In this previous discussion, it was demonstrated that the standard Least Common Multiple sequence $\text{lcm}(1, 2, \dots, n)$ is not a subset of the highly abundant numbers. In analytic number theory, the Least Common Multiple of the first $n$ integers is fundamentally linked to the second Chebyshev function $\psi(n)$ and the distribution of prime powers through the identity:
$$\text{lcm}(1, 2, \dots, n) = e^{\psi(n)} = \prod_{p \leq n} p^{\lfloor \log_p n \rfloor}$$
I defined and I am investigating an extension of this structure, which I denote as $\Xi(n)$. This function is defined by a hierarchical $p$-adic valuation that accounts for the harmonic contribution of all $k$-th roots of $n$:
$$\Xi(n) = \prod_{p \leq n} p^{\sum_{k=1}^{\infty} \lfloor \frac{1}{k} \log_p n \rfloor}$$
To see how this relates to the standard LCM, we can expand the exponent $E_p(n) = \sum_{k=1}^{\infty} \lfloor \log_p (n^{1/k}) \rfloor$ and distribute the product over each $k$. By applying the property $\log_p(n^{1/k}) = \frac{1}{k}\log_p n$, the development of the product is as follows:
$$\Xi(n) = \prod_{p \leq n} p^{\lfloor \log_p n \rfloor + \lfloor \frac{1}{2} \log_p n \rfloor + \lfloor \frac{1}{3} \log_p n \rfloor + \lfloor \frac{1}{4} \log_p n \rfloor + \dots}$$
$$\Xi(n) = \left( \prod_{p \leq n} p^{\lfloor \log_p n \rfloor} \right) \cdot \left( \prod_{p \leq n^{1/2}} p^{\lfloor \log_p n^{1/2} \rfloor} \right) \cdot \left( \prod_{p \leq n^{1/3}} p^{\lfloor \log_p n^{1/3} \rfloor} \right) \cdot \left( \prod_{p \leq n^{1/4}} p^{\lfloor \log_p n^{1/4} \rfloor} \right) \cdots$$
By identifying each term as an LCM of decreasing order, we obtain the following explicit equivalence:
$$\Xi(n) = \text{lcm}(1, \dots, n) \cdot \text{lcm}(1, \dots, \lfloor n^{1/2} \rfloor) \cdot \text{lcm}(1, \dots, \lfloor n^{1/3} \rfloor) \cdot \text{lcm}(1, \dots, \lfloor n^{1/4} \rfloor) \cdots$$Which can be written compactly as:$$\Xi(n) = \prod_{k=1}^{\infty} \text{lcm}(1, 2, \dots, \lfloor n^{1/k} \rfloor)$$
I believe firmly, through my research, that this construction "injects" additional density into the exponents of smaller primes, effectively weighting the prime factors to improve the sequence $\text{lcm}(1, 2, \dots, n)$ and makes it "Highly Abundant" completely. Therefore, I have come to the following improved conjecture:
The Conjecture: For every $n \in \mathbb{N}$, $\Xi(n)$ is a Highly Abundant number ($\sigma(\Xi(n)) > \sigma(m)$ for all $m < \Xi(n)$).
