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Let $\operatorname{LCM}(x_1,x_2,\ldots,x_n)$ be the least common multiple of the integers $x_i$.

How can one find the asymptotics of $\operatorname{LCM}(f(1),f(2),\dots,f(n))$ as $n$ approaches infinity if one knows the asymptotics of the strictly increasing function on the integers $f(n)$?

Edit: Is there some result if one assumes $f(n)$ has natural density in the primes ie is prime with probability $1/\ln(f(n))$, has an average of $ln(f(n))$ factors, $ln(ln(f(n))$ prime factors, and a $6/\pi^2$ probability to be squarefree.

And how to prove these average properties ?

Edit2: Instead of above estimates use: For every strictly increasing $f(n)$, consider instead the question for $g(f(n))$, which is a uniformly random integer in the range $(0.99f(n),1.01f(n))$ say, alternatively $(f(n)-1000,f(n)+1000)$.
Then im looking for asymptotics of $LCM[g(f(1)),g(f(2))...g(f(x))]$ as $x\rightarrow\infty$, given $f(n)$.

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Asymptotics for $\mathrm{LCM}(1,2,3...,n)$ is already interesting! –  GEdgar Oct 4 '11 at 18:43
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Here a reference for $\mathrm{LCM}(1,2,3...,n)$: mathforum.org/kb/message.jspa?messageID=287886&tstart=0 –  lhf Oct 4 '11 at 20:33
    
Experiment: $(1,2,3\ldots,n)$ $(1,2,3...,n)$ These do look different from each other. That's why \ldots exists. (I know of a philosophy professor's web site where the advice to students about grammar, spelling, usage, punctuation, etc., says to put spaces between the dots in an ellipsis. Same thing here, but it's built in.) –  Michael Hardy Oct 5 '11 at 0:56
    
So in that reference, LCM($1,2,\cdots,n)$ is asymptotically greater than $e^{ax}$ if $a<1$ and asymptotically smaller than $e^{ax}$ if $a>1$. But that is not yet an asymptotic estimate for this! –  GEdgar Oct 6 '11 at 18:28
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Generally sequences with the same asymptotic growth can have wildly different prime factor frequencies. If you added the condition that it is a (perhaps strong) divisibility sequence, then I believe there would be more interesting answers... –  anon Oct 6 '11 at 19:00

2 Answers 2

up vote 5 down vote accepted
+100

The paper The least common multiple of a quadratic sequence by Javier Cilleruelo, Compositio Mathematica (2011), 147: 1129-1150 gives the answer when $f$ is a quadratic polynomial. From the abstract:

For any irreducible quadratic polynomial $f(x)$ in $\mathbb{Z}[x]$ we obtain the estimate $\log\mathrm{LCM} (f(1),...,f(n)) = n\log n + B\,n + o(n)$ where $B$ is a constant depending on $f$.

This is a link to the paper in Arxix.

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Konowing asymptotics of $f$ is not enough. Consider functions $f_1(n)=2^n$ and $f(n)=p_1p_2\ldots p_n\;$, where $p_n$ is the $n$-th prime number.

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I don't understand your example. Those functions have different asymptotics and LCM(f(1),f(2),...,f(n)) has a different asymptotic for each of them. –  Oscar Cunningham Oct 4 '11 at 19:05
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Consider $f_1(n) = 2^n$ and $f_2(n) = 2^n-1$, then. Then the LCM of the second sequence grows much faster than that of the first. –  Michael Lugo Oct 4 '11 at 19:10

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