# Constant LCM of $n$ consecutive numbers

Let $f:\Bbb N\setminus\{0,1\}\to\Bbb N$ be a function defined by $$f(n)=\operatorname{lcm}[1,2,\ldots,n]$$ Prove that for all $n\ge2$ there exis $n$ consecutive numbers for which $f$ is constant.

Find the greatest number of elements of a set of consecutive integers on which $f$ is strictly increasing and determine all sets for which this maximum is realized.

I tried with $n!$ but the idea didn't work.

Any help will be truly appreciated.

• It has something to do with factorials. – TheRandomGuy Mar 23 '16 at 6:50

Hint :

The function $f(n)$ strictly increases at $n$ if and only if $n$ is a prime power $p^k$ with $p$ prime and $k\ge 1$.

(a) let $$n$$ be given. Let the first $$n$$ primes be $$p_1$$ to $$p_n$$.

By Chinese remainder theorem, there exist some $$x$$ such that $$\begin{eqnarray} x &=& 0 \pmod {p_1} \\ x &=& -1 \pmod {p_2} \\ ... \\ x &=& -(n-1) \pmod{p_n} \end{eqnarray}$$

Then $$\{x, x+1,..., x+(n-1)\}$$ are $$n$$ consecutive numbers which are multiples of the primes $$p_1$$ to $$p_n$$.

So $$LCM[1,2,...,x] = LCM[1,2,...,x,x+1] = ... = LCM[1,2,...,x+(n-1)]$$

$$\implies f(x) = f(x+1) = ... = f(x+n-1)$$

(b) suppose $$f(n) = x$$. Consider $$f(n+1)$$.

if $$n+1$$ is prime, $$f(n+1) > f(n)$$

if $$n+1$$ is a prime power, then also $$f(n+1) > f(n)$$

if $$n+1$$ is not either, then it is composite, and $$n+1 = ab$$ where $$a,b \in S_n$$. In this case, $$f(n+1) = f(n)$$

So we have that $$f(n+1) > f(n) \iff n+1$$ is a prime or prime power.

Now the longest strictly increasing sequence. We have: $$f(2) = 2, f(3) = 6, f(4) = 12, f(5) = 60$$ is a strictly increasing sequence of $$4$$ numbers.

Any other consecutive sequence of $$4$$ or more numbers will have $$2$$ even numbers. It is not possible for both to be powers of $$2$$ (since only $$1$$ of them can be a multiple of $$4$$). So $$f(n)$$ cannot be strictly increasing on that sequence of numbers.

• what about $f(6) = 60, f(7)=420, f(8)=840, f(9)=2520$? – Merk Zockerborg Oct 6 '18 at 15:24