4 Ramanujan factor recipe

I will start off with the simplest type, $$d(n) \leq \sqrt{3 n}$$ and $$d(n) \leq 24 \left(\frac{n}{315}\right)^{1/3}$$ The first one has equality at $n = 12,$ second at $n =2520.$ Instead of continuing with fractional powers $1/k$ the better results switch to logarithms. Reference is a paper by J. L. Nicolas in a book called Ramanujan Revisited.

With equality at $n = 6983776800 = 2^5 \cdot 3^3 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19$ and $d(n) = 2304,$ $$d(n) \leq n^{ \left( \frac{\log 2}{\log \log n} \right) \left( 1.5379398606751... \right)}$$

With equality at a number $n$ near $6.929 \cdot 10^{40},$ $$d(n) \leq n^{ \left( \frac{\log 2}{\log \log n} \right) \left( 1 + \frac{1.934850967971...}{\log \log n} \right)}$$

With equality at a number $n$ near $3.309 \cdot 10^{135},$ $$d(n) \leq n^{ \left( \frac{\log 2}{\log \log n} \right) \left( 1 + \frac{1}{\log \log n} + \frac{4.762350121177...}{\left(\log \log n \right)^2} \right)}$$

Just to fill in one blank, the special integers $n$ here are "superior highly composite numbers" using Ramanujan's original recipe for prime factorization, which I like to write, with $\delta > 0,$ as $$N_\delta = \prod_p \; p^{\left\lfloor \frac{1}{p^\delta - 1} \right\rfloor }$$ So $$N_{1/2} = 12, \; N_{1/3} = 2520, \; N_{0.23} = 6983776800, \; N_{0.155} \approx 6.929 \cdot 10^{40}, \; N_{0.1218} \approx 3.309 \cdot 10^{135}.$$

3 factor and 2304

I will start off with the simplest type, $$d(n) \leq \sqrt{3 n}$$ and $$d(n) \leq 24 \left(\frac{n}{315}\right)^{1/3}$$ The first one has equality at $n = 12,$ second at $n =2520.$ Instead of continuing with fractional powers $1/k$ the better results switch to logarithms. Reference is a paper by J. L. Nicolas in a book called Ramanujan Revisited.

With equality at $6983776800,$ n = 6983776800 = 2^5 \cdot 3^3 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19$and$d(n) = 2304,$$$d(n) \leq n^{ \left( \frac{\log 2}{\log \log n} \right) \left( 1.5379398606751... \right)}$$ With equality at a number$n$near$6.929 \cdot 10^{40},$$$d(n) \leq n^{ \left( \frac{\log 2}{\log \log n} \right) \left( 1 + \frac{1.934850967971...}{\log \log n} \right)}$$ With equality at a number$n$near$3.309 \cdot 10^{135},$$$d(n) \leq n^{ \left( \frac{\log 2}{\log \log n} \right) \left( 1 + \frac{1}{\log \log n} + \frac{4.762350121177...}{\left(\log \log n \right)^2} \right)}$$ 2 added 500 characters in body It is going to take me a while to typeset this, so I will start off with the simplest type(but not the best), $$d(n) \leq \sqrt{3 n}$$ and $$d(n) \leq 24 \left(\frac{n}{315}\right)^{1/3}$$ The first one has equality at$n = 12,$second at$n =2520.$Instead of continuing with fractional powers$1/k $1/k$ the better results switch to logarithms, I will get to that. Reference is a paper by J. L. Nicolas in a book called Ramanujan Revisited.

More soom

With equality at $6983776800,$ $$d(n) \leq n^{ \left( \frac{\log 2}{\log \log n} \right) \left( 1.5379398606751... \right)}$$

With equality at a number $n$ near $6.929 \cdot 10^{40},$ $$d(n) \leq n^{ \left( \frac{\log 2}{\log \log n} \right) \left( 1 + \frac{1.934850967971...}{\log \log n} \right)}$$

With equality at a number $n$ near $3.309 \cdot 10^{135},$ $$d(n) \leq n^{ \left( \frac{\log 2}{\log \log n} \right) \left( 1 + \frac{1}{\log \log n} + \frac{4.762350121177...}{\left(\log \log n \right)^2} \right)}$$

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