Number of non-primes in the set of units modulo n
aa r X i v : . [ m a t h . G M ] J u l NUMBER OF NON-PRIMES IN THE SET OF UNITS MODULO n ABHIJIT A J, A. SATYANARAYANA REDDY
Abstract.
In this work, we studied various properties of arithmetic function ˜ ϕ , where ˜ ϕ ( n ) = |{ m ∈ N | ≤ m ≤ n, ( m, n ) = 1 , m is not a prime }| . Key Words : Greatest common divisor, Arithmetic function, Euler-totient function.
AMS(2010) : 11A05,11A25. 1.
Introduction
For a fixed positive integer n , let U n = { k : 1 ≤ k ≤ n, gcd( k, n ) = 1 } . If | S | denotes the cardinality ofthe set S , then | U n | = ϕ ( n ), the well known Euler-totient function . Let E n = { m ∈ N | ≤ m ≤ n, ( m, n ) =1 , m is not a prime } . That is E n = { m ∈ U n | m is not a prime } . It is known that ϕ ( n ) = n − n isprime, hence E n = { m ∈ U n | ϕ ( m ) = m − } . Let ˜ ϕ ( n ) = | E n | . It is clear that ˜ ϕ is an arithmetic function. Firsttwenty values of ˜ ϕ are given in the following tables. n E n ˜ ϕ ( n )1 { } { } { } { } { , } { } { , , } { } { , , } { , } n E n ˜ ϕ ( n )11 { , , , , , } { } { , , , , , , } { , } { , , , } { , , } { , , , , , , , , , } { } { , , , , , , , , , , } { , } ϕ function. Recall that for n ∈ N , π ( n ) , ω ( n )denotes the number of primes less than or equal to n , the number of distinct primes divisors of n respectively.We denote N j = p p · · · p j , product of first j primes, for example N = 30 . It would be interesting to note that √ n < ϕ ( n ) < n − √ n and by prime number theorem, for a large enough n , π ( n ) ≈ nlog ( n ) . But, ω ( n ) does not haveany ordinary behaviour. In fact for any natural number k , one can find a subsequence { n i } of natural numberssuch that ω ( n i ) = k for all i . To know more properties of π ( n ) , ω ( n ) , ϕ ( n ) and U n refer any one of [1, 2, 4, 5]. Proposition 0. If p k denotes the k th prime, then ˜ ϕ ( p k ) = p k − k. In general ˜ ϕ ( n ) = ϕ ( n ) − π ( n ) + ω ( n ) . Proof.
First part follows from the observation that E p k = U p k \ { p , p , . . . , p k − } . Let n = q α q α · · · q α k k . We claim U n ∪ { q , q , . . . , q k } = E n ∪ { p , p , . . . , p π ( n ) } Suppose x ∈ U n ∪ { q , q , . . . , q k } . Now if x is prime, then then x ∈ { p , p , . . . , p π ( n ) } and if x is not prime, then x ∈ E n . Thus x ∈ E n ∪{ p , p , . . . , p π ( n ) } . Suppose y ∈ E n ∪ { p , p , . . . , p π ( n ) } . If y ∈ E n , then y ∈ U n if y ∈ { p , p , . . . , p π ( n ) } then y ∈ U n or y ∈{ q , q , . . . , q k } depending on y ∤ n or y | n respectively. By definition, E n ∩{ p , p , . . . , p π ( n ) } = U n ∩{ q , q , . . . , q k } = ∅ . Hence ˜ ϕ ( n ) + π ( n ) = | E n | + |{ p , p , . . . , p π ( n ) }| = | U n | + |{ q , q , . . . , q k }| = ϕ ( n ) + ω ( n ) . (cid:3) roposition 1. (1) If n is the square free part of n then ˜ ϕ ( n ) ≥ ˜ ϕ ( n ) . (2) Let n, k ∈ N and n ≥ . Then ˜ ϕ ( n k ) < ˜ ϕ ( n k +1 ) . (3) Let n, p, q ∈ N where p, q primes with p < q and ( p, n ) = 1 . Then ˜ ϕ ( np ) ≤ ˜ ϕ ( nq ) . (4) If a ∈ N and ω ( a ) = i , then ˜ ϕ ( N i ) ≤ ˜ ϕ ( a ) . Proof.
