An Extension of the Euler Phi-function to Sets of Integers Relatively Prime to 30

Let n ≥ 1 be an integer and let S = {1, 7, 11, 13, 17, 19, 23, 29}, the set of integers which are both less than and relatively prime to 30. Define φ3(n) to be the number of integers x, 0 ≤ x ≤ n − 1, for which gcd(30n, 30x + i) = 1 for all i ∈ S . In this note we show that φ3 is multiplicative, that is, if gcd(m, n) = 1, then φ3(mn) = φ3(m)φ3(n).We make a conjecture about primes generated by S.


Introduction
Let n ≥ 1 be an integer.In (Mothebe & Samuel, 2015) we define ϕ 2 (n) to be the number of integers x, 1 ≤ x ≤ n, for which both 6x − 1 and 6x + 1 are relatively prime to 6n.We proved that the function is multiplicative and thereby obtained a formula for its evaluation.
Let S = {1, 7, 11, 13, 17, 19, 23, 29}, the set of integers which are both less than and relatively prime to 30.Define ϕ 3 (n) to be the number of integers x, 0 ≤ x ≤ n − 1, for which gcd(30n, 30x + i) = 1 for all i ∈ S .In this note we draw analogy with our study of ϕ 2 and show that ϕ 3 is multiplicative.In the same vain we obtain a formula for evaluating ϕ 3 .Our study motivates a conjecture to the effect that there are infinitely many integers x for which there is a set of the form, {30x + 1, 30x + 7, 30x + 11, 30x + 13, 30x + 17, 30x + 19, 30x + 23, 30x + 29}, that contains seven primes.We illustrate with some computations.The conjecture in turn implies cases of Alphonse de Polignac's conjecture that for every number k, there are infinitely many prime pairs p and p ′ such that p ′ − p = 2k.
Let 2, 3, 5, p t , . . ., p k be first k consecutive primes in ascending order.One may generalise the above definition and define ϕ k (n) to be the number of integers x, 0 ≤ x ≤ n − 1, for which gcd(( ∏ k t=1 p t )n, ( ∏ k t=1 p t )x + i) = 1 for all integers i which are less than 2.3.5 . . .p k and relatively prime to each of the primes 2, 3, 5, p t , . . .p k .These functions may also be shown to be multiplicative.
We now proceed to show that we can evaluate ϕ 3 (n) from the prime factorization of n.Our arguments are based on those used by Burton in (Burton, 2002), to show that the Euler phi-function is multiplicative.We first note: Theorem 1.Let k and s be nonnegative numbers and let p ≥ 13 be a prime number.Then: Proof.We shall only consider the cases (iii) and (iv) as (i) and (ii) are easy to verify.
We recall that: From the proof of Theorem 1 it is clear that for all non-negative integers k, s, t, r : We now show that the function ϕ 3 is multiplicative.This will enable us to obtain a formula for ϕ 3 (n) based on a factorization of n as a product of primes.
We require the following results.
We note also the following standard result.
Theorem 2. The function Proof.The result holds if either m or n equals 1.Further ϕ 3 (mn) = ϕ 3 (m)ϕ 3 (n) = 0 if m or n is equal to 7. We shall therefore assume neither m nor n equals 1 or 7.For each integer x denote the set {30x + i | i ∈ S } by {30x + i}.Arrange the sets {30x + i}, 1 ≤ x ≤ mn, in an n × m array as follows: We know that ϕ 3 (mn) is equal to the number of sets {30x + i} in this matrix for which all elements of {30x + i} are relatively prime to 30mn.By virtue of Lemma 1 this is the same as the number of sets {30x + i} in the same matrix for which all elements of {30x + i} are relatively prime to each of 30m and 30n.We first note, by virtue of Lemma 3, that for all i ∈ S and all x, 0 ≤ x ≤ m − 1, and q, 0 ≤ q ≤ n − 1 : Therefore each of the sets {30(qm + x) + i} in the x th column contains elements which are all relatively prime to 30m if and only if the set {30x + i} contains elements all of which are relatively prime to 30m.Therefore only ϕ 3 (30m) columns contain sets {30x + i} for which every element is relatively prime to 30m and every other set in the column will constitute of integers all of which are relatively prime to 30m.
The problem now is to show that in each of these ϕ 3 (30m) columns there are exactly ϕ 3 (30n) sets {30x + i} all of whose elements are relatively prime to 30n, for then altogether there would be ϕ 3 (30m)ϕ 3 (30n) sets in the table for which every element is relatively prime to both 30m and 30n.
The sets that are in the x th column, 0 ≤ x ≤ m − 1, (where it is assumed gcd(30x + i, 30m) = 1 for all i) are: There are n sets in this sequence and for no two distinct sets in the sequence can we have 30(qm + x) + i ≡ 30( jm + x) + i (mod n) for any i ∈ S , since otherwise we would arrive at a contradiction q ≡ j (mod n).
Thus the terms of the sequence, x, m + x, 2m + x, . . ., (n − 1)m + x are congruent modulo n to 0, 1, 2, . . ., n − 1 in some order.Now suppose t is congruent modulo n to qm+x.Then the elements of the set {30(qm+x)+i} are all relatively prime to 30n if and only if the elements of the set {30t +i} are all relatively prime to 30n.The implication is that the x th column contains as many sets, all of whose elements are relatively prime to 30n, as does the collection {S , {30+i}, {2.30+i} . . ., {30(n−1)+i}}, namely ϕ 3 (30n) sets.Thus the number of sets {30x + i} in the matrix each of whose elements is relatively prime to 30m and 30n is ϕ 3 (30m)ϕ 3 (30n).This completes the proof of the theorem.
For each integer x ≥ 0 the set {30x + i} has an element which is divisible by 7.This is the case since for each element j of the integers modulo 7 there is an element i ∈ S such that i ≡ j (mod 7).For some integers x the sets {30x + i} contain seven primes.This in is the case when, for instance, x = 0, 1, 2, 49, 62, 79, 89, 188.We shall call such sets seven-prime sets.Given an integer n a sieve method (a modified version of the sieve of Eratosthenes) may be employed to find the number of integers x ≤ n − 1 for which {30x + i} is a seven-prime set.This motivates the following conjecture which is in the same mould as the Twin Prime Conjecture.
Conjecture 1.There are infinitely many seven-prime sets.
The following table gives the values of π 7 (10 k ) for k ≤ 10 8 .