Arithmetic Functions of Balancing Numbers
aa r X i v : . [ m a t h . N T ] M a r ARITHMETIC FUNCTIONS OF BALANCING NUMBERS
MANASI KUMARI SAHUKAR AND G.K.PANDA
Abstract.
Two inequalities involving the Euler totient function and the sum of the k -thpowers of the divisors of balancing numbers are explored. Key words:
Balancing numbers, Euler totient function, Arithmetic functions introduction
For any positive integer n , the Euler totient function φ ( n ) is defined as number of positiveintegers less than n and relatively prime to n , and σ k ( n ) denote the sum of the k -th powerof divisors of n . If k = 0, σ k ( n ) reduces to the function τ ( n ), which counts the numberof positive divisors of n . For many centuries, mathematicians were more concerned on thearithmetic functions of natural numbers and solved many Diophantine equations concerningthese functions. Subsequently, some researchers focus their attention on study of arithmeticfunctions relating to binary recurrence sequences such as Fibonacci sequence, Lucas sequence,Pell sequence and associated Pell sequence.In 1997, Luca [5] showed that the Euler totient function for the homogeneous binary recur-rence sequences { u n } n ≥ satisfy the inequality φ ( | u n | ) ≥ | u φ ( n ) | for those binary recurrenceswith characteristic equations having real roots and the inequality is not valid for those re-currences with characteristic equations having complex roots. In [6], he proved that the n -thFibonacci number satisfies σ k ( F n ) ≤ F σ k ( n ) and τ ( F n ) ≥ F τ ( n ) for all n ≥
1. Motivatedby these works, we study two similar inequalities involving arithmetic functions of balancingnumbers.Recall that a natural number B is a balancing number with balancer R if the pair ( B, R )satisfies the Diophantine equation 1 + 2 + · · · + ( B −
1) = ( B + 1) + · · · + ( B + R ). If B isa balancing number then 8 B + 1 is a perfect square and its positive square root is called aLucas balancing number. The n -th balancing number is denoted by B n while the n -th Lucas-balancing number is denoted by C n . The balancing numbers satisfy the binary recurrence B n +1 = 6 B n − B n − , B = 0, B = 1 which holds for n ≥
1, while the Lucas-balancingnumbers satisfy a binary recurrence identical with that of balancing numbers, however withinitial values C = 1, C = 3. The characteristic equation of these recurrences is given by x − x + 1 = 0 whose roots are α = 3 + 2 √ β = 3 − √
2. The Binet forms of balancingand Lucas-balancing numbers are given by B n = α n − β n α − β , C n = α n + β n A >
2, the sequence arising out of the class of binary recurrence x n +1 = Ax n − x n − with initial terms x = 0, x = 1 is known as a balancing-like sequencebecause the case A = 6 corresponds to balancing sequence [9]. It is interesting to note that hen A = 2, the above recurrence relation generates the sequence of natural numbers. Further,when A = 3, the corresponding balancing-like sequence coincides with the sequence of evenindexed Fibonacci numbers. The balancing-like sequences (and hence the balancing sequence)satisfy certain identities in which they behave like natural numbers [8, 9] and hence thesesequences are considered as generalization of the sequence of natural numbers.2. Auxiliary results
To establish the inequalities concerning arithmetic functions of balancing numbers, we needthe following results. Some results of this section are new and hence we provide proofs of suchresults.The following lemma presents some basic properties of balancing numbers.
Lemma 2.1. ([8], Theorem 2.5, [10], Theorem 5.2.6) If m and n are natural numbers then (1) B m + n = B m C n + C m B n . (2) 5 n − < B n < n − for n ≥ . The following two lemmas deal with the divisibility property of balancing numbers.
Lemma 2.2. ([8], Theorem 2.8) If m and n are natural numbers then B m divides B n if andonly if m divides n . Lemma 2.3. ([8], Theorem 2.13) If m and n are natural numbers then ( B m , B n ) = B ( m,n ) ,where ( x, y ) denotes the greatest common divisor of x and y . Given any two nonzero integers A and B , we consider the second order linear recurrencesequence { w n } n ≥ defined by w n +1 = Aw n + Bw n − with initial terms w = 0 and w = 1.If A + 4 B > x − Ax − B = 0 has distinct real roots α = A + √ A +4 B , β = A −√ A +4 B and the Binet form is given by w n = α n − β n α − β . A prime p iscalled as primitive divisor of w n if p divides w n but does not divide w m for 0 < m < n .The following two lemmas deal with the existence of primitive divisors of the sequence { w n } n ≥ described in the last paragraph and the balancing sequence { B n } n ≥ . Lemma 2.4. ([13], Theorem 1) . If the roots α and β are real and n = 1 , , , , then w n contains at least one primitive divisor. Lemma 2.5.