Proof of Part 1: Since n and n have same set of prime factors, E n ⊆ E n . Proof of Part 2: It is easy to see that E n k ⊆ E n k +1 . Sufficient to show that E n k +1 \ E n k = ∅ . Since n ≥ ℓ ∈ { , , . . . } such that ( n − ℓ − ≤ n k < ( n − ℓ . Consequently n k < ( n − ℓ < ( n − ℓ − · n ≤ n k +1 . Thus we produced an element ( n − ℓ in E n k +1 \ E n k . Proof of Part 3: Let x ∈ E np and q m || x for some m ∈ N ∪ { } . Then x = q m y for some y ∈ N . Further that( p m y, nq ) = 1 as ( p m , n ) = 1 , ( y, q ) = 1 . Now we define a map f : E np → E nq as f ( x ) = p m y. It is easy to see that f is one-one. Hence the result follows.Proof of Part 4 Let q , q , . . . , q i be all distinct prime divisors of a and s = q q · · · q i . Then from the proof ofPart 1 we have ˜ ϕ ( s ) ≤ ˜ ϕ ( a ) . Hence the result follows from the facts that N i ≤ s and E N i \ { q , q , . . . , q i } ⊆ E s \ { p , p , . . . , p i } . (cid:3) Main results
It is easy to see that ˜ ϕ ( n ) = 1 if and only if n ∈ { , , , , , , , , , } . That is ˜ ϕ ( ℓ ) > ℓ ≥ . In general we prove.
Theorem 2.
For every n ∈ N , there exists N ∈ N such that ˜ ϕ ( m ) > n , for all m > N. Lemma 3.
Let n ∈ N . There exists an M n ∈ N such that ˜ ϕ ( N k ) > n for all k ∈ N , k ≥ M n . Proof.
Let p i denote the i th prime. For i ≥ , we define P i = { m ∈ N | ϕ ( m ) = m − , m ∈ U N i } , Q i = { r ∈ P i | rp i +1 < N i } . From Bertrand’s Postulate p i +1 ∈ Q i . Suppose Q i = { p i +1 , p i +2 , . . . , p h } then | Q i | = h − i. Further p i +2 < p i +1 and p h +2 < p h +1 < p h . Thus p i +2 p h +2 < p i +1 p h < N i < N i +1 as p i +1 > . Hence { p i +2 , . . . , p h +2 } ⊆ Q i +1 consequently | Q i | < | Q i +1 | . Now { rp i +1 | r ∈ Q i } ⊂ E N i . Thus ˜ ϕ ( N i ) > | Q i | . Hence theresult follows from the fact that | Q i | increases with i. (cid:3) Since | Q | = 4, therefore for all n ≥ | Q n | ≥ n and ˜ ϕ ( N k ) > k when k ≥
4. Hence M n ≤ max { , n } . Further,it is easy to see that ˜ ϕ is an increasing function on { N , N , N , . . . } . Lemma 4.
Let a, b ∈ N and A ( a, b ) = { n ∈ N | ˜ ϕ ( n ) = a, ω ( n ) = b } . Then A ( a, b ) is finite.Proof. Let p be the ( b + ℓ ) th prime, where a < ℓ ( ℓ +1)2 . Let n ∈ N with ω ( n ) = b and n > p . Since there are atleast ℓ primes less than p ,which are co-prime to n , we can choose two such primes q , q . Then q q < p < n. Thus q q ∈ E n as a consequence ˜ ϕ ( n ) > ℓ ( ℓ +1)2 > a. Hence the result follows. (cid:3)
Now we will prove Theorem 2.
Proof.
