Another Criterion For The Riemann Hypothesis

Let’s deﬁne δ ( x ) = ( (cid:80) q ≤ x 1 q − log log x − B ), where B ≈ 0 . 2614972128 is the Meissel-Mertens constant. The Robin theorem states that δ ( x ) changes sign inﬁnitely often. Let’s also deﬁne S ( x ) = θ ( x ) − x , where θ ( x ) is the Chebyshev function. It is known that S ( x ) changes sign inﬁnitely often. We deﬁne the another function (cid:36) ( x ) = (cid:16)(cid:80) q ≤ x 1 q − log log θ ( x ) − B (cid:17) . We prove that when the inequality (cid:36) ( x ) ≤ 0 is satisﬁed for some number x ≥ 3, then the Riemann Hypothesis should be false. The Riemann Hypothesis is also false when the inequalities δ ( x ) ≤ 0 and S ( x ) ≥ 0 are satisﬁed for some number x ≥ 3 or when (cid:82) ∞ x S ( y ) × (1 + log y ) y 2 × log 2 y dy ≥ S ( x ) 2 x 2 × log x is satisﬁed for some number x ≥ 121.


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 [1]. Let N n = 2 × 3 × 5 × 7 × 11 × · · · × p n denotes a primorial number of order n such that p n is the n th prime number. Say Nicolas(p n ) holds provided q|N n q q − 1 > e γ × log log N n .
The constant γ ≈ 0.57721 is the Euler-Mascheroni constant, log is the natural logarithm, and q | N n means the prime number q divides to N n . The importance of this property is: [3]. Nicolas(p n ) holds for all prime numbers p n > 2 if and only if the Riemann Hypothesis is true.
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 know this: It is a known result that: . S (x) changes sign infinitely often.
In addition, the Mertens second theorem states that: Besides, Rosser and Schoenfeld derived a remarkable identity: We define another function: Putting all together yields the proof that the inequality (x) > u(x) is satisfied for a number x ≥ 3 if and only if Nicolas(p) holds, where p is the greatest prime number such that p ≤ x. In this way, we introduce another criterion for the Riemann Hypothesis based on the Nicolas criterion and deduce some of its consequences.

Results
Theorem 2.1. The inequality (x) > u(x) is satisfied for a number x ≥ 3 if and only if Nicolas(p) holds, where p is the greatest prime number such that p ≤ x.
Proof. We start from the inequality: which is equivalent to Let's add the following formula to the both sides of the inequality, q≤x log( q q − 1 ) − 1 q and due to the theorem 1.5, we obtain that Let's distribute it and remove B from the both sides: If we apply the exponentiation to the both sides of the inequality, then we have that which means that Nicolas(p) holds, where p is the greatest prime number such that p ≤ x. The same happens in the reverse implication. Proof. This is a direct consequence of theorems 1.1 and 2.1.
Theorem 2.3. If the inequality (x) ≤ 0 is satisfied for some number x ≥ 3, then the Riemann Hypothesis should be false.
Proof. This is an implication of theorems 1.6, 2.1 and 2.2.
Theorem 2.4. If the inequalities δ(x) ≤ 0 and S (x) ≥ 0 are satisfied for some number x ≥ 3, then the Riemann Hypothesis should be false.
Proof. If the inequalities δ(x) ≤ 0 and S (x) ≥ 0 are satisfied for some number x ≥ 3, then we obtain that (x) ≤ 0 is also satisfied, which means that the Riemann Hypothesis should be false according to the theorem 2.3.
Proof. We know that lim x→∞ (x) = 0 for the limits lim x→∞ δ(x) = 0 and lim x→∞ θ(x) x = 1. In this way, this is a consequence from the theorems 1.8 and 1.2.