When the Riemann Hypothesis might be false

Robin criterion states that the Riemann Hypothesis is true if and only if the inequality σ ( n ) < e γ × n × loglog n holds for all natural numbers n > 5040, where σ ( n ) is the sum-of-divisors function and γ ≈ 0 . 57721 is the Euler-Mascheroni constant. Let q 1 = 2 , q 2 = 3 ,..., q m denote the ﬁrst m consecutive primes, then an integer of the form ∏ mi = 1 q a i i with a 1 ≥ a 2 ≥ ··· ≥ a m ≥ 0 is called an Hardy-Ramanujan integer. If the Riemann Hypothesis is false, then there are inﬁnitely many Hardy-Ramanujan integers n > 5040 such that Robin inequality does not hold and n < ( 4 . 48311 ) m × N m , where N m = ∏ mi = 1 q i is the primorial number of order m .


Introduction
In mathematics, the Riemann Hypothesis is a conjecture that the Riemann zeta function has its zeros only at the negative even integers and complex numbers with real part 1 2 [4]. As usual σ (n) is the sum-of-divisors function of n [2]: The constant γ ≈ 0.57721 is the Euler-Mascheroni constant, and log is the natural logarithm. The importance of this property is: If the Riemann Hypothesis is false, then there are infinitely many natural numbers n > 5040 such that Robins(n) does not hold [4].
We recall that an integer n is said to be square free if for every prime divisor q of n we have q 2 n [2]. Robins(n) holds for all natural numbers n > 5040 that are square free [2]. In addition, we show that Robins(n) holds for some n > 5040 when π 2 6 × log log n ≤ log log n such that n is the square free kernel of the natural number n. Let q 1 = 2, q 2 = 3, . . . , q m denote the first m consecutive primes, then an integer of the form ∏ m i=1 q a i i with a 1 ≥ a 2 ≥ · · · ≥ a m ≥ 0 is called an Hardy-Ramanujan integer [2]. Based on the theorem 1.1, we know this result: Theorem 1.2 If the Riemann Hypothesis is false, then there are infinitely many natural numbers n > 5040 which are an Hardy-Ramanujan integer and Robins(n) does not hold [2].
We prove if the Riemann Hypothesis is false, then there are infinitely many Hardy-Ramanujan integers n > 5040 such that Robins(n) does not hold and n < (4.48311) m × N m , where N m = ∏ m i=1 q i is the primorial number of order m.

A Central Lemma
These are known results: Lemma 2.1 [2]. For n > 1: The following is a key lemma. It gives an upper bound on f (n) that holds for all natural numbers n. The bound is too weak to prove Robins(n) directly, but is critical because it holds for all natural numbers n. Further the bound only uses the primes that divide n and not how many times they divide n.
Lemma 2.3 Let n > 1 and let all its prime divisors be q 1 < · · · < q m . Then, Proof We use that lemma 2.1: Then by lemma 2.2, Putting this together yields the proof:

A Particular Case
We can easily prove that Robins(n) is true for certain kind of numbers: Lemma 3.1 Robins(n) holds for n > 5040 when q ≤ 5, where q is the largest prime divisor of n.
Proof Let n > 5040 and let all its prime divisors be q 1 < · · · < q m ≤ 5, then we need to prove f (n) < e γ × log log n that is true when However, we know for n > 5040 e γ × log log(5040) < e γ × log log n and therefore, the proof is complete when q 1 < · · · < q m ≤ 5.

Helpful Lemmas
For every prime number p n > 2, we define the sequence Y n = e 1 2×log(pn) . Lemma 4.1 For every prime number p n > 2, the sequence Y n is strictly decreasing.
Proof For every real value x ≥ 3, we state the function which is equivalent to 2×log(x) and h(u) = u u−1 for u = log(x). We know that g(x) decreases as x ≥ 3 increases, Moreover, we note that h(u) decreases as u > 1 increases where u = log(x) > 1 for x ≥ 3. In conclusion, we can see that the function f (x) is monotonically decreasing for every real value x ≥ 3 and therefore, the sequence Y n is monotonically decreasing as well. In addition, Y n is essentially a strictly decreasing sequence, since there is not any natural number n > 1 such that Y n = Y n+1 .
In mathematics, the Chebyshev function θ (x) is given by where p ≤ x means all the prime numbers p that are less than or equal to x.
We will prove another important inequality: Lemma 4.4 Let q 1 , q 2 , . . . , q m denote the first m consecutive primes such that q 1 < q 2 < · · · < q m and q m > 286. Then Proof From the theorem 4.2, we know that ) × q m .
In this way, we can show that ) .
We know that .
Consequently, we obtain that ).

Due to the theorem 4.3, we prove that
when q m > 286.

Proof of Main Theorems
The next theorem implies that Robins(n) holds for a wide range of natural numbers n > 5040.
Theorem 5.1 Let π 2 6 × log log n ≤ log log n for some n > 5040 such that n is the square free kernel of the natural number n. Then Robins(n) holds.
Proof Let n be the square free kernel of the natural number n. Let n be the product of the distinct primes q 1 , . . . , q m . By assumption we have that × log log n ≤ log log n.
By the previous lemma 2. 3: Suppose by way of contradiction that Robins(n) fails. Then f (n) ≥ e γ × log log n.
We claim that Since otherwise we would have a contradiction. This shows that This is a contradiction since f (n ) is equal to (q 1 + 1) × · · · × (q m + 1) q 1 × · · · × q m .
Theorem 5.2 If the Riemann Hypothesis is false, then there are infinitely many Hardy-Ramanujan integers n > 5040 such that Robins(n) does not hold and n < (4.48311) m × N m , where N m = ∏ m i=1 q i is the primorial number of order m.
Proof Let ∏ m i=1 q a i i be the representation of some natural number n > 5040 as a product of primes q 1 < · · · < q m with natural numbers as exponents a 1 , . . . , a m . The primes q 1 < · · · < q m must be the first m consecutive primes and a 1 ≥ a 2 ≥ · · · ≥ a m ≥ 0 since the natural number n > 5040 could be an Hardy-Ramanujan integer. We assume that Robins(n) does not hold. Indeed, we know there are infinitely many Hardy-Ramanujan integers such as n > 5040 when the Riemann Hypothesis is false according to the theorem 1.2. From the lemma 4.4, we know that when q m > 286. In this way, if Robins(n) does not hold, then n < N Y m m since by the lemma 2.1 we have that That is the same as n < N Y m −1 m × N m . We can check that q Y m −1 m is monotonically decreasing for all primes q m > 286 due to the lemma 4.1. Certainly, the function complies that its derivative is lesser than zero for all real numbers x > 286. Indeed, a function g(x) of a real variable x is monotonically decreasing in some interval if the derivative of g(x) is lesser than zero and the function g(x) is continuous over that interval [1]. We know that q m could comply with q m ≥ 1000000! for infinitely many Hardy-Ramanujan integers n > 5040 such that Robins for all primes q m ≥ 1000000!. Moreover, we would obtain that for every integer 1 ≤ j < m. Finally, we can state that n < (4.48311) m × N m since N Y m −1 m < (4.48311) m when n > 5040 could be any of the infinitely many Hardy-Ramanujan integers such that Robins(n) does not hold and q m ≥ 1000000!.