Sums of Squares and Products of Jacobsthal Numbers

Transcription

1 Journal of Integer Sequences, Vol , Article Sums of Squares and Products of Jacobsthal Numbers Zvonko Čerin Department of Mathematics University of Zagreb Bijenička 0 Zagreb 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 + 2J n 2 for n 2, and j 0 = 2, j 1 = 1, j n = j n 1 + 2j 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 A from [] while the numbers j k is A The properties of these numbers are summarized in [7]. For the convenience of the reader we shall now explicitly define these sequences. 1

2 The first eleven terms of the sequence J k are 0, 1, 1,, 5, 11, 21, 4, 85, 171 and 41. It is given by the formula J n = 2n 1 n. The first eleven terms of the sequence j k are 2, 1, 5, 7, 17, 1, 65, 127, 257, 511 and 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,, 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 = 24n+2 1 = J 4n+2, γ n = 24n+2 +1 = j 4n+2 and τ 5 5 n = 24n 1 = J 4n Lemma 1. For every m 0 and k 0 the following equality holds β n+1 γ n+1 β n γ n J 2 2k + 8 β n+1 τ n+1 β n τ n J 2k = J 2 2k+4n+4 + J 2 2k+4n+2 J 2 4n+4 J 2 4n 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 k + 1 AB n+2 k n A + 1C k k + 1 B n 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 P. 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. 2

3 In P 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 = j 4n+4. Lemma 2. For every m 0 and k 0 the following equality holds Proof. P τ n+2 η n+1 τ n+1 η n J 2 2k + 8 β n+1 τ n+2 β n τ n+1 J 2k = J 2 2k+4n+6 + J 2 2k+4n+4 J 2 4n+6 J 2 4n k + 1 AB n+2 k n A + 1C 1 4 k k + 1 B n n+4 k. Theorem 1. For every m 0 and k 0 the following equalities hold m J2k+2i 2 = m J2i 2 + β n J 2 k [γ n J 2k + 8τ n ], 2. if m = 2n and n = 0, 1, 2,... and m J2k+2i 2 = m J2i 2 + τ n+1 J 2 k [η n J 2k + 8β n ], 2.4 if m = 2n + 1 and n = 0, 1, 2,... Proof of 2.. P. The proof is by induction on n. When n = 0 we obtain J 2 2k = J β 0 J 2k [γ 0 J 2k + 8τ 0 ] = J 2k J 2k = J 2 2k, because J 0 = 0, β 0 = 1, γ 0 = 1 and τ 0 = 0. Assume that the relation 2. is true for n = r. Then J 2 2k+2i = J 2 2k+4r+2 + J 2 2k+4r+4 + 2r 2r J 2 2i + β r J 2 k [γ r J 2k + 8τ r ] = J 2 2k+2i = J 2 2k+4r+2 + J 2 2k+4r+4+ J 2 2i + β r+1 J 2 k [γ r+1 J 2k + 8τ r+1 ], where the last step uses Lemma 1 for n = r + 1. Hence, 2. is true for n = r + 1 and the proof is completed.

4 Proof of 2.4. P. When n = 0 we obtain J 2 2k + J 2 2k+2 = J J τ 1 J 2k [η 0 J 2k + 8β 0 ] = 17J 2 2k + 8J 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 J2k+2 2 = 16J2 2k + 8J 2k + 1 = 4J 2k which is true because it follows from the relation J n+2 = 4J n + 1. Assume that the relation 2.4 is true for n = r. Then +1 J 2 2k+2i = J 2 2k+4r+4 + J 2 2k+4r+6 + J 2 2i + τ r+1 J 2 k [η r J 2k + 8β r ] = J 2 2k+2i = J 2 2k+4r+4 + J 2 2k+4r J 2 2i + τ r+2 J 2 k [η r+1 J 2k + 8β r+1 ], where the last step uses Lemma 2 for n = r + 1. Jacobsthal odd squares The initial step in an inductive proof of the first part of our second theorem uses the following lemma. Lemma. For every k 0 the following identity holds J 2 2k+1 = 16J 2k J 2k 2 + 8J 2k + 1. Proof. P2. Let M = 2 k and P = 1 k. The difference of the left and the right hand side P 1P+1 is equal to 8M 2 5P 2 + = 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 2 2k+1 + J 2 2k+ = J 2k 4J 2k Proof. P2. 10 P 1 P + 1 4M 2 P 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 = 24n+1 1 = J 4n+1. Lemma 5. For every m 0 and k 0 the following equality holds 8J 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 2 2k+4n+5 + J 2 2k+4n+ J 2 4n+5 J 2 4n+..1 4