A primitive prime factor of B n exists if n > .Proof. In Section 1, we have seen that the characteristic roots α = 3 + 2 √ β = 3 − √ B n has a primitive divisor for all n ∈ Z except possibly n ∈ { , , , } . Butone can easily check that B = 6, B = 6930 and B = 271669860 have primitive divisors 3,11 and 1153 respectively. (cid:3) Lemma 2.6. ([11], Theorem 3.2) If p is a prime of the form x ± then p divides B p − ,further if the prime p is of the form x ± then p divides B p +1 . The following lemma provides bounds for ratios of two consecutive balancing numbers.
Lemma 2.7.
For any natural number n , B n +1 B n > α . roof. Using the fact that αβ = 1, we get B n +1 − αB n = α n +1 − β n +1 α − β − α α n − β n α − β = β n − − β n +1 α − β = β n − (1 − β ) β − β = β n > (cid:3) The following corollary is a direct consequence of Lemma 2.7.
Corollary 2.8.
For all natural number n ≥ , B n > α n − . The following Lemma provides an upper bound for the n -th balancing number. Lemma 2.9.
For all natural number n ≥ , B n < α n .Proof. It follows from the Binet formula for balancing numbers that B n = α n − β n α − β < α n √ <α n . (cid:3) The following Lemma gives a comparison of the ( m + n )-th and ( m − n )-th balancing numberswith the product and ratio of the m -th and n -th balancing numbers, respectively. Lemma 2.10. If m and n are two natural numbers, then B m + n > B m B n and B m − n < B m B n .Proof. Let m and n be natural numbers. Since by Lemma 2.1, B m + n = B m C n + C m B n andfrom the definition of Lucas-balancing numbers B n < C n , it follows that B m + n > B m B n .Since B m = B ( m − n )+ n > B m − n B n , the inequality B m − n < B m B n follows. (cid:3) The next lemma gives a comparison of n k -th balancing number with the k -th power of n -thbalancing number. Lemma 2.11. B n k > B nk f or n ≥ and k ≥ . Proof.
Let m , n and k be natural numbers. Since B m ≥ B n whenever m ≥ n , and n k ≥ nk ,for all n ≥
2, it follows that B n k ≥ B nk . Now, using Lemma 2.10 and simple mathematicalinduction, it is easy to see that B nk > B nk . (cid:3) The following lemma gives certain bounds involving the arithmetic functions. For the proofof this lemma the readers are advised to go through [5] and [12].
Lemma 2.12. ([5], Lemma 3)
Let m and n be natural numbers. (1) If n ≥ · , then φ ( n ) > n log n . (2) If ≤ n < · , then φ ( n ) > n . (3) If m ≥ and k ≥ , then mφ ( m ) > σ k ( m ) m k . (4) If n is not prime, then n − φ ( n ) ≥ √ n . If n is not prime, then σ k ( n ) − n k ≥ √ n k . The following lemma deals with an inequality involving the Euler totient function of bal-ancing numbers.
Lemma 2.13.
For any natural numbers n , φ ( B n ) ≥ B φ ( n ) and equality holds only if n = 1 .Proof. Consider the binary recurrence sequence { w n } n ≥ defined just after Lemma 2.3. Luca[5] proved that if the characteristic roots α and β are real then φ ( | w n | ) ≥ | w φ ( n ) | . Since thecharacteristic roots α = 3 + 2 √ β = 3 − √ φ ( B n ) ≥ B φ ( n ) holds for all n ≥ (cid:3) The following lemma will play a very crucial role while proving an important result of thispaper.
Lemma 2.14. If n is an odd prime and B n = p γ · · · p γ t t is the canonical decomposition of B n then p i ≥ n − for i = 1 , . . . , t . Further, if the inequality σ k ( B n ) > B σ k ( n ) is satisfied for allnatural numbers k and n ≥ , then t > n −
1) log 5 .Proof.
Let p be any odd prime. By virtue of Lemma 2.6, p | B p +1 or p | B p − . Since p + 1 and p − p −
1, it follows from Lemma 2.2 that both B p − and B p +1 divide B p − and hence p | B p − . If p is one of the primes p , p , . . . , p t then p | B n and hence p | ( B p − , B n ).Since ( B p − , B n ) = B ( p − ,n ) , by virtue of Lemma 2.3, it follows that p | B ( p − ,n ) . If n ∤ p − p − , n ) = 1 and then p | B = 1 which is not possible. Thus, n | p − n isa prime, n | p + 1 or n | p − p ≡ ± n ). Clearly p = n ± p and n areboth primes and n >
2. Hence p ≥ n −
1. This proves the first part.We next prove the second part assuming that the inequality σ k ( B n ) ≥ B σ k ( n ) holds for allnatural numbers k and n ≥
2. Since B n φ ( B n ) = B n B n Q ti =1 (cid:0) − p i (cid:1) = t Y i =1 (cid:18) p i − (cid:19) , using Lemma 2.7, Lemma 2.11 and Lemma 2.12, we get t Y i =1 (cid:18) p i − (cid:19) = B n φ ( B n ) > σ k ( B n ) B nk > B σ k ( n ) B nk > B n k B nk ≥ B n k B n k > α > t X i =1 log (cid:18) p i − (cid:19) > log 5 . Since log(1 + x ) < x for all x >
0, we conclude that t X i =1 p i − > log 5 (2.2)In view of first part of the lemma, it follows that t n − > log 5 (2.3)which is equivalent to t > n −
1) log 5. (cid:3) MAIN RESULTS
In this section, we provide two important theorems dealing with arithmetic functions of thebalancing sequence. In the first theorem, we establish an inequality concerning the sum of k -th powers of divisors of balancing numbers. Theorem 3.1.
