}^{2} = ∑_{i=0}^{m} J_{2i}^{2} + τ_{n+1} J_{2 k} [η_{n} J_{2k} + 8 β_{n}] ,  (2.4)

if m = 2n + 1 and n = 0, 1, 2, . . ..

Proof of (2.3). (P3). The proof is by induction on n. When n = 0 we obtain

J_{2k}^{2} = J_{0}^{2} + β_{0} J_{2k} [γ_{0} J_{2k} + 8 τ_{0}] = J_{2k} J_{2k} = J_{2k}^{2},

because J_{0} = 0, β_{0} = 1, γ_{0} = 1 and τ_{0} = 0.

Assume that the relation (2.3) is true for n = r. Then

∑_{i=0}^{2(r+1)} J_{2k+2i}^{2} = J_{2k+4r+2}^{2} + J_{2k+4r+4}^{2} + ∑_{i=0}^{2r} J_{2k+2i}^{2} = J_{2k+4r+2}^{2} + J_{2k+4r+4}^{2}+

∑_{i=0}^{2r} J_{2i}^{2} + β_{r} J_{2 k} [γ_{r} J_{2k} + 8 τ_{r}] = ∑_{i=0}^{2(r+1)} J_{2i}^{2} + β_{r+1} J_{2 k} [γ_{r+1} J_{2k} + 8 τ_{r+1}] ,

where the last step uses Lemma 1 for n = r + 1. Hence, (2.3) is true for n = r + 1 and the proof is completed.

Proof of (2.4). (P3). When n = 0 we obtain

J_{2k}^{2} + J_{2k+2}^{2} = J_{0}^{2} + J_{2}^{2} + τ_{1} J_{2k} [η_{0} J_{2k} + 8 β_{0}] = 17 J_{2k}^{2} + 8 J_{2k} + 1,

since J_{0} = 0, J_{1} = 1, β_{0} = 1, η_{0} = 17 and τ_{1} = 1. The above equality is equivalent to J_{2k+2}^{2} = 16 J_{2k}^{2} + 8 J_{2k} + 1 = (4 J_{2k} + 1)^{2} which is true because it follows from the relation J_{n+2} = 4 J_{n} + 1.

Assume that the relation (2.4) is true for n = r. Then

∑_{i=0}^{2(r+1)+1} J_{2k+2i}^{2} = J_{2k+4r+4}^{2} + J_{2k+4r+6}^{2} + ∑_{i=0}^{2r+1} J_{2k+2i}^{2} = J_{2k+4r+4}^{2} + J_{2k+4r+6}^{2}+

∑_{i=0}^{2r+1} J_{2i}^{2} + τ_{r+1} J_{2 k} [η_{r} J_{2k} + 8 β_{r}] = ∑_{i=0}^{2(r+1)+1} J_{2i}^{2} + τ_{r+2} J_{2 k} [η_{r+1} J_{2k} + 8 β_{r+1}] ,

where the last step uses Lemma 2 for n = r + 1.

3 Jacobsthal odd squares

The initial step in an inductive proof of the first part of our second theorem uses the following lemma.

Lemma 3. For every k ⩾ 0 the following identity holds

J_{2k+1}^{2} = 16 J_{2k} J_{2k−2} + 8 J_{2k} + 1.

Proof. (P2). Let M = 2^{k} and P = (−1)^{k}. The difference of the left and the right hand side is equal to \frac{(P −1)(P +1)}{3} (8 M^{2} − 5 P^{2} + 3) = 0.

The initial step in an inductive proof of the second part of our second theorem uses the following lemma.

Lemma 4. For every k ⩾ 0 the following identity holds

J_{2k+1}^{2} + J_{2k+3}^{2} = 10 + 8 J_{2k} (34 J_{2k−2} + 15).

Proof. (P2). 10 (P − 1) (P + 1) (4 M^{2} − 3 P^{2} + 1).

The following lemma is needed to accomplish the inductive step in the proof of the first part of our second theorem.

For any n = 0, 1, 2, … let π_{n} = \frac{2^{4n+1}−1}{3} = J_{4n+1}.

Lemma 5. For every m ⩾ 0 and k ⩾ 0 the following equality holds

8 J_{2k} [2 (β_{n+1} γ_{n+1} − β_{n} γ_{n}) J_{2k−2} + β_{n+1} π_{n+1} − β_{n} π_{n}]

= J_{2k+4n+5}^{2} + J_{2k+4n+3}^{2} − J_{4n+5}^{2} − J_{4n+3}^{2}. (3.1)

Proof. (P1).

\frac{5440}{9} ((−1)^{1+2 k} + 1) AB + \frac{80}{9} ((−1)^{4 n+2 k+1} + (−1)^{4 n}) (A + 1)C

+ \frac{1088}{9} (4 (−1)^{4 k} + 5 (−1)^{1+2 k} + 1) B + \frac{2}{9} ((−1)^{8 n} + (−1)^{8 n+4 k+1}) .

The following lemma is needed to accomplish the inductive step in the proof of the second part of our second theorem.

For any n = 0, 1, 2, … let σ_{n} = 5 J_{4n+3}. Lemma 6. For every m ⩾ 0 and k ⩾ 0 the following equality holds

8 J_{2k} [2 (τ_{n+2} η_{n+1} − τ_{n} η_{n}) J_{2k−2} + τ_{n+2} σ_{n+1} − τ_{n+1} σ_{n}]

= J_{2k+4n+7}^{2} + J_{2k+4n+5}^{2} − J_{4n+7}^{2} − J_{4n+5}^{2}. (3.2)

Proof. (P1).

\frac{87040}{9} ((−1)^{1+2 k} + 1) AB + \frac{320}{9} ((−1)^{4 n+2 k+1} + (−1)^{4 n}) (A + 1)C

+ \frac{17408}{9} (4 (−1)^{4 k} + 5 (−1)^{1+2 k} + 1) B + \frac{2}{9} ((−1)^{8 n} + (−1)^{8 n+4 k+1}) .

Theorem 2. For every m ⩾ 0 and k ⩾ 0 the following equalities hold

∑_{i=0}^{m} J_{2k+2i+1}^{2} = ∑_{i=0}^{m} J_{2i+1}^{2} + 8 β_{n} J_{2 k} [2 γ_{n} J_{2k−2} + π_{n}] ,  (3.3)

if m = 2n and n = 0, 1, 2, … and

∑_{i=0}^{m} J_{2k+2i+1}^{2} = ∑_{i=0}^{m} J_{2i+1}^{2} + 8 τ_{n+1} J_{2 k} [2 η_{n} J_{2k−2} + σ_{n}] ,  (3.4)

if m = 2n + 1 and n = 0, 1, 2, . . .. Proof of (3.3). (P3). When n = 0 we obtain

J_{2k+1}^{2} = J_{1}^{2} + 8 β_{0} J_{2k} [2 γ_{0} J_{2k−2} + π_{0}] = 1 + 8 J_{2k} [2 J_{2k−2} + 1],

because J_{1} = 1, β_{0} = 1, γ_{0} = 1 and π_{0} = 1. But, this equality is true by Lemma 3. Assume that the relation (3.3) is true for n = r. Then

∑_{i=0}^{2(r+1)} J_{2k+2i+1}^{2} = J_{2k+4r+5}^{2} + J_{2k+4r+3}^{2} + ∑_{i=0}^{2r} J_{2k+2i+1}^{2} = J_{2k+4r+5}^{2} + J_{2k+4r+3}^{2} + ∑_{i=0}^{2r} J_{2i+1}^{2}

+ 8 β_{n} J_{2 k} [2 γ_{n} J_{2k−2} + π_{n}] = ∑_{i=0}^{2(r+1)} J_{2i+1}^{2} + 8 β_{n+1} J_{2 k} [2 γ_{n+1} J_{2k−2} + π_{n+1}] ,

where the last step uses Lemma 5 for n = r + 1.

Proof of (3.4). (P3). The proof is by induction on n. When n = 0 we obtain

J_{2k+1}^{2} + J_{2k+3}^{2} = J_{1}^{2} + J_{3}^{2} + 8 τ_{1} J_{2 k} [2 η_{0} J_{2k−2} + σ_{0}] = 10 + 8 J_{2 k} [34 J_{2k−2} + 15] , since J_{1} = 1, J_{3} = 3, τ_{1} = 1, η_{0} = 17 and σ_{0} = 15. The above equality is true by Lemma 4.

Assume that the relation (3.4) is true for n = r. Then

∑_{i=0}^{2(r+1)+1} J_{2k+2i+1}^{2} = J_{2k+4r+7}^{2} + J_{2k+4r+5}^{2} + ∑_{i=0}^{2r+1} J_{2k+2i+1}^{2} =

J_{2k+4r+7}^{2} + J_{2k+4r+5}^{2} + ∑_{i=0}^{2r+1} J_{2i+1}^{2} + 8 τ_{r+1} J_{2 k} [2 η_{r} J_{2k−2} + σ_{r}]

= ∑_{i=0}^{2(r+1)+1} J_{2i+1}^{2} + 8 τ_{r+2} J_{2 k} [2 η_{r+1} J_{2k−2} + σ_{r+1}] ,

where the last step uses Lemma 6 for n = r + 1.

4 Jacobsthal products

For the first two steps in a proof by induction of our next theorem we require the following lemma. Lemma 7. For every k ⩾ 0 the following equalities hold

J_{2k+1} = 8 J_{2k−2} + 3.  (4.1)
J_{2k} J_{2k+1} + J_{2k+2} J_{2k+3} = 3 + J_{2k} (136 J_{2k−2} + 55).  (4.2)

Proof of (4.1). By the formula J_{k} = \frac{2^{k}−(−1)^{k}}{3} we have

J_{2k+1} = \frac{2^{2k+1} − (−1)^{2k+1}}{3} = \frac{2^{2k+1} + 1}{3} = 8 · \frac{2^{2k−2} − (−1)^{2k−2}}{3} + 3 = 8 J_{2k−2} + 3.

Proof of (4.2). (P2). \frac{(P −1)(P +1)}{3} (55 M^{2} − 46 P^{2} + 9) .

With the following lemma we shall make the inductive step in the proof of the first part of our third theorem. Lemma 8. For every m ⩾ 0 and k ⩾ 0 the following equality holds

J_{2k} [8 (β_{n+1} γ_{n+1} − β_{n} γ_{n}) J_{2k−2} + β_{n+1} μ_{n+1} − β_{n} μ_{n}] = J_{2k+4n+5} J_{2k+4n+4} + J_{2k+4n+3} J_{2k+4n+2} − J_{4n+5} J_{4n+4} − J_{4n+3} J_{4n+2}. (4.3)

Proof. (P1).

\frac{2720}{9} ((−1)^{1+2 k} + 1) AB + \frac{20}{9} ((−1)^{4 n+2 k} + (−1)^{4 n+1}) (A + 1)C+

\frac{544}{9} (4 (−1)^{4 k} + 5 (−1)^{1+2 k} + 1) B + \frac{2}{9} ((−1)^{8 n+1} + (−1)^{8 n+4 k}) .

With the following lemma we shall make the inductive step in the proof of the second part of our third theorem. Lemma 9. For every m ⩾ 0 and k ⩾ 0 the following equality holds

J_{2k} [8 (τ_{n+2} η_{n+1} − τ_{n+1} η_{n}) J_{2k−2} + τ_{n+2} ν_{n+1} − τ_{n+1} ν_{n}] =

J_{2k+4n+5} J_{2k+4n+4} + J_{2k+4n+7} J_{2k+4n+6} − J_{4n+5} J_{4n+4} − J_{4n+7} J_{4n+6}. (4.4)

Proof. (P1).

\frac{43520}{9} ((−1)^{1+2 k} + 1) AB + \frac{80}{9} ((−1)^{4 n+2 k} + (−1)^{4 n+1}) (A + 1) C

+ \frac{8704}{9} (4 (−1)^{4 k} + 5 (−1)^{1+2 k} + 1) B + \frac{2}{9} ((−1)^{8 n+1} + (−1)^{8 n+4 k}) .

For any n = 0, 1, 2, … let μ_{n} = \frac{2^{4n+3}+1}{3} = J_{4n+3} and ν_{n} = \frac{5(2^{4n+5}+1)}{3} = 5 J_{4n+5}. Theorem 3. For every m ⩾ 0 and k ⩾ 0 the following equalities hold

∑_{i=0}^{m} J_{2k+2i} J_{2k+2i+1} = ∑_{i=0}^{m} J_{2i} J_{2i+1} + β_{n} J_{2 k} [8 γ_{n} J_{2k−2} + μ_{n}] ,  (4.5)

if m = 2n and n = 0, 1, 2, … and

∑_{i=0}^{m} J_{2k+2i} J_{2k+2i+1} = ∑_{i=0}^{m} J_{2i} J_{2i+1} + τ_{n+1} J_{2 k} [8 η_{n} J_{2k−2} + ν_{n}] ,  (4.6)

if m = 2n + 1 and n = 0, 1, 2, . . .. Proof of (4.5). (P3). For n = 0 the relation (4.5) is

J_{2k} J_{2k+1} = J_{0} J_{1} + β_{0} J_{2k}(8 γ_{0} J_{2k−2} + μ_{0}) = J_{2k}(8 J_{2k−2} + 3)

which is true since J_{2k+1} = 8 J_{2k−2} + 3 by the relation (4.1) in Lemma 7.

Assume that the relation (4.5) is true for n = r. Then

∑_{i=0}^{2(r+1)} J_{2k+2i} J_{2k+2i+1} = J_{2k+4r+4} J_{2k+4r+5} + J_{2k+4r+2} J_{2k+4r+3} + ∑_{i=0}^{2r} J_{2k+2i} J_{2k+2i+1} =

J_{2k+4r+4} J_{2k+4r+5} + J_{2k+4r+2} J_{2k+4r+3} + ∑_{i=0}^{2r} J_{2i} J_{2i+1} + β_{n} J_{2 k} [8 γ_{n} J_{2k−2} + μ_{n}]

= ∑_{i=0}^{2(r+1)} J_{2i} J_{2i+1} + β_{n+1} J_{2 k} [8 γ_{n+1} J_{2k−2} + μ_{n+1}] ,

where the last step uses Lemma 8 for n = r + 1. Proof of (4.6). (P3). For n = 0 the relation (4.6) is

J_{2k} J_{2k+1} + J_{2k+2} J_{2k+3} = J_{0} J_{1} + J_{2} J_{3} + τ_{1} J_{2k}(8 η_{0} J_{2k−2} + ν_{0}) = 3 + J_{2k}(136 J_{2k−2} + 55) which is true by (4.2) in Lemma 7.

Assume that the relation (4.6) is true for n = r. Then

∑_{i=0}^{2(r+1)+1} J_{2k+2i} J_{2k+2i+1} = J_{2k+4r+4} J_{2k+4r+5} + J_{2k+4r+6} J_{2k+4r+7} + ∑_{i=0}^{2r+1} J_{2k+2i} J_{2k+2i+1} =

J_{2k+4r+4} J_{2k+4r+5} + J_{2k+4r+6} J_{2k+4r+7} + ∑_{i=0}^{2r+1} J_{2i} J_{2i+1} + τ_{n+1} J_{2 k} [8 η_{n} J_{2k−2} + ν_{n}]

= ∑_{i=0}^{2(r+1)+1} J_{2i} J_{2i+1} + τ_{n+2} J_{2 k} [8 η_{n+1} J_{2k−2} + ν_{n+1}] ,

where the last step uses Lemma 9 for n = r + 1.

5 Alternating Jacobsthal even squares

In this section we look for formulas that give closed forms for alternating sums of squares of Jacobsthal numbers with even indices. Lemma 10. For every k ⩾ 0 we have

J_{2k} = 4 J_{2k−2} + 1.  (5.1)

Proof. By the formula J_{k} = \frac{2^{k}−(−1)^{k}}{3} we get

4 J_{2k−2} + 1 = 4 · \frac{2^{2k−2} − (−1)^{2k−2}}{3} + 1 = \frac{2^{2k} − (−1)^{2k}}{3} = J_{2k}.

Lemma 11. For every k ⩾ 0 we have

J_{2k+2}^{2} − J_{2k}^{2} = 1 + J_{2k} (60 J_{2k−2} + 23).  (5.2)

Proof. (P2). \frac{(P −1)(P +1)}{3} (23 M^{2} − 20 P^{2} + 3) .

Let τ_{0}^{∗} = 1 and τ_{n+1}^{∗} − τ_{n}^{∗} = 2^{4n+3} (50 · 2^{4n} − 1), for n = 0, 1, 2, . . .. For any n = 0, 1, 2, … let γ_{n}^{∗} = \frac{2^{8n+4}+1}{17} = \frac{j_{8n+4}}{17} .

Lemma 12. For every k ⩾ 0 and every n ⩾ 0 we have

J_{2k} (4 (γ_{n+1}^{∗} − γ_{n}^{∗}) J_{2k−2} + τ_{n+1}^{∗} − τ_{n}^{∗}) = J_{2k+4n+4}^{2} − J_{2k+4n+2}^{2} − J_{4n+4}^{2} + J_{4n+2}^{2}. (5.3)

Proof. (P1).

\frac{400}{3} ((−1)^{1+2 k} + 1) AB + \frac{8}{3} ((−1)^{4 n+2 k} − 1) AC+

\frac{80}{3} (4 (−1)^{4 k} + 5 (−1)^{1+2 k} + 1) B + \frac{8}{3} ((−1)^{2 k} + (−1)^{4 n+1}) C.

Let β_{0}^{∗} = 23 and β_{n+1}^{∗} − β_{n}^{∗} = 2^{4n+5} (200 · 2^{4n} − 1) for any n = 0, 1, 2, . . .. For the same values of n, let η_{n}^{∗} = \frac{2^{8n+8}−1}{17} = \frac{3}{17} J_{8n+8}.

