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A Ces\`aro Average of Goldbach numbers

Abstract

Let Λ\Lambda be the von Mangoldt function and (rG(n)=m1+m2=nΛ(m1)Λ(m2))(r_G(n) = \sum_{m_1 + m_2 = n} \Lambda(m_1) \Lambda(m_2)) be the counting function for the Goldbach numbers. Let N2N \geq 2 be an integer. We prove that nNrG(n)(1n/N)kΓ(k+1)=N2Γ(k+3)2ρΓ(ρ)Γ(ρ+k+2)Nρ+1+ρ1ρ2Γ(ρ1)Γ(ρ2)Γ(ρ1+ρ2+k+1)Nρ1+ρ2+Ok(N1/2),\begin{align} &\sum_{n \le N} r_G(n) \frac{(1 - n/N)^k}{\Gamma(k + 1)} = \frac{N^2}{\Gamma(k + 3)} - 2 \sum_\rho \frac{\Gamma(\rho)}{\Gamma(\rho + k + 2)} N^{\rho+1}\\ &\qquad+ \sum_{\rho_1} \sum_{\rho_2} \frac{\Gamma(\rho_1) \Gamma(\rho_2)}{\Gamma(\rho_1 + \rho_2 + k + 1)} N^{\rho_1 + \rho_2} + \mathcal{O}_k(N^{1/2}), \end{align} for k>1k > 1, where ρ\rho, with or without subscripts, runs over the non-trivial zeros of the Riemann zeta-function ζ(s)\zeta(s).Comment: submitte

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