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.

  • $\begingroup$ It has something to do with factorials. $\endgroup$ – 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.

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

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