5 Proof. P k AB n+2 k n A + 1C k k + 1 B n 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 = 5J 4n+. Lemma 6. For every m 0 and k 0 the following equality holds Proof. P 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 ] k = J 2 2k+4n+7 + J 2 2k+4n+5 J 2 4n+7 J 2 4n+5..2 AB n+2 k n A + 1C k k + 1 B n n+4 k+1. Theorem 2. For every m 0 and k 0 the following equalities hold m m J2k+2i+1 2 = J2i β n J 2 k [2γ n J 2k 2 + π n ],. if m = 2n and n = 0, 1, 2,... and m m J2k+2i+1 2 = J2i τ n+1 J 2 k [2η n J 2k 2 + σ n ],.4 if m = 2n + 1 and n = 0, 1, 2,... Proof of.. P. When n = 0 we obtain J 2 2k+1 = J β 0 J 2k [2γ 0 J 2k 2 + π 0 ] = 1 + 8J 2k [2J 2k 2 + 1], because J 1 = 1, β 0 = 1, γ 0 = 1 and π 0 = 1. But, this equality is true by Lemma. Assume that the relation. is true for n = r. Then J 2 2k+2i+1 = J 2 2k+4r+5 + J 2 2k+4r+ + 2r + 8β n J 2 k [2γ n J 2k 2 + π n ] = J 2 2k+2i+1 = J 2 2k+4r+5 + J 2 2k+4r+ + where the last step uses Lemma 5 for n = r r J 2 2i+1 J 2 2i+1 + 8β n+1 J 2 k [2γ n+1 J 2k 2 + π n+1 ], 5

6 Proof of.4. P. The proof is by induction on n. When n = 0 we obtain J 2 2k+1 + J 2 2k+ = J J 2 + 8τ 1 J 2 k [2η 0 J 2k 2 + σ 0 ] = J 2 k [4J 2k ], since J 1 = 1, J =, τ 1 = 1, η 0 = 17 and σ 0 = 15. The above equality is true by Lemma 4. Assume that the relation.4 is true for n = r. Then +1 J 2 2k+2i+1 = J 2 2k+4r+7 + J 2 2k+4r+5 + J 2 2k+4r+7 + J 2 2k+4r+5 + = J 2 2k+2i+1 = J 2 2i+1 + 8τ r+1 J 2 k [2η r J 2k 2 + σ r ] +1 J 2 2i+1 + 8τ r+2 J 2 k [2η r+1 J 2k 2 + σ r+1 ], where the last step uses Lemma 6 for n = r 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 = 8J 2k J 2k J 2k+1 + J 2k+2 J 2k+ = + J 2k 16J 2k Proof of 4.1. By the formula J k = 2k 1 k we have J 2k+1 = 22k+1 1 2k+1 = 22k = 8 22k 2 1 2k 2 + = 8J 2k 2 +. Proof of 4.2. P2. P 1P+1 55M 2 46P 2 +. 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+ J 2k+4n+2 J 4n+5 J 4n+4 J 4n+ J 4n

7 Proof. P k + 1 AB n+2 k n+1 A + 1C k k + 1 B n 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. For every m 0 and k 0 the following equality holds Proof. P 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 k AB n+2 k n+1 A + 1C k k + 1 B n n+4 k. For any n = 0, 1, 2,... let µ n = 24n+ +1 = J 4n+ and ν n = 524n+5 +1 = 5J 4n+5. Theorem. For every m 0 and k 0 the following equalities hold m J 2k+2i J 2k+2i+1 = m J 2i J 2i+1 + β n J 2 k [8γ n J 2k 2 + µ n ], 4.5 if m = 2n and n = 0, 1, 2,... and m J 2k+2i J 2k+2i+1 = m J 2i J 2i+1 + τ n+1 J 2 k [8η n J 2k 2 + ν n ], 4.6 if m = 2n + 1 and n = 0, 1, 2,... Proof of 4.5. P. For n = 0 the relation 4.5 is J 2k J 2k+1 = J 0 J 1 + β 0 J 2k 8γ 0 J 2k 2 + µ 0 = J 2k 8J 2k 2 + which is true since J 2k+1 = 8J 2k 2 + by the relation 4.1 in Lemma 7. 7

