10
$\begingroup$

Suppose $n$ be a given positive integer. Then the Diophantine equation $x=n$ has only $1$ solution. Just by inspection, I found that the Diophantine equation $x+2y=n$ has $\left\lfloor \dfrac{n}{2}+1\right\rfloor$ non-negative solutions for $(x,y).$
Also, according to this post the Diophantine equation $x+2y+3z=n$ has $\left\lfloor \dfrac{n^2}{12}+\dfrac{n}{2}+1 \right\rfloor$ non-negative solutions for $(x,y).$

Is there any closed form for the number of non-negative integer solutions for $(x_1,x_2,\cdots,x_k)$ of $$x_1+2x_2+3x_3+\cdots+kx_k=n$$ for a given $k\in\Bbb{N}$?

How can I prove these formulas rigorously?

EDIT
After a very tedious calculation I found that the equation $w+2x+3y+4z=n$ has $\left\lfloor \dfrac{n^3}{144}+\dfrac{5n^2}{48}+\dfrac{(15+(-1)^n)n}{32}+1 \right\rfloor$ solutions.
This solution completely agree with the approximation given by Rus May.
However still I believe that we can do something more than this.

$\endgroup$
3
  • 2
    $\begingroup$ You are asking for the number of partitions of $n$ into parts of sizes $\leq k$. Don't think there is a closed formula holding for all $n, k$. But for each fixed $k$, the number can be computed as the $t^n$-coefficient of the formal power series $\dfrac{1}{\left(1-t\right)\left(1-t^2\right) \cdots\left(1-t^k\right)}$, which should fall prey to some standard methods (it satisfies a linear recurrence, if I remember correctly?). $\endgroup$
    – darij grinberg
    Commented Dec 11, 2016 at 2:50
  • 1
    $\begingroup$ It can be proved that the multinomial of $x_i$, $1\leq i \leq k$ is equal the $n$th term of generalized Fibonacci numbers, as shown $$ \sum_{(x_1,x_2,\cdots,x_k)} \left( \begin{array}{c} x_1+\cdots+x_k \\ x_1,\cdots , x_k \end{array} \right)=f_n\, . $$ where the summation is over non-negative integers satisfying $$ x_1+2\, x_2+3\, x_3+\cdots + k\, x_k=n \, . $$ and $f_n$ is the generalized Fibonacci numbers, as follows $$ f_n=\sum_{i=1}^k f_{n-i} \quad , \quad (f_0,\cdots,f_{k-2},f_{k-1})=(0,\cdots,0,1) \, . $$ $\endgroup$
    – Amin235
    Commented Dec 15, 2016 at 15:40
  • $\begingroup$ References are generalized Fibonacci sequence and Combinatorial method. $\endgroup$
    – Amin235
    Commented Dec 15, 2016 at 15:41

2 Answers 2

Reset to default
3
$\begingroup$

As Darij Grinberg says, there's not a nice closed formula for this. There is, however, a really neat approximation via Schur's theorem in combinatorics. It goes like this.

The singularities of the generating function $\frac1{(1-t)(1-t^2)\cdots(1-t^k)}$ all lie on the unit circle in the complex plane. The partial fractions decomposition of the generating function has terms of the form $\frac\alpha{(1-x/\omega)^{1+\ell}}$, where $\alpha$ is a constant, $\omega$ is a root of unity, and $\ell$ is a natural number less than $k$. The coefficient of such a term is $\alpha\binom{n+\ell}{\ell}/\omega^\ell$, so the term with the highest multiplicity makes the greatest contribution. In this case, it is the singularity at 1 with $\ell=k-1$. Then the coefficient of $t^n$ in the generating function is approximately \begin{eqnarray*} [t^n]\frac1{(1-t)\cdots(1-t^k)}&=&\alpha\binom{n+k-1}{k-1}+o(n^{k-1})\\ &=&\alpha\frac{n^{k-1}}{(k-1)!}+o(n^{k-1}). \end{eqnarray*} To evaluate the constant $\alpha$, just multiply the generating function and the partial fractions decomposition by $(1-t)^k$ and take the limit at 1, resulting in $\alpha=1/k!$. Then, Schur's approximation to the number of solutions of $x_1+2x_2+\cdots+kx_k=n$ is $$\frac{n^{k-1}}{(k-1)!\,k!} .$$

$\endgroup$
3
  • $\begingroup$ Is there any relationship between this equation and the partition function? If there any such relation, could you please explain it? $\endgroup$
    – Bumblebee
    Commented Jan 5, 2017 at 17:42
  • 2
    $\begingroup$ If by "partition function" you mean partitions of integers, then your problem is a finite version of partitions of integers. The generating function that marks the number of partitions of an integer would be $\frac{1}{(1-t)(1-t^2)(1-t^3)\cdots}$. $\endgroup$
    – Rus May
    Commented Jan 11, 2017 at 14:41
  • $\begingroup$ If we substitute $x_k=y_1$, $x_k+x_{k-1}=y_2$ and finally $x_k+\cdots+x_2+x_1=y_k,$ then $n=y_1+y_2+\cdots+y_k,$ where $0\le y_1\le y_2\le \cdots\le y_k.$ Therefore actually we need is number of ways that $n$ can written as sum of $k$ non-negative integers without considering the order. Am I correct? Is there any way to do this without using generating functions? $\endgroup$
    – Bumblebee
    Commented Jan 12, 2017 at 5:42
0
$\begingroup$

Read my answer here; the number of partitions of $n$ with largest part $k$ is the same as the number of partitions into $k$ parts, which is $p(n,k)$. Using this, the number of solutions of the equation $\sum i x_{i} = n$ is equal to :

$$p(n, 1) + p(n, 2) + \cdots + p(n, k).$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .