Global Journal of Mathematical Sciences: Theory and Practical. ISSN 0974-3200 Volume 5, Number 2 (2013), pp. 117-122 © International Research Publication House http://www.irphouse.com
On Arithmetic Functions
Bhabesh Das
Department of Mathematics, Gauhati University, Guwahati-781014, Assam, India E-mail: [email protected]
Abstract
In this paper, we intend to establish some inequalities on Jordan totient function. Using arithmetic functions like the Dedekind’s function , the divisors function , Euler’s totient function and their compositions we gave some results on perfect numbers and generalized perfect numbers.
Mathematics Subject Classification: 11A25, 11A41, 11Y70
Keywords: perfect number, 2-hyperperfect number, super-perfect number, super-hyperperfect number.
1. Introduction Let ()n denote the sum of divisors of n i.e. ()n ∑ | , where (1) 1. It is well- known that n is said to be a perfect number if(n ) 2 n . Euclid and Euler showed that even perfect numbers are of the form n 2p1 (2 p 1) , where p and 2p 1 are primes. The prime of the form 2p 1 is called Mersenne prime and at this moment there are exactly 48 such primes, which mean that there are 48 even perfect numbers [11]. On the other hand, no odd perfect number is known, but no one has yet been able to prove conclusively that none exist. In [5] the author provided detailed information on perfect numbers. Positive integer n with the property(n ) 2 n 1 is called almost perfect number, while that of(n ) 2 n 1, quasi-perfect number. For many result and conjectures on this topic, see [4]. D.Suryanarayana [10] introduced the notation of super-perfect numbers. A positive integer n is said to be super- perfect number if ( (n )) 2 n and even super-perfect numbers are of the form 2 p1 such that 2p 1 are Mersenne primes. It is asked in this paper and still unsolved whether there are odd super-perfect numbers. Miloni and Bear [8] introduced the concept of
118 Bhabesh Das k hyperperfect number and they conjecture that there are k hyperperfect numbers for everyk . A. Beg and K. Fogarasi gave a table of k hyperperfect numbers in [1] and defined positive integer n as k hyperperfect number if ()n + .They proved that if n 3k1 (3 k 2) , where 3k 2 is prime, then n is a 2 hyperperfect number. Perfect numbers are 1 hyper- perfect numbers. If ( (n )) + , then n is called super-hyperperfect number. In [1] authors conjectured that all super-hyperperfect numbers are of the formn 3p1 , where p and are primes. The Jordan’s totient function and Dedekind’s function are respectively defined as 1 1 J( n ) nk (1 ) and (n ) n (1 ) , where p runs through the distinct k pk p prime divisors ofn . Let by convention Jk (1) 1, (1) 1. Euler’s totient function is 1 defined as (n ) n (1 ) i.e. J()() n n . All these functions are multiplicative, p 1 i.e. satisfy the functional equation f()()() mn f m f n for g.c.d(m , n ) 1. In [9] the author defines -perfect, -perfect, -perfect numbers, where "" denotes composition.
Basic symbols and notations: ()n Sum of divisors of n , ()n Dedekind’s arithmetic function,
Jk () n Jordan totient function, ()n Euler’s totient function, a b a divides b, a Œbb a does not divide b, pr{} n set of distinct prime divisors of n .
Main Results k Proposition2.1. Jk()() ab a J k b , for any a, b 2 , k , with equality only if, pr{}{} a pr b , where pr{} a denotes the set of distinct prime factors of a Proof: Let 1 1 1 J() ab ak b k (1 )(1 )(1 ) ab p.. q r , then k k k k p a,, pŒŒ bq a, q b r a r b p q r 1 1 1 Since (1 ) 1, so J( ab ) ak b k (1 ) (1 ) a k J ( b ) Thus, pk kqk r k k k Jk()() ab a J k b , for any a, b 2 . If pr{}{} a pr b , then ab p. q and p a, p b qŒ a, q b k hence Jk()() ab a J k b .
On Arithmetic Functions 119
k Proposition2.2. If pr{}{} a pr b , then Jk( ab ) ( a 1) J k ( b ) for any a, b 2 , k .