8 Assume that the relation 4.5 is true for n = r. Then J 2k+2i J 2k+2i+1 = J 2k+4r+4 J 2k+4r+5 + J 2k+4r+2 J 2k+4r+ + J 2k+4r+4 J 2k+4r+5 + J 2k+4r+2 J 2k+4r+ + = where the last step uses Lemma 8 for n = r r Proof of 4.6. P. For n = 0 the relation 4.6 is 2r J 2k+2i J 2k+2i+1 = J 2i J 2i+1 + β n J 2 k [8γ n J 2k 2 + µ n ] J 2i J 2i+1 + β n+1 J 2 k [8γ n+1 J 2k 2 + µ n+1 ], J 2k J 2k+1 + J 2k+2 J 2k+ = J 0 J 1 + J 2 J + τ 1 J 2k 8η 0 J 2k 2 + ν 0 = + J 2k 16J 2k which is true by 4.2 in Lemma 7. Assume that the relation 4.6 is true for n = r. Then +1 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 + = +1 where the last step uses Lemma for n = r Alternating Jacobsthal even squares J 2k+2i J 2k+2i+1 = J 2i J 2i+1 + τ n+1 J 2 k [8η n J 2k 2 + ν n ] J 2i J 2i+1 + τ n+2 J 2 k [8η n+1 J 2k 2 + ν n+1 ], 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 Proof. By the formula J k = 2k 1 k we get J 2k = 4J 2k J 2k = 4 22k 2 1 2k = 22k 1 2k = J 2k. 8

9 Lemma 11. For every k 0 we have Proof. P2. J 2 2k+2 J 2 2k = 1 + J 2k 60J 2k P 1P+1 2M 2 20P 2 +. Let τ0 = 1 and τn+1 τn = 2 4n n 1, for n = 0, 1, 2,... For any n = 0, 1, 2,... let γn = 28n+4 +1 = j 8n Lemma 12. For every k 0 and every n 0 we have J 2k 4 γ n+1 γ nj 2k 2 + τ n+1 τ n = J 2 2k+4n+4 J 2 2k+4n+2 J 2 4n+4 + J 2 4n Proof. P k + 1 AB n+2 k 1 AC k k + 1 B k n+1 C. Let β0 = 2 and βn+1 βn = 2 4n n 1 for any n = 0, 1, 2,... For the same values of n, let ηn = 28n+8 1 = J n+8. Lemma 1. For every k 0 and every n 0 we have J 2k 4 η n+1 η nj 2k 2 + β n+1 β n = J 2 2k+4n+6 J 2 2k+4n+4 + J 2 4n+4 J 2 4n Proof. P k + 1 AB n+2 k 1 AC k k + 1 B k n+1 C. Theorem 4. For every m 0 and k 0 the following equalities hold m 1 i J2k+2i 2 = m 1 i J2i 2 + J 2 k [4γn J 2k 2 + τn], 5.5 if m = 2n and n = 0, 1, 2,... and m 1 i J2k+2i 2 = m 1 i J2i 2 J 2 k [4ηn J 2k 2 + βn], 5.6 if m = 2n + 1 and n = 0, 1, 2,...

