BdMO Online Forum

Find a closed formula for:
\[ \displaystyle \sum\limits_{i,j=0}^\infty\binom{n}{3i}\binom{n-3i}{3j} \]

I am sorry but I couldn't understand what you are trying to say? Please explain it.

To calculate this sum!

yo79 wrote:Find a closed formula for:
\[ \displaystyle \sum\limits_{i,j=0}^\infty\binom{n}{3i}\binom{n-3i}{3j} \]

Define $n$ at first. It has been used without any definition. Or do you want a formula for that sum involving $n$?

I am guessing, $n\in\mathbb{N}$....

n is a natural number! The answer must be something simple like: $ \binom{n-1}{2n+5} $ or something like $ \frac{n^2+n \binom{n}{2n}}{45} $ (these are not the answers but they are some examples of answers) or other form, something like these!

Note that:
$(a+\omega )^n+(a+\omega ^2)^n+(a+1)^n=3\sum_{i=0}^{\infty}\binom{n}{3i}a^{n-3i}$
Where $1+\omega+\omega^2=0$ and $\omega^3=1$

Now,
$\sum_{i=0,j=0}^{\infty}\binom{n}{3i}\binom{n-3i}{3j}=\sum_{i=0}^{\infty}\binom{n}{3i}\left [ \sum_{j=0}^{\infty}\binom{n-3i}{3j}\right ]$

$=\sum_{i=0}^{\infty}\binom{n}{3i}\frac{1}{3}\left [(1+\omega )^{n-3i} +(1+\omega ^2)^{n-3i}+(1+1)^{n-3i}\right ]$

$=\sum_{i=0}^{\infty}\binom{n}{3i}\frac{1}{3}\left [(-\omega )^{n-3i} +(-\omega ^2)^{n-3i}+2^{n-3i}\right ]$

$=\frac{1}{9} [ \left \{ (-\omega ^2+\omega)^n+(-\omega^2+\omega^2)^n+(-\omega^2+1)^n \right \}+$
$\left \{ (-\omega+\omega)^n +(-\omega+\omega^2)^n+(-\omega+1)^n\right \} +$
$\left \{ (2+\omega)^n+(2+\omega^2)^n+(2+1)^n \right \} ]$

$=\frac{1}{9}\left [ (-\omega ^2+\omega )^n+(\omega ^2-\omega )^n+2(-\omega ^2+1)^n+2(-\omega +1)^n+3^n \right ]$

I don't know how to simplify more. I will be grateful to you if you find any other suitable form.

Thank you!