Proof: Let a p q and b p r , where p are the common prime factors, and q pr{} a are such that q pr{} b , so 1 and , , 0 and 1 1 1 1 pr{}{} a pr b . J() ab ak b k (1 )(1 )(1 ) a k (1 )() J b kpk q k r k q k k 1 1 1 1 1 But, 1 1 1 1 (1 ) so, k k k k k a() p q q q q 1 1 Jaba()k (1 )() Jba k (1 )()( Jba k 1)() Jb This implies, k qk k a k k k k Jk( ab ) ( a 1) J k ( b ) , for any a, b 2 .
Proposition2.3. Let n 3, and suppose that n is -deficient number, then k k Jk ( ( n )) (2 1) n .
Proof: If n is a -deficient number, then (n ) 2 n . Again for any n 3, ()n is an even number. Using Proposition 2.1, we get 1 J(2) a ak J (2) a k 2(1 k ) a k (2 k 1) Let u 2 a , an even number, then k k 2k uk { (n )}k (2n )k J( u ) (2k 1) .Therefore, J( ( n )) (2k 1) (2k 1) (2 k 1)n k k 2k k 2k 2k .
k Proposition 2.4. For any n 3, Jk ( ( n )) n
Proof: First remark that for any n 3, ()n is even number and ()n n . Let (n ) 2 u , an even integer. Proceeding as Proposition 2.3, we obtain (2k 1) (2 k 1) J(2 u ) uk (2 k 1) , i.e. J(()){()}. n nk n k n k k k 2k 2 k
k k Proposition2.5. If n is an even perfect number, then Jk ( ( n )) (2 1) n .
Proof: If n is an even perfect number, then n 2p1 (2 p 1), where 2p 1 is a k k k Mersenne prime and (n ) 2 n , so Jk(()) n j k (2) n n j k (2) (2 1) n .
Proposition2.6. If n be the product of distinct even perfect numbers n1, n 2 ,..... nr , then 2 a. (a) ()n ()r1 ( n )( n ).....( n ) 3 1 2 r
120 Bhabesh Das b. (n ( n )) 2r n2 . r1 c. ( (n )) 2 ( ( n1 )) ( ( n 2 ))..... ( ( nr ))
pi1 p i Proof: Let n n1 n 2... nr , where ni 2 (2 1) are distinct perfect numbers with
pi Mersenne primes 2 1( pi primes, i1,2,...... , r )
2 pi 2 (2p11) (2 p 2 1).....(2 pr 2) a. (ni ) 3.2 for each i , so (n ) 3.2 2 ()r 1 (n )( n ).....( n ) . 3 1 2 r
pi1 p i 1 r1 b. (ni ) 2 (2 1) for each i , and clearly (n ) 2 ( n1 ). ( n 2 )...... ( nr ) , so (n ) 2r1 n .Therefore (n ( n )) ( n ) ( n ) 2r n2 .
2(pi 1) (2p11) (2 p 2 1) .....(2 pr 2) ( (ni )) 2 , for eachi , so( (n )) (3.2 ) = 2(2p11) (2 p 2 1) .....(2 pr 2) r1 = 2 ( (n1 ))( ( n 2 )).....( ( nr ))
Proposition 2.7. If n be the product of distinct 2-hyperperfect numbers n1, n 2 ,..... nr , then 3 a. ()n ()r1 ( n )( n ).....( n ) 2 1 2 r 3 b. ()n ()r1 ( n )( n ).....( n ) 4 1 2 r
ki1 k i Proof: Let n n1 n 2... nr , where ni 3 (3 2) distinct 2-hyperperfect numbers with primes are 3ki 2 ( i1,2,3,..... r ).
ki1 k i 1 a. (ni ) 2.3 (3 1) , for eachi .