The balancing numbers satisfy σ k ( B n ) ≤ B σ k ( n ) for all n ≥ . Equality holdsonly if n = 1 .Proof. Since for k ≥ σ k ( B ) = σ k (1) = B σ k (1) , the assertion of the theorem holds for n = 1and all k ≥
1. For n ≥
2, assume to the contrary that σ k ( B n ) > B σ k ( n ) (3.1)for some k ≥ n ≥
2. Firstly, we show that Inequality (3.1) holds only if n is prime.Assume that Inequality (3.1) holds for some composite number n ≥ Case 1:
Suppose that B n < · . It is only possible when n <
13. From Lemmas 2.10,2.11, 2.12 and Inequality (3.1), it follows that B = 6 > B n φ ( B n ) > σ k ( B n ) B nk > B σ k ( n ) B n k > B σ k ( n ) − n k (3.2)which implies that 2 > σ k ( n ) − n k . Since n is not prime, it follows from Lemma 2.12 that2 > √ n k . (3.3)One can easily check that Inequality (3.3) does not hold for any composite number n . Case 2:
Suppose that B n ≥ · . Then certainly n ≥
14. From Lemmas 2.10, 2.11, 2.12and Inequality (3.1), it follows thatlog B n > B n φ ( B n ) > σ k ( B n ) B nk ≥ B σ k ( n ) B n k > B σ k ( n ) − n k . (3.4)Since α n > B n > α n − by Lemmas 2.8, 2.9 and σ k ( n ) − n k ≥ √ n k , it follows that n log α > log B n > B σ k ( n ) − n k > B √ n k ≥ α √ n k − . (3.5)Further, since √ n k ≥ n for k ≥
2, Inequality (3.5) gives n log α > α n − , (3.6)which holds only when n = 1, contradicting n ≥
14. Thus, the only possibility left is k = 1.But k = 1 implies n log α > α √ n − which is true for n <
5, which again contradicts n ≥
14. Hence Inequality (3.1) doesn’t holdfor any composite number. Hence n is prime. 5et n be any odd prime. From Lemma 2.14, it follows that n log α > log B n ≥ t X i =1 log p i ≥ t log(2 n − > n − n − n log α n − n − − log 5 > n . Hence σ k ( B n ) ≤ B σ k ( n ) for all natural numbers k and odd primes n . For n = 2, we need to show that σ k ( B ) = σ k (6) = 1+ 2 k + 3 k + 6 k ≤ B k .It is sufficient to prove that 4 · k < B k . Since 2 k + 2 ≤ k for all natural number k ≥
3, itfollows that 4 · k = 2 k +2 k < α k +2 ≤ α k < B k and for k = 1 ,
2, one can easily check that σ k ( B ) < B k . This completes the proof. (cid:3) In the following theorem, we present an inequality involving another arithmetic functionnamely the tau function of balancing numbers. We denote the number of distinct primedivisors of B n by ω ( B n ). Theorem 3.2.
For any natural number n , τ ( B n ) > B (cid:4) τ ( n )3 (cid:5) , where ⌊·⌋ denote the floor func-tion.Proof. Let n be a natural number. By virtue of Lemma 2.2, corresponding to each divisor m of n , there exist a primitive divisor of B m which divides B n and hence the number of distinctprime divisors of B n is at least the total number of divisors of n , i.e, ω ( B n ) ≥ τ ( n ) for n > n , it is easy to see that τ ( n ) ≥ ω ( n ) . Thus, τ ( B n ) ≥ ω ( B n ) ≥ τ ( n ) . (3.7)Since for each natural number n , B n ≤ n − < n − = 2 n − , it follows that B (cid:4) n (cid:5) < n − , (3.8)Now, from Inequality (3.7), we have τ ( B n ) ≥ τ ( n ) > τ ( n ) − > B (cid:4) τ ( n )3 (cid:5) . (3.9)This completes the proof. (cid:3) ACKNOWLEDGMENTSIt is our pleasure to thank the anonymous referee for his comments and suggestion thatsignificantly improved the accuracy and presentation of this paper.6 eferences [1] A. Behera and G. K. Panda,
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MSC 2010: 11B39.
Department of Mathematics, National Institute of Technology, Rourkela, Odisha, India
E-mail address : [email protected] Department of Mathematics, National Institute of Technology, Rourkela, Odisha, India
E-mail address : gkpanda [email protected] [email protected]