Almost divisible by all

First note that we can rule out the option of $n$ being a perfect cube $k^{3}$ with $k \gt 2,$ since then $n$ could not be divisible by $k - 1.$ (With $n = k = 1$ there are no positive integers less than $\sqrt[3]{n},$ and with $n = 8, k = 2,$ while we do have $n$ divisible by all integers less than $k$ we will be able to find larger $n$ satisfying the requirements.) This follows from the fact that

$(k - 1)|(k^{3} - 1),$ and thus $n = k^{3} \equiv 1 \pmod{(k - 1)}.$

Now let $m = \lfloor \sqrt[3]{n} \rfloor.$ Then for $n$ (not a perfect cube) to be divisible by all positive integers $\le m$ we require it to be a multiple of $L_{m} = lcm(1,2,....,(m-1),m).$

Now $L_{8} = 2^{3}*3*5*7 = 840 \gt 9^{3},$ so any such $n$ would also have to be divisible by $9.$ Thus we will require that $n \lt 8^{3} = 512.$ Next, we have that $L_{7} = 2^{2}*3*5*7 = 420 \gt 7^{3} = 343,$ and since this is the greatest integer $n$ less than $8^{3}$ and greater than $7^{3}$ that is divisible by all positive integers less than $\sqrt[3]{n}$ we can conclude that $\boxed{420}$ is the number we are seeking.

Definition: $\begin{aligned} g:&&\mathbb{N}&\rightarrow\mathbb{N},&&&g(n)&:=\operatorname{lcm}\left(1;\:\ldots;\:\left\lceil\sqrt[3](n)\right\rceil-1)\right) \end{aligned}$

We have to find the largest $n\in\mathbb{N}$ such that $n=cg(n),\quad c\in\mathbb{N}$ . Obviously, $g(n)$ only increases after every perfect cube $n=m^3$ - lets take a look at the first few values: $\begin{array}{r | r r r r r r r r r} m & 5 & 6 & 7 & 8 & 9 & 10&11&\ldots\\[.5em] \hline a_m:=m^3 + 1& 126 & 217 & 344 & 513 & \red{730} &1001&1332\\[.5em] g(a_m) & 60 & 60 & 420 & \red{840} & 2520&2520&27720 \end{array}$

It seems to be the case that $\red{g(a_m) > a_{m+1}}$ for $m\geq 8$ . If we can prove that fact, there cannot be a solution $n\geq a_8$ ! Using link Betrand's Postulate $(*)$ , we can be sure to have a prime factor $8\cdot 2^k<p_k<8\cdot 2^{k+1},\qquad k\geq 0$

These prime factors are the reason that $g(a_m)$ grows very fast for $m\geq 8$ $\begin{aligned} \frac{ g(a_{8\cdot 2^{k+1}}) }{ g(a_{8\cdot 2^k}) }&=\frac{ \operatorname{lcm}(1;\ldots;8\cdot 2^{k+1}) }{ \operatorname{lcm}(1;\ldots;8\cdot 2^{k}) }\underset{(*)}{\geq}p_k>8\cdot 2^k\geq 8&\Rightarrow && g(a_{8\cdot 2^k})&<g(a_8)\cdot 8^k\\[1.5em] \frac{a_{2m+1}}{a_{m+1}}&=\frac{ (2m+\green{1})^3+\orange{1} }{ (m+1)^3+\blue{1} }\leq \frac{ (2m+\green{2})^3+\orange{8} }{ (m+1)^3+1 } \leq 8&\Rightarrow&& a_{8\cdot 2^{k}+1}&\leq a_9\cdot 8^k< g(a_8)\cdot 8^k =g(a_{8\cdot 2^k}),&&k\geq 0 \end{aligned}$

Both sequences are monotonically increasing, so there can not be a solutiom in any interval $[a_{8\cdot 2^k};\:a_{8\cdot 2^{k+1}})$ and $k\geq 0$ . Using the table, we find the largest solution to $n=cg(n)<a_8=513$ to be $\boxed{n=420}$

Rem.: It is important that both sequences are monotonic: We only checked them on the borders of intervals and assumed nothing bad happens in between. The fact that both sequences are increasing makes sure there is no point in between where the lower sequence shortly has greater values than the other!

we're looking for where the floor[{n=LCM(1,2,3...x)}^1/3]=x which happens at x=7 n=420