10 Proof of 5.5. P. For n = 0 the relation 5.5 is J 2 2k = J J 2k 4γ 0 J 2k 2 + τ 0 = J 2k 4J 2k 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 2r 1 i J2k+2i 2 = Q 2 2k+4r+4 Q 2 2k+4r i J2k+2i 2 2r = Q 2 2k+4r+4 Q 2 2k+4r i J2i 2 + J 2 k [4γn J 2k 2 + τn] = where the last step uses Lemma 12. Proof of 5.6. P. For n = 0 the relation 5.6 is 1 i J 2 2i + J 2 k [ 4γ n+1 J 2k 2 + τ n+1], J 2 2k J 2 2k+2 = J 2 0 J 2 2 J 2 k [4η 0 J 2k 2 + β 0] = 1 J 2 k [60J 2k 2 + 2], which is true by Lemma 11. Assume that the relation 5.6 is true for n = r. Then +1 1 i J 2 2k+2i = J 2 2k+4r+4 J 2 2k+4r+6 + = J 2 2k+4r+4 J 2 2k+4r+6 + where the last step uses Lemma 1. = 1 i J 2 2k+2i 1 i J 2 2i J 2 k [4η n J 2k 2 + β n] +1 6 Alternating Jacobsthal odd squares 1 i J 2 2i J 2 k [ 4η n+1 J 2k 2 + β n+1], 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 2 2k+1 = 1 + 8J 2k [2J 2k 2 + 1]. 6.1 Proof. P2. P 1P+1 8M 2 5P

11 Lemma 15. For every k 0 we have Proof. P2. J 2 2k+ J 2 2k+1 = 8 + 8J 2k 0J 2k P 1P+1 1M 2 10P 2 +. Let τ 0 = 1 and τ n+1 τ n = n n+ 2 4n 1 for n = 0, 1, 2,... Lemma 16. For every k 0 and every n 0 we have 8J 2k 2 γ n+1 γ nj 2k 2 + τ n+1 τ n = J 2 2k+4n+5 J 2 2k+4n+ J 2 4n+5 + J 2 4n+. 6. Proof. P k + 1 AB n+2 k k k + 1 AC+ B k n C. Let β 0 = 1 and β n+1 β n = n n+5 2 4n 1 for any n = 0, 1, 2,... Lemma 17. For every k 0 and every n 0 we have 8J 2k 2 η n+1 η nj 2k 2 + β n+1 β n = J 2 2k+4n+7 J 2 2k+4r+5 + J 2 4n+5 J 2 4r Proof. P k AB n+2 k k k + 1 AC+ B k n C. Theorem 5. For every m 0 and k 0 the following equalities hold m 1 i J2k+2i+1 2 = m 1 i J 2 2i+1 + 8J 2 k [2γ n J 2k 2 + τ n ], 6.5 if m = 2n and n = 0, 1, 2,... and m 1 i J2k+2i+1 2 = m 1 i J 2 2i+1 8J 2 k [2η n J 2k 2 + β n ], 6.6 if m = 2n + 1 and n = 0, 1, 2,... 11

12 Proof of 6.5. P. For n = 0 the relation 6.5 is J 2 2k+1 = J J 2k 2γ 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 2r 1 i J2k+2i+1 2 = J2k+4r+5 2 J2k+4r i J2k+2i+1 2 2r = J2k+4r+5 2 J2k+4r i J2i J 2 k [2γn J 2k 2 + τn ] where the last step uses Lemma 16. = 1 i J 2 2i+1 + 8J 2 k [ 2γ n+1 J 2k 2 + τ n+1], Proof of 6.6. P. The proof is again by induction on n. For n = 0 the relation 6.6 is J2k+1 2 J2 2k+ = J2 1 J 2 8J 2 k [2η0 J 2k 2 + β0 ] which is true by Lemma 15 since η0 = 15 and β0 = 1. Assume that the relation 6.6 is true for n = r. Then +1 1 i J 2 2k+2i+1 = J 2 2k+4r+5 J 2 2k+4r+7 + = J 2 2k+4r+5 J 2 2k+4r+7 + where the last step uses Lemma 17. = 1 i J 2 2k+2i+1 1 i J 2 2i+1 8J 2 k [2η n J 2k 2 + β n ] +1 1 i J 2 2i+1 8J 2 k [ 2η n+1 J 2k 2 + β n+1], 7 Alternating Jacobsthal products Lemma 18. For every k 0 we have Proof. P2. J 2k+ J 2k+2 J 2k+1 J 2k = + J 2k [120J 2k 2 + 4]. 7.1 P 1P+1 4M 2 40P 2 +. Lemma 1. For every k 0 and every n 0 we have J 2k 8 γ n+1 γnj 2k 2 + τn+1 τn = J 2k+4n+5 J 2k+4n+4 J 2k+4n+ J 2k+4n+2 J 4n+5 J 4n+4 + J 4n+ J 4n