Lemma 13. For every k ⩾ 0 and every n ⩾ 0 we have

J_{2k} (4 (η_{n+1}^{∗} − η_{n}^{∗}) J_{2k−2} + β_{n+1}^{∗} − β_{n}^{∗}) = J_{2k+4n+6}^{2} − J_{2k+4n+4}^{2} + J_{4n+4}^{2} − J_{4n+6}^{2}. (5.4)

Proof. (P1).

\frac{6400}{3} ((−1)^{1+2 k} + 1) AB + \frac{32}{3} ((−1)^{4 n+2 k} − 1) AC+

\frac{1280}{3} (4 (−1)^{4 k} + 5 (−1)^{1+2 k} + 1) B + \frac{32}{3} ((−1)^{2 k} + (−1)^{4 n+1}) C.

Theorem 4. For every m ⩾ 0 and k ⩾ 0 the following equalities hold

∑_{i=0}^{m} (−1)^{i} J_{2k+2i}^{2} = ∑_{i=0}^{m} (−1)^{i} J_{2i}^{2} + J_{2 k} [4 γ_{n}^{∗} J_{2k−2} + τ_{n}^{∗}] ,  (5.5)

if m = 2n and n = 0, 1, 2, … and

∑_{i=0}^{m} (−1)^{i} J_{2k+2i}^{2} = ∑_{i=0}^{m} (−1)^{i} J_{2i}^{2} − J_{2 k} [4 η_{n}^{∗} J_{2k−2} + β_{n}^{∗}] ,  (5.6)

if m = 2n + 1 and n = 0, 1, 2, . . ..

Proof of (5.5). (P3). For n = 0 the relation (5.5) is

J_{2k}^{2} = J_{0}^{2} + J_{2k} (4 γ_{0}^{∗} J_{2k−2} + τ_{0}^{∗}) = J_{2k} (4 J_{2k−2} + 1)

(i. e., the relation (5.1) multiplied by J_{2k}) which is true by Lemma 10.

Assume that the relation (5.5) is true for n = r. Then

∑_{i=0}^{2(r+1)} (−1)^{i} · J_{2k+2i}^{2} = Q_{2k+4r+4}^{2} − Q_{2k+4r+2}^{2} + ∑_{i=0}^{2r} (−1)^{i} · J_{2k+2i}^{2}

= Q_{2k+4r+4}^{2} − Q_{2k+4r+2}^{2} + ∑_{i=0}^{2r} (−1)^{i} · J_{2i}^{2} + J_{2 k} [4 γ_{n}^{∗} J_{2k−2} + τ_{n}^{∗}] =

∑_{i=0}^{2(r+1)} (−1)^{i} · J_{2i}^{2} + J_{2 k} [4 γ_{n+1}^{∗} J_{2k−2} + τ_{n+1}^{∗}] ,

where the last step uses Lemma 12.

Proof of (5.6). (P3). For n = 0 the relation (5.6) is

J_{2k}^{2} − J_{2k+2}^{2} = (J_{0}^{2} − J_{2}^{2}) − J_{2 k} [4 η_{0}^{∗} J_{2k−2} + β_{0}^{∗}] = −1 − J_{2 k} [60 J_{2k−2} + 23] ,

which is true by Lemma 11.

Assume that the relation (5.6) is true for n = r. Then

∑_{i=0}^{2(r+1)+1} (−1)^{i} · J_{2k+2i}^{2} = J_{2k+4r+4}^{2} − J_{2k+4r+6}^{2} + ∑_{i=0}^{2r+1} (−1)^{i} · J_{2k+2i}^{2}

= J_{2k+4r+4}^{2} − J_{2k+4r+6}^{2} + ∑_{i=0}^{2r+1} (−1)^{i} · J_{2i}^{2} − J_{2 k} [4 η_{n}^{∗} J_{2k−2} + β_{n}^{∗}]

= ∑_{i=0}^{2(r+1)+1} (−1)^{i} · J_{2i}^{2} − J_{2 k} [4 η_{n+1}^{∗} J_{2k−2} + β_{n+1}^{∗}] ,

where the last step uses Lemma 13.

6 Alternating Jacobsthal odd squares

In this section we look for formulas that give closed forms for alternating sums of squares of Jacobsthal numbers with odd indices.

Lemma 14. For every k ⩾ 0 we have

J_{2k+1}^{2} = 1 + 8 J_{2k} [2 J_{2k−2} + 1] .  (6.1)

Proof. (P2). \frac{(P −1)(P +1)}{3} (8 M^{2} − 5 P^{2} + 3) .

