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I am interested in an asymptotic formula $f(m,k)$ as $k\rightarrow\infty$. Starting points include the following two previous Stackechange questions

as well as the following sequence corresponding to $m=3$: https://oeis.org/A000100. For $m=2$, it involves Fibonacci numbers and an exact formula is easy to obtain. Exact formulas (based on recursions) are available for all $m$ but cumbersome. It seems that asymptotically, we have:

$f(m,k) = a_1 \cdot r_1^k + a_2 \cdot r_2^k + \cdots$

where $r_1 > r_2 > \cdots$ are the (real?) roots of $r^m=1+r+r^2 + \cdots + r^{m-1}$. Is this correct, and if yes, what are the coefficients $a_1, a_2$ and so on? At least, what is the dominant coefficient $a_1$?

Vincent Granville
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1 Answers1

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$r^m=1+r+r^2 + \cdots + r^{m-1}$ is correct. If $c(k,m)$ is the number of compositions of $k$ with parts of at most $m$, for $k \gt m$ the recurrence is $c(k,m)=c(k-1,m)+c(k-2,m)+\ldots c(k-m,m)$ because you can make a composition of $k$ by starting with one of $k-m$ to $k-1$ and adding the proper number. The usual technique of finding the characteristic polynomial for a recursion relation gives the equation you gave.

This gives $$r^m=\frac {r^m-1}{r-1}\\ r^{m+1}-r^m=r^m-1\\r^{m+1}-2r^m+1=0$$ If $m$ is large the dominant root will be just less than $2$. If we write $r=2-a$ with $a \ll 1$ we have $$2(2-a)^m-(2-a)^{m+1}=1\\ a(2-a)^m=1$$ and to first order $a\approx 2^{-m}$ We could write $a=\frac 1{(2-a)^m}$ and iterate to convergence to get a better value. By the time $m=15$ we get three place accuracy with just $a\approx 2^{-m}$

I just fed the $m=10$ case to Alpha. The dominant root is near $r=2$ as expected. The rest seem to be roughly on the circle with radius $1$ in the complex plane. The same thing happened for $[m=15][2]$. I would expect it to be general. If $r^m=1, r^{m+1}-2r^m+1 =r-1$ and we don't have to perturb $r$ much to make it work. enter image description here

Ross Millikan
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