Landau's function

In mathematics, Landau's function g(n), named after Edmund Landau, is defined for every natural number n to be the largest order of an element of the symmetric group Sn. Equivalently, g(n) is the largest least common multiple (lcm) of any partition of n, or the maximum number of times a permutation of n elements can be recursively applied to itself before it returns to its starting sequence.

For instance, 5 = 2 + 3 and lcm(2,3) = 6. No other partition of 5 yields a bigger lcm, so g(5) = 6. An element of order 6 in the group S5 can be written in cycle notation as (1 2) (3 4 5). Note that the same argument applies to the number 6, that is, g(6) = 6. There are arbitrarily long sequences of consecutive numbers n, n + 1, ..., n + m on which the function g is constant.

The integer sequence g(0) = 1, g(1) = 1, g(2) = 2, g(3) = 3, g(4) = 4, g(5) = 6, g(6) = 6, g(7) = 12, g(8) = 15, ... is named after Edmund Landau, who proved in 1902 that
 * $$\lim_{n\to\infty}\frac{\ln(g(n))}{\sqrt{n \ln(n)}} = 1$$

(where ln denotes the natural logarithm). Equivalently (using little-o notation), $$g(n) = e^{(1+o(1))\sqrt{n\ln n}}$$.

More precisely,
 * $$\ln g(n)=\sqrt{n\ln n}\left(1+\frac{\ln\ln n-1}{2\ln n}-\frac{(\ln\ln n)^2-6\ln\ln n+9}{8(\ln n)^2}+O\left(\left(\frac{\ln\ln n}{\ln n}\right)^3\right)\right).$$

If $$\pi(x)-\operatorname{Li}(x)=O(R(x))$$, where $$\pi$$ denotes the prime counting function, $$\operatorname{Li}$$ the logarithmic integral function with inverse $$\operatorname{Li}^{-1}$$, and we may take $$R(x)=x\exp\bigl(-c(\ln x)^{3/5}(\ln\ln x)^{-1/5}\bigr)$$ for some constant c > 0 by Ford, then
 * $$\ln g(n)=\sqrt{\operatorname{Li}^{-1}(n)}+O\bigl(R(\sqrt{n\ln n})\ln n\bigr).$$

The statement that
 * $$\ln g(n)<\sqrt{\mathrm{Li}^{-1}(n)}$$

for all sufficiently large n is equivalent to the Riemann hypothesis.

It can be shown that
 * $$g(n)\le e^{n/e}$$

with the only equality between the functions at n = 0, and indeed
 * $$g(n) \le \exp\left(1.05314\sqrt{n\ln n}\right).$$