Lemma 15. For every k ⩾ 0 we have

J_{2k+3}^{2} − J_{2k+1}^{2} = 8 + 8 J_{2k} (30 J_{2k−2} + 13).  (6.2)

Proof. (P2). \frac{(P −1)(P +1)}{3} (13 M^{2} − 10 P^{2} + 3) .

Let τ_{0}^{∗∗} = 1 and τ_{n+1}^{∗∗} − τ_{n}^{∗∗} = 101 · 2^{4n+1} + 25 · 2^{4n+3} (2^{4n} − 1) for n = 0, 1, 2, . . .. Lemma 16. For every k ⩾ 0 and every n ⩾ 0 we have

8 J_{2k} (2 (γ_{n+1}^{∗} − γ_{n}^{∗}) J_{2k−2} + τ_{n+1}^{∗∗} − τ_{n}^{∗∗}) = J_{2k+4n+5}^{2} − J_{2k+4n+3}^{2} − J_{4n+5}^{2} + J_{4n+3}^{2}. (6.3) Proof. (P1).

\frac{1600}{3} ((−1)^{1+2 k} + 1) AB + \frac{16}{3} ((−1)^{4 n+2 k+1} + 1) AC+

\frac{320}{3} (4 (−1)^{4 k} + 5 (−1)^{1+2 k} + 1) B + \frac{16}{3} ((−1)^{2 k+1} + (−1)^{4 n}) C.

Let β_{0}^{∗∗} = 13 and β_{n+1}^{∗∗} − β_{n}^{∗∗} = 401 · 2^{4n+3} + 100 · 2^{4n+5} (2^{4n} − 1) for any n = 0, 1, 2, . . .. Lemma 17. For every k ⩾ 0 and every n ⩾ 0 we have

8 J_{2k} (2 (η_{n+1}^{∗} − η_{n}^{∗}) J_{2k−2} + β_{n+1}^{∗∗} − β_{n}^{∗∗}) = J_{2k+4n+7}^{2} − J_{2k+4r+5}^{2} + J_{4n+5}^{2} − J_{4r+7}^{2}. (6.4) Proof. (P1).

\frac{25600}{3} ((−1)^{1+2 k} + 1) AB + \frac{64}{3} ((−1)^{4 n+2 k+1} + 1) AC+

\frac{5120}{3} (4 (−1)^{4 k} + 5 (−1)^{1+2 k} + 1) B + \frac{64}{3} ((−1)^{2 k+1} + (−1)^{4 n}) C.

Theorem 5. For every m ⩾ 0 and k ⩾ 0 the following equalities hold

∑_{i=0}^{m} (−1)^{i} J_{2k+2i+1}^{2} = ∑_{i=0}^{m} (−1)^{i} J_{2i+1}^{2} + 8 J_{2 k} [2 γ_{n}^{∗} J_{2k−2} + τ_{n}^{∗∗}] ,  (6.5)

if m = 2n and n = 0, 1, 2, … and

∑_{i=0}^{m} (−1)^{i} J_{2k+2i+1}^{2} = ∑_{i=0}^{m} (−1)^{i} J_{2i+1}^{2} − 8 J_{2 k} [2 η_{n}^{∗} J_{2k−2} + β_{n}^{∗∗}] ,  (6.6)

if m = 2n + 1 and n = 0, 1, 2, . . ..

J_{2k+1}^{2} = J_{1}^{2} + 8 J_{2k}(2 γ_{0}^{∗} J_{2k−2} + τ_{0}^{∗∗})

(i. e., the relation (6.1)) which is true by Lemma 14.

Assume that the relation (6.5) is true for n = r. Then

∑_{i=0}^{2(r+1)} (−1)^{i} J_{2k+2i+1}^{2} = J_{2k+4r+5}^{2} − J_{2k+4r+3}^{2} + ∑_{i=0}^{2r} (−1)^{i} J_{2k+2i+1}^{2}

= J_{2k+4r+5}^{2} − J_{2k+4r+3}^{2} + ∑_{i=0}^{2r} (−1)^{i} J_{2i+1}^{2} + 8 J_{2 k} [2 γ_{n}^{∗} J_{2k−2} + τ_{n}^{∗∗}]

= ∑_{i=0}^{2(r+1)} (−1)^{i} J_{2i+1}^{2} + 8 J_{2 k} [2 γ_{n+1}^{∗} J_{2k−2} + τ_{n+1}^{∗∗}] ,

where the last step uses Lemma 16.

