Let $\text{lcm}(x)$ be the least common multiple of $\{1,2,3,\dots,x\}$

Let $x\#$ be the the primorial for $x$.

It occurs to me that for $x \ge 10$:

$$\frac{\text{lcm}(x^2+x)}{\text{lcm}(x^2)} < 2^x\frac{(x^2+x)\#}{(x^2)\#}$$

Am I right?

Here is my thinking:

(1) $\dfrac{\text{lcm}(x+1)}{\text{lcm}(x)} > 1$ if and only if $x+1$ is a power of a prime.

(2) If a prime $x^2 \ge p > x$, then $p^2 \ge (x+1)^2 > x^2 + x$ so it will divide out in the ratio.

(3) If a prime $x \ge p \ge (\sqrt{x}+1)$, then $p^2 \le x^2$ and $p^4 \ge (\sqrt{x}+1)^4 \ge {4\choose0}x^2 + {4\choose1}x^{3/2} + {4\choose2}x + {4\choose3}x^{1/2} + 1 > x^2+x$

(4) But there can only be one such $x \ge p \ge \sqrt{x} + 1$ such that $p^3 | \frac{(x^2+x)!}{(x^2)!}$ since $(p+2)^3 - p^3 \ge {3\choose1}2p^2 + {3\choose2}4p + 1 > 2x$

(5) Using Hanson's result that $\text{lcm}(x) < 3^x$, we have:

$$\frac{\text{lcm}(x^2 + x)}{\text{lcm}(x^2)} < 3^{\sqrt{x}+1}x\frac{(x^2+x)\#}{(x^2)\#}$$

(6) $2^x > 3^{\sqrt{x}+1}x$ since $2^{10} = 1024 > 969>3^{\sqrt{10}+1}\times10$ and $2^{\sqrt{x}} > 3(3x)^{1/\sqrt{x}}$

Edit: Forgot the case where a prime might be between $\sqrt{x}$ and $\sqrt{x}+1$

So, I've updated my argument.


HINT $1$

Note that $$r(x)={\large \dfrac {\mathrm{lcm}(x^2+x)}{\mathrm{lcm}(x^2)}\cdot{\dfrac {(x^2)\#}{(x^2+x)\#}} = \prod_{p_k<x+1} p_k^{[\log_{p_k}(x^2+x)]-[\log_{p_k}(x^2)]}}.$$ Calculations using Wolfram Alpha of function $$\bar r(x) = \left(\dfrac45\right)^x\prod_{p_k<x+1} p_k^{[\log_{p_k}(x^2+x)]-[\log_{p_k}(x^2)]}$$ gives the plot of


which shows that $\bar r(x) \leq 1,$ and that leads to the empiric estimation $$r(x)\leq \left(\dfrac54\right)^x,\quad x\in\mathbb N.$$ The reason is that all primes greater than $x,$ contains both in the ratio for lcm and the ratio for primorial. Therefore, correction of $r(m)$ happens at least for the squares of primes.

Let $x=11,$ then the interval $[x^2+1,x^2+x] = [122,132]$ contains new degrees $2^7=128$ and $5^3=125.$ So $r(11)=2\cdot5=10.$

This means that complex consideration leads to the better estimation.

HINT $2$

For $x>>1$ $$\dfrac{(x^2+x)\#}{x^2\#}\approx \left(x^2+\frac12x\right)^d,$$ where $$d=\pi(x^2+x)-\pi(x^2)\approx \int\limits_{x^2}^{x^2+x}\dfrac{\mathrm dt}{\log t} = x^2\int\limits_{x^2}^{x^2+x}\dfrac1{\log t}\mathrm d\dfrac t{x^2} = x^2\int\limits_{1}^{1+1/x}\dfrac{\mathrm du}{\log(x^2u)}$$ $$ = \dfrac{x^2}{2\log x}\int\limits_{1}^{1+1/x}\dfrac{\mathrm du}{1+\dfrac{\log u}{2\log x}} \approx \dfrac{x^2}{2\log x}\int\limits_{1}^{1+1/x}\left(1-\dfrac{\log u}{2\log x}\right)\,\mathrm du$$ $$ = \dfrac{x^2}{2\log x}\left(u - \dfrac{u\log u -1}{2\log x}\right)\bigg|_1^{1+\frac1x} = \dfrac{x^2}{2\log x}\left(\dfrac1x-\dfrac{(1+\frac1x)\log (1+\frac1x)}{2\log x}\right),$$ $$d\approx\dfrac x{2\log x} - \dfrac{x+1}{4\log^2x}.$$

  • $\begingroup$ Thanks for your analysis! Using my equation, I noticed that for $x \ge 63, (5/4)^x > 3^{\sqrt{x}+1}x$. It is very interesting that the number can go so significantly lower than $2^x!$. $\endgroup$ – Larry Freeman Dec 8 '18 at 18:47
  • $\begingroup$ @LarryFreeman This anaysis has significant modeling part... $\endgroup$ – Yuri Negometyanov Dec 8 '18 at 18:50
  • $\begingroup$ @LarryFreeman You are welcome! I like real analysis. $\endgroup$ – Yuri Negometyanov Dec 8 '18 at 22:09

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