2 Then (n ) .3(1)(1).....(1)k1 k 2 kr (3 k 1 1 1)(3 k 2 1 1)...... (3 k r 1 1) 3 3 ( )r1 (n )( n )...... ( n ) (b) (n ) 22 3ki2 (3 k i 1) , for each i . 2 1 2 r i 4 3 (n ) .3(k1 1) ( k 2 1) .... ( kr 1) (3 k 1 1)(3 k 2 1)...... (3 k r 1) ()r1 (n )( n ).....( n ) . 3 4 1 2 r
( (n )) Proposition2.8. If n is a super-hyperperfect number, then( (n )) . 3
Proof: From definition of super-hyperperfect number we know that n 3p1 , where p 3p 1 4n and are primes. (n ) 4.3 p2 , so ( (n )) 4.3p3 and (n ) 2.3p2 , so 2 9
On Arithmetic Functions 121
4n ( (n )) ( (n )) 4.3p2 .Therefore ( (n )) . 3 3
Proposition2.9. If n be the product of distinct super-hyperperfect numbers n1, n 2 ,..... nr , then 3r2 a. (()) n .( n )( n )...... ( n ) 2r 1 1 2 r 3r 1 b. (()) n .()( n n ).....( n ) 22r 3 1 2 r
pi 1 Proof: Let n n1 n 2... nr , where ni 3 are distinct super-hyperperfect numbers with 3pi 1 p and are primes (i 1,2,3,..... r .). Again (n ) 2.3pi 2 and (n ) 4.3pi 2 . i 2 i i
(p 1) ( p 1) ...... ( p 1) 2 (p 1) ( p 1) ...... ( p 1) (p 1) ( p 1) ...... ( p 2) a. (n ) (31 2 r ) .3 1 2 r 2.3 1 2 r , so, 3 r2 2 (p 1) ( p 1) ...... ( p 1) 3 ( (n )) .3 1 2 r . (n ) ( n )...... ( n ) . 9 2r1 1 2 r
(p 1) ( p 1) ...... ( p 1) 4 (p 1) ( p 1) ...... ( p 1) (p 1) ( p 1) ...... ( p 2) (n ) (31 2 r ) .3 1 2 r 4.3 1 2 r , so b. 3 8 3r1 ( (n ) .3(p1 1) ( p 2 1) ...... ( pr 1) . (n ) ( n )..... ( n ) . 3 22r 3 1 2 r
Proposition2.10. (a) There are infinitely many n such that (J2 ( n )) J 2 ( ( n )) . (b) J(()) n There are infinitely many n such that (J ( n )) 2 . (c) There are infinitely 2 2 2 many n such that (J2 ( n )) J 2 ( ( n )) 2 n . (d) There are infinitely many n such that 2 J2( ( n )) n ( J 2 ( n )) .
a a2 a 2 Proof: (a) is valid for n 2 for any a 1, since J 2 (2 ) 3.2 , 3 3n2 (J (2a )) 3.22 a 1 n 2 and (2a ) 3.2 a1 , J ( (2a )) 3.22 a 1 . This implies 2 2 2 2 a b a b2 a 1 2 b 1 (J2 ( n )) J 2 ( ( n )) . For (b) put n 2 3 , (a 1, b 1) . Then J2 (2 3 ) 2 3 , 4 ((23))Ja b (22 a 1 3 2 b 1 )2 2 a 2 3 2 b 1 n 2 . Again (2a 3 b ) 2 a1 3 b , 2 3 8 J(()) n J( (2a 3 b )) 22 a 3 3 2 b 1 n 2 , so (J ( n )) 2 . To prove (c) put n 2a 7 b , 2 3 2 2 576 (a , b 1) J (2a 7 b ) 32 .2 2 a 2 7 2 b 2 , (J (2a 7 b )) n2 , and (2a 7 b ) 3.2 a2 7 b 1 , 2 2 343
122 Bhabesh Das
4608 J( (2a 7 b )) n2 , So (J ( n )) J ( ( n )) 2 n2 . 2 2401 2 2 288 To prove (d), put n5a ,( a 1) J(5a ) 24.52 a 2 , ( J (5 a )) n 2 , 2 2 125 576 (5a ) 6.5 a1 ,J ( (5 a )) n 2 , 2 625 2 So, J2( ( n )) n ( J 2 ( n )) .
Proposition2.11. (a) There are infinitely many n such that (J2 ( n )) J 2 ( ( n )) (())J n (b) There are infinitely many n such that J( ( n )) 2 . 2 3
Proof: (a) is valid for n 3a for any a 1 . (b) is valid for n 2a 3 b for any a1, b 1 . Then an easy computation shows that 2 2 (())J n J( ( n )) n2 and (())J n n2 , so J( ( n )) 2 . 2 27 2 9 2 3
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