13 Proof. P k + 1 AB n+2 k 1 AC k k + 1 B k n+1 C. Let β 0 = 4 and β n+1 β n = 7 2 4n n+ 2 4n 1 for any n = 0, 1, 2,... Lemma 20. For every k 0 and every n 0 we have Proof. P J 2k 8 η n+1 η nj 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 k AB n+2 k 1 AC k k + 1 B k n+1 C. Theorem 6. For every m 0 and k 0 the following equalities hold m m 1 i J 2k+2i J 2k+2i+1 = 1 i J 2i J 2i+1 + J 2 k [8γn J 2k 2 + τn ], 7.4 if m = 2n and n = 0, 1, 2,... and m 1 i J 2k+2i J 2k+2i+1 = if m = 2n + 1 and n = 0, 1, 2,... m Proof of 7.4. P. For n = 0 the relation 7.4 is 1 i J 2i J 2i+1 J 2 k [8η n J 2k 2 + β n ], 7.5 J 2k J 2k+1 = J 0 J 1 + J 2 k [8γ 0 J 2k 2 + τ 0 ] = J 2 k [8J 2k 2 + ] which is true by the relation 4.1 in Lemma 7 since J 0 = 0, J 1 = 1, γ 0 = 1 and τ 0 =. Assume that the relation 7.4 is true for n = r. Then 2r 1 i J 2k+2i J 2k+2i+1 = 1 i J 2k+2i J 2k+2i+1 + J 2k+4r+4 J 2k+4r+5 J 2k+4r+2 J 2k+4r+ 2r = J 2k+4r+4 J 2k+4r+5 J 2k+4r+2 J 2k+4r+ + 1 i J 2i J 2i+1 + J 2 k [8γn J 2k 2 + τ where the last step uses Lemma 1. = n ] 1 i J 2i J 2i+1 + J 2 k [ 8γ n+1 J 2k 2 + τ n+1], 1

14 Proof of 7.5. P. For n = 0 the relation 7.5 is J 2k J 2k+1 J 2k+2 J 2k+ = J 0 J 1 J 2 J i. e., the relation 7.1 which is true by Lemma 18. Assume that the relation 7.5 is true for n = r. Then J 2k [8η 0 J 2k 2 + β 0 ] = J 2k [120J 2k 2 + 4], +1 1 i J 2k+2i J 2k+2i+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 i J 2i J 2i+1 +1 [ J 2 k [8ηn J 2k 2 + βn ] = 1 i J 2i J 2i+1 J 2 k 8η n+1 J 2k 2 + βn+1], where the last step uses Lemma 20. References [1] Z. Čerin, Properties of odd and even terms of the Fibonacci sequence, Demonstratio Math., 2006, [2] Z. Čerin, On sums of squares of odd and even terms of the Lucas sequence, Proccedings of the 11th Fibonacci Conference, to appear. [] Z. Čerin, Some alternating sums of Lucas numbers, Central European J. Math. 2005, 1 1. [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, , # A15. Available at [6] Z. Čerin and G. M. Gianella, Formulas for sums of squares and products of Pell numbers, Acc. Sc. Torino - Atti Sci. Fis , to appear. [7] A. F. Horadam, Jacobsthal representation numbers, Fib. Quart. 4 16, [8] V. Rajesh and G. Leversha, Some properties of odd terms of the Fibonacci sequence, Math. Gazette, , [] N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences,

15 2000 Mathematics Subject Classification: Primary 11B; Secondary 11Y55, 05A1. Keywords: Jacobsthal numbers, Jacobsthal-Lucas numbers, integer sequences, sum of squares, combinatorial identity, Maple. Concerned with sequences A and A Received May ; revised version received January Published in Journal of Integer Sequences, January Return to Journal of Integer Sequences home page. 15