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 S_n. 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.^[1]

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, ... (sequence A000793 in the OEIS) is named after Edmund Landau, who proved in 1902^[2] that

${\displaystyle \lim _{n\to \infty }{\frac {\ln(g(n))}{\sqrt {n\ln(n)}}}=1}$

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

More precisely,^[3]

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

If ${\displaystyle \pi (x)-\operatorname {Li} (x)=O(R(x))}$, where ${\displaystyle \pi }$ denotes the prime counting function, ${\displaystyle \operatorname {Li} }$ the logarithmic integral function with inverse ${\displaystyle \operatorname {Li} ^{-1}}$, and we may take ${\displaystyle R(x)=x\exp {\bigl (}-c(\ln x)^{3/5}(\ln \ln x)^{-1/5}{\bigr )}}$ for some constant c > 0 by Ford,^[4] then^[3]

${\displaystyle \ln g(n)={\sqrt {\operatorname {Li} ^{-1}(n)}}+O{\bigl (}R({\sqrt {n\ln n}})\ln n{\bigr )}.}$

The statement that

${\displaystyle \ln g(n)<{\sqrt {\mathrm {Li} ^{-1}(n)}}}$

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

It can be shown that

${\displaystyle g(n)\leq e^{n/e}}$

with the only equality between the functions at n = 0, and indeed

${\displaystyle g(n)\leq \exp \left(1.05314{\sqrt {n\ln n}}\right).}$^[5]