A variant of Landau's function

Question

For any integer $n\geq 1$ Landau's function is defined as $$g(n):=\max\{ \operatorname{lcm}(a_1, \ldots, a_k) \mid n = a_1 + \ldots + a_k \mbox{ for some $k$}\},$$ the least common multiple of all partitions of $n$. It is the maximal order of an element in the symmetric group $S_n$.

Now I consider the following variant: $$h(n):=\max\{ \operatorname{lcm}(a_1, \ldots, a_k) \mid n = a_1 + \ldots + a_k \mbox{ for some $k$ and $a_1,\ldots,a_k$ odd}\}.$$ I.e., the only difference is that all parts of the partition have to be odd numbers.

Clearly, we have $h(n)\leq g(n)$, and I am wondering how much faster does $g(n)$ grow compared to $h(n)$. Specifically, is it true that $\limsup g(n)/h(n)=+\infty$? Is it also true that $\liminf g(n)/h(n)=+\infty$?

The following picture shows the ratio $g(n)/h(n)$ for $n=1,\ldots,140$, and based on the picture it seems that the answer to both of my questions could be 'yes'. Here are some thoughts: It is easy to see that the maximum for $g(n)$ is attained when the $a_i$ are distinct powers of primes and 1s. Similarly, the maximum for $h(n)$ is attained when the $a_i$ are distinct powers of odd primes and 1s (i.e., powers of 2 are forbidden). Furthermore, in the maximizing partitions for $g(n)$, all small primes appear with larger and larger powers (and these powers are all of similar size), and in $h(n)$ powers of 2 are forbidden. This should force the ratio to grow.

Answer 1

We have $\lim g(n)/h(n)=\infty$.

To prove this, take large $n$ and a partition $n=a_1+\ldots+a_k$ with odd $a_i$'s such that $\operatorname{lcm}(a_1, \ldots, a_k)=h(n)$. Choose $a_i$ which is divisible by a maximal power of $3$, say $a_1=3^sb$ for $b$ non divisible by 3. Let $2^r\in [3^{s-1},2\cdot 3^{s-1}]$ be a power of 2, then $a_1/3+2^r\leqslant a_1$, thus $n$ has a partition $n=a_1/3+2^r+a_2+\ldots+a_k+$(several 1's) with the least common multiple at least $2^rh(n)/3\geqslant 3^{s-2}h(n)$. Therefore $$g(n)/h(n)\geqslant 3^{s-2}.\quad\quad(1)$$ Now choose the maximal prime power $p^A$ which divides one of $a_i$'s, say $a_j=p^Ac$. Choose maximal powers of $2$ and $3$ not exceeding $p^A/2$, denote them $2^u$ and $3^v$, so $2^u\geqslant p^A/4$ and $3^v\geqslant p^A/6$. Take a partition of $n$ in which $a_j$ is replaced to $2^uC$ and $3^v$ (and several 1's). It's lcm is at least $p^A3^{-s} h(n)/24$ (indeed, we possibly remove $p^A$ but add $3^{v-s}$ and $2^u$). Therefore $$g(n)/h(n)\geqslant p^A3^{-s}/24.\quad\quad(2)$$ Multiplying (1) and (2) and taking the square root we get $$ g(n)/h(n)\geqslant \frac{p^{A/2}}{6\sqrt{4}}.\quad\quad(3) $$ Since $p^A$ clearly tends to infinity with $n$, (3) proves that $\lim g(n)/h(n)=\infty$.

Answer 2

Let $k = \nu_2(g(n))$. Then $g(n)$ is attained at $n = 2^k + \texttt{1s and powers of odd primes}$, implying that $g(n) = h(n-2^k)2^k$.

Hence, $h(n) = \frac{g(n+2^k)}{2^k}$ for some $k$ satisfying $k=\nu_2(g(n+2^k))$. It also follows that $$h(n) ~~\geq~~\min_{l\geq 0} \frac{g(n+2^l)}{2^l},$$ which can be used to derive a lower asymptotic bound for $h(n)$. Based on the asymptotic of $g(n)$, this minimum seems to be achieved at $l\approx \frac{\log_2n}2$.