Let n ∈ N . There exists k ∈ N such that if ω ( t ) > k , then ˜ ϕ ( t ) ≥ ˜ ϕ ( N ω ( t ) ) > n. Now from Lemma 4 foreach i ∈ N , i ≤ k there exists m i ∈ N such that for all t ∈ N with ω ( t ) = i and t > m i , then ˜ ϕ ( t ) > n. Thus if N = max { m , m , . . . , m n } and ℓ > N, then ˜ ϕ ( ℓ ) > n. (cid:3) Future work
The following question is natural.Let k ∈ N be fixed. Find all n such that ˜ ϕ ( n ) = k. One can see the following result by calculating the appropriate bounds and checking accordingly. emma 5. Let s ( k ) = { n ∈ N | ˜ ϕ ( n ) = k } . Then the set of values k ∈ { , , . . . , } such that s ( k ) is empty set is { , , } . Conjecture: If M = { ˜ ϕ ( n ) | n ∈ N } , then N \ M contains infinite number of elements.Also recall the Carmichael conjecture [3]: if N ( m ) = |{ n ∈ N | ϕ ( n ) = m }| , then N ( m ) = 1 . This conjecture nolonger true for ˜ ϕ as ˜ ϕ ( n ) = 16 only for n = 144 . If s ( k ) is as defined in the Lemma 5, then then the values of k ∈ { , . . . , } such that s ( k ) contains a single element is { , , , , , , , , , , , , } . The following tables and observations may be useful in answering the above questions.Given a positive integer k , the following tables provides the smallest value of n such that ˜ ϕ ( n ) = k. For example˜ ϕ (1) = ˜ ϕ (2) = ˜ ϕ (3) = ˜ ϕ (4) = 1 , hence the smallest value n such that ˜ ϕ ( n ) = 2 is 5 . k 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15n 1 5 7 15 26 11 13 38 102 17 19 25 ?? 23 35k 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30n 144 74 198 29 31 75 57 104 94 37 55 69 41 43 118k 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45n ?? 47 81 128 87 134 53 93 480 146 77 59 61 117 111k 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60n 166 172 67 250 91 71 73 350 194 129 202 79 206 212 83The following observations are useful for the question posed in the beginning of the section for the case k ≤ . (1) ˜ ϕ ( n ) = 1 ⇔ n ∈ { , , , , , , , , } . (2) ˜ ϕ ( n ) = 2 ⇔ n ∈ { , , , , , } . (3) ˜ ϕ ( n ) = 3 ⇔ n ∈ { , , , , , } . (4) ˜ ϕ ( n ) = 4 ⇔ n ∈ { , , , } . (5) ˜ ϕ ( n ) = 5 ⇔ n ∈ { , , , } . (6) ˜ ϕ ( n ) = 6 ⇔ n ∈ { , , , , , , } . (7) ˜ ϕ ( n ) = 7 ⇔ n ∈ { , , } . (8) ˜ ϕ ( n ) = 8 ⇔ n ∈ { , , , } . (9) ˜ ϕ ( n ) = 9 ⇔ n ∈ { , , } . (10) ˜ ϕ ( n ) = 10 ⇔ n ∈ { , , , , , , } . (11) ˜ ϕ ( n ) = 11 ⇔ n ∈ { , , , } . (12) ˜ ϕ ( n ) = 12 ⇔ n ∈ { , , , } . (13) ˜ ϕ ( n ) = 14 ⇔ n ∈ { , , , , , } . (14) ˜ ϕ ( n ) = 15 ⇔ n ∈ { , , , , } . (15) ˜ ϕ ( n ) = 16 ⇔ n = 144 . (16) ˜ ϕ ( n ) = 17 ⇔ n ∈ { , , , } . (17) ˜ ϕ ( n ) = 18 ⇔ n ∈ { , , } . (18) ˜ ϕ ( n ) = 19 ⇔ n ∈ { , , , , , , } . (19) ˜ ϕ ( n ) = 20 ⇔ n ∈ { , , , } . Acknowledgment:
We would like to thank referee for valuable comments.
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