1
$\begingroup$

How many solutions of equation $x_1+x_2+x_3+x_4=n$ in $N_0$ such that $x_1\leq x_2\leq x_3 \leq x_4$?

I found solutions of $x_1+x_2+x_3=n$ in $N_0$ , $x_1\leq x_2\leq x_3 $ in the following way :

Let $S$ set of all solutions and $A_k$ set of all solutions of equation $x_1+x_2+x_3+x_4=n$, for which is $k=x_1\leq x_2\leq x_3 $, $k\in 1,2,..., [\dfrac {n}{3} ]$. Sets $A_0,...,A_{[\dfrac{n}{3}]}$ are disjoint.
$a=x_2-k$, $b=x_3-k$. $|A_k|$ is number of pairs $(a,b)$ such that $a+b=n-3k$ and $a\leq a \leq b$
$|A_k|=[\dfrac{n-3k}{2}]+1$ and $|S|=\sum _{k=0}^{[\dfrac{n}{3}]}|A_k|$. It is complicated for equation $x_1+x_2+x_3+x_4=n$ . I need another way to solved it.

$\endgroup$
  • $\begingroup$ I assume that you want the $x_i$'s to be positive from the statement about $k$. $\endgroup$ – Michael Burr Apr 15 '15 at 13:20
  • $\begingroup$ Positive or zero. $\endgroup$ – JJMM Apr 15 '15 at 13:38
  • $\begingroup$ Let $y_1=x_1,y_2=x_2-x_1,y_3=x_3-x_2,y_4=x_4-x_3$ and solve for $y_i\geq 0$ the equation $4y_1+3y_2+2y_3+y_4=n$ $\endgroup$ – Lozenges Apr 15 '15 at 13:48
0
$\begingroup$

Sometimes a little research can help. The formula you posted was discovered by Jon Perry in 2003.

The generating function for this problem is:

$$g(x) = \frac{1}{(1-x) \left(1-x^2\right) \left(1-x^3\right) \left(1-x^4\right)} $$

There does not seem to be something simple for your question but Michael Somos comes up with

$$a(n)=\text{Round}\left[\frac{1}{288} \left(2 (n+5)^3-3 (n+5) \left(5+3 (-1)^{n+5}\right)\right)\right] $$

For a whole lot more:

A001400

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.