Proof of (6.6). (P3). The proof is again by induction on n. For n = 0 the relation (6.6) is J_{2k+1}^{2} − J_{2k+3}^{2} = J_{1}^{2} − J_{3}^{2} − 8 J_{2 k} [2 η_{0}^{∗} J_{2k−2} + β_{0}^{∗∗}] which is true by Lemma 15 since η_{0}^{∗} = 15 and β_{0}^{∗∗} = 13.

Assume that the relation (6.6) is true for n = r. Then

∑_{i=0}^{2(r+1)+1} (−1)^{i} J_{2k+2i+1}^{2} = J_{2k+4r+5}^{2} − J_{2k+4r+7}^{2} + ∑_{i=0}^{2r+1} (−1)^{i} J_{2k+2i+1}^{2}

= J_{2k+4r+5}^{2} − J_{2k+4r+7}^{2} + ∑_{i=0}^{2r+1} (−1)^{i} J_{2i+1}^{2} − 8 J_{2 k} [2 η_{n}^{∗} J_{2k−2} + β_{n}^{∗∗}]

= ∑_{i=0}^{2(r+1)+1} (−1)^{i} J_{2i+1}^{2} − 8 J_{2 k} [2 η_{n+1}^{∗} J_{2k−2} + β_{n+1}^{∗∗}] ,

where the last step uses Lemma 17.

7 Alternating Jacobsthal products

Lemma 18. For every k ⩾ 0 we have

J_{2k+3} J_{2k+2} − J_{2k+1} J_{2k} = 3 + J_{2k} [120 J_{2k−2} + 49] .  (7.1)

Proof. (P2). \frac{(P −1)(P +1)}{3} (49 M^{2} − 40 P^{2} + 9) .

Lemma 19. For every k ⩾ 0 and every n ⩾ 0 we have

J_{2k} (8 (γ_{n+1}^{∗} − γ_{n}^{∗}) J_{2k−2} + τ_{n+1}^{∗∗∗} − τ_{n}^{∗∗∗}) =

J_{2k+4n+5} J_{2k+4n+4} − J_{2k+4n+3} J_{2k+4n+2} − J_{4n+5} J_{4n+4} + J_{4n+3} J_{4n+2}. (7.2)

Proof. (P1).

\frac{800}{3} ((−1)^{1+2 k} + 1) AB + \frac{4}{3} ((−1)^{4 n+2 k} − 1) AC+

\frac{160}{3} (4 (−1)^{4 k} + 5 (−1)^{1+2 k} + 1) B + \frac{4}{3} ((−1)^{2 k} + (−1)^{4 n+1}) C.

Let β_{0}^{∗∗∗} = 49 and β_{n+1}^{∗∗∗} − β_{n}^{∗∗∗} = 799 · 2^{4n+4} + 25 · 2^{4n+9} (2^{4n} − 1) for any n = 0, 1, 2, . . .. Lemma 20. For every k ⩾ 0 and every n ⩾ 0 we have

J_{2k} (8 (η_{n+1}^{∗} − η_{n}^{∗}) J_{2k−2} + β_{n+1}^{∗∗∗} − β_{n}^{∗∗∗}) =

J_{2k+4n+6} J_{2k+4n+7} − J_{2k+4n+4} J_{2k+4n+5} + J_{4n+5} J_{4n+4} − J_{4n+6} J_{4n+7}. (7.3)

Proof. (P1).

\frac{12800}{3} ((−1)^{1+2 k} + 1) AB + \frac{16}{3} ((−1)^{4 n+2 k} − 1) AC+

\frac{2560}{3} (4 (−1)^{4 k} + 5 (−1)^{1+2 k} + 1) B + \frac{16}{3} ((−1)^{2 k} + (−1)^{4 n+1}) C.

Theorem 6. For every m ⩾ 0 and k ⩾ 0 the following equalities hold

∑_{i=0}^{m} (−1)^{i} J_{2k+2i} J_{2k+2i+1} = ∑_{i=0}^{m} (−1)^{i} J_{2i} J_{2i+1} + J_{2 k} [8 γ_{n}^{∗} J_{2k−2} + τ_{n}^{∗∗∗}] ,  (7.4)

if m = 2n and n = 0, 1, 2, … and

∑_{i=0}^{m} (−1)^{i} J_{2k+2i} J_{2k+2i+1} = ∑_{i=0}^{m} (−1)^{i} J_{2i} J_{2i+1} − J_{2 k} [8 η_{n}^{∗} J_{2k−2} + β_{n}^{∗∗∗}] ,  (7.5)

if m = 2n + 1 and n = 0, 1, 2, . . .. Proof of (7.4). (P3). For n = 0 the relation (7.4) is

J_{2k} J_{2k+1} = J_{0} J_{1} + J_{2 k} [8 γ_{0}^{∗} J_{2k−2} + τ_{0}^{∗∗∗}] = J_{2 k} [8 J_{2k−2} + 3]

which is true by the relation (4.1) in Lemma 7 since J_{0} = 0, J_{1} = 1, γ_{0}^{∗} = 1 and τ_{0}^{∗∗∗} = 3.

