I asked the question in Math StackExchange. Link: https://math.stackexchange.com/questions/4765476/approximation-of-partial-sum-over-prime-omega-function

I haven't got any response yet. Here are the details:

The prime omega function $\omega(n)$ counts the number of distinct prime factors of a natural number $n$, and can be defined as $\omega(n)=\sum_{p \mid n}1$. Let $S(N)=\sum_{n=1}^{N}n\omega(n)$. Let $\pi(n)$ denote the number of primes up to $n$.

$$\eqalign{S(N) &=\dfrac{1}{2}\sum_{\scriptstyle p{\text{ prime}} \atop \scriptstyle p\leq N} p\left(\left\lfloor\dfrac{N}{p}\right\rfloor ^2+\left\lfloor\dfrac{N}{p}\right\rfloor\right)\cr &\approx \dfrac{N\pi(N)}{2}+\dfrac{N^2}{2}\sum_{\scriptstyle p{\text{ prime}} \atop \scriptstyle p\leq N} \dfrac{1}{p}\cr &\approx \dfrac{N\pi(N)}{2}+\dfrac{N^2}{2}\left[\log \log N+M+o(1)\right],\cr}$$ where $M$ is the Meissel–Mertens constant.

The error term of this approximation is quite large. Is it possible to find a better approximation for this sum?

A better approximation for the sum of reciprocal primes (see https://arxiv.org/pdf/1703.08032.pdf) does not help to improve the overall sum.

The following approximation seems to work well. $$\tilde{S}(N,\epsilon)=\left[\dfrac{N\pi(N)}{2}+\dfrac{N^2}{2}\left(\log \log N+M\right)\right]\cdot \left(1+\dfrac{1}{(\log N)^{1+\epsilon}}\right)^{-1}$$, where $\epsilon \in (0,1)$. Define the percentage error as $E(N,\epsilon)=\left |\dfrac{S(N)-\tilde{S}(N,\epsilon)}{S(N)}\right |\times 100$. The following table gives an idea as to how well the approximation works for $\epsilon=\dfrac{2}{5}$.

$N$ | $S(N)$ | $E\left(N,\frac{2}{5}\right)$ |
---|---|---|

$10^3$ | $1104960$ | $0.192894816907$ |

$10^4$ | $124547620$ | $0.0961202670514$ |

$10^5$ | $13558020500$ | $0.0234415251444$ |

$10^6$ | $1446379298344$ | $0.00741488319826$ |

$10^7$ | $152305298038685$ | $0.0118235266249$ |

$10^8$ | $15895296473923817$ | $0.00880016331673$ |

$10^9$ | $1648198262567978862$ | $0.00327960320611$ |

$10^{10}$ | $170070011690546177056$ | $0.00270740071241$ |

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