Assume that the relation (7.4) is true for n = r. Then

∑_{i=0}^{2(r+1)}(−1)^{i}J_{2k+2i}J_{2k+2i+1} = ∑_{i=0}^{2r}(−1)^{i}J_{2k+2i}J_{2k+2i+1} + J_{2k+4r+4}J_{2k+4r+5} − J_{2k+4r+2}J_{2k+4r+3}

= J_{2k+4r+4}J_{2k+4r+5} − J_{2k+4r+2}J_{2k+4r+3} + ∑_{i=0}^{2r}(−1)^{i}J_{2i}J_{2i+1} + J_{2 k} [8 γ_{n}^{∗} J_{2k−2} + τ_{n}^{∗∗∗}]

= ∑_{i=0}^{2(r+1)}(−1)^{i}J_{2i}J_{2i+1} + J_{2 k} [8 γ_{n+1}^{∗} J_{2k−2} + τ_{n+1}^{∗∗∗}] ,

where the last step uses Lemma 19.

J_{2k} J_{2k+1} − J_{2k+2} J_{2k+3} = J_{0} J_{1} − J_{2} J_{3}− J_{2k} [8 η_{0}^{∗} J_{2k−2} + β_{0}^{∗∗∗}] = −3 − J_{2k} [120 J_{2k−2} + 49] ,

(i. e., the relation (7.1)) which is true by Lemma 18. Assume that the relation (7.5) is true for n = r. Then

∑_{i=0}^{2(r+1)+1} (−1)^{i} J_{2k+2i} J_{2k+2i+1} = ∑_{i=0}^{2r+1} (−1)^{i} J_{2k+2i} J_{2k+2i+1} + J_{2k+4r+4} J_{2k+4r+5}−

J_{2k+4r+6} J_{2k+4r+7} = J_{2k+4r+4} J_{2k+4r+5} − J_{2k+4r+6} J_{2k+4r+7} + ∑_{i=0}^{2r+1} (−1)^{i} J_{2i} J_{2i+1}−

J_{2 k} [8 η_{n}^{∗} J_{2k−2} + β_{n}^{∗∗∗}] = ∑_{i=0}^{2(r+1)+1} (−1)^{i} J_{2i} J_{2i+1} − J_{2 k} [8 η_{n+1}^{∗} J_{2k−2} + β_{n+1}^{∗∗∗}] ,

where the last step uses Lemma 20.

References

[1] Z. Čerin, Properties of odd and even terms of the Fibonacci sequence, Demonstratio Math., 39 (2006), 55–60.

[2] Z. Čerin, On sums of squares of odd and even terms of the Lucas sequence, Proccedings of the 11th Fibonacci Conference, to appear.

[3] Z. Čerin, Some alternating sums of Lucas numbers, Central European J. Math. 3 (2005), 1–13.

[4] Z. Čerin, Alternating sums of Fibonacci products, Atti del Seminario Matematico e Fisico dell’Università di Modena e Reggio Emilia, (to appear).

[5] Z. Čerin and G. M. Gianella, On sums of squares of Pell-Lucas numbers, INTEGERS: Electronic Journal of Combinatorial Number Theory, 6 (2006), # A15. Available at

   http://www.integers-ejcnt.org/vol6.html.

[6] Z. Čerin and G. M. Gianella, Formulas for sums of squares and products of Pell numbers, Acc. Sc. Torino - Atti Sci. Fis. 140 (2006), to appear.

[7] A. F. Horadam, Jacobsthal representation numbers, Fib. Quart. 34 (1996), 40–54.

[8] V. Rajesh and G. Leversha, Some properties of odd terms of the Fibonacci sequence, Math. Gazette, 88 (2004), 85–86.

[9] N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences, 2006.

2000 Mathematics Subject Classification: Primary 11B39; Secondary 11Y55, 05A19. Keywords: Jacobsthal numbers, Jacobsthal-Lucas numbers, integer sequences, sum of squares, combinatorial identity, Maple.

(Concerned with sequences \underline{A001045} and \underline{A014551}.)

Received May 17 2006; revised version received January 19 2007. Published in Journal of Integer Sequences, January 19 2007.

Return to Journal of Integer Sequences home page.