Number of partitions of $12$

Question

How can I calculate the coefficient of $x^{12}$ in the expression $$(1+x^2+x^4+x^6+x^8+x^{10}+x^{12})\cdot (1+x^4+x^8+x^{12})\cdot$$ $$ \cdot (1+x^6+x^{12})\cdot (1+x^8)\cdot (1+x^{10})\cdot (1+x^{12})$$

Answer 1

Since we have $\left(\sum\limits_{a=0}^6 x^{2a}\right)\left(\sum\limits_{b=0}^3 x^{4b}\right)\left(\sum\limits_{c=0}^2 x^{6c}\right)\left(\sum\limits_{d=0}^1 x^{8d}\right)\left(\sum\limits_{e=0}^1 x^{10e}\right)\left(\sum\limits_{f=0}^1 x^{12f}\right)$

It is equivalent to finding the number of integer solutions of $$2a+4b+6c+8d+10e+12f=12$$

In this case the constraints on $c,d,e,f$ are so strong that it can easily be done manually.

rem: when not specified, all other variables are $0$

• $4)\qquad$ $f=1\quad$ or $\quad c=2\quad$ or $\quad b=3\quad$ or $\quad a=6$
• $1)\qquad$ $e=1$ then $a=1$
• $2)\qquad$ $d=1$ then $a=2$ or $b=1$
• $2)\qquad$ $c=1$ then $a=3$ or $a=b=1$
• $2)\qquad$ finally $2a+4b=12$ gives two solutions $a=b=2$ or $a=4,b=1$

In total $4+1+2+2+2=11$ is the coefficient of $x^{12}$.

Answer 2

HINT:

$$ 1+x+x^2+x^3+\ldots =\frac{x}{x-1} $$

for formal sums.

Your first factor is

$$ 1+x^2+x^4+\ldots=\frac{x}{1-x^2} $$

you can ignore the fact that this is an infinite sum since you are interested in a certain coefficient therefore can ignore the rest.

For more hints check this link out.

Hope this helps

Answer 3

Note that every term in the product comes from a product of one term per polynomial. There are two cases to consider

Case $1$: The term in question involves a non-$1$ value from one of the last $4$ polynomials.

Note that there can only be one such value, since all of the non-$1$ terms are at least $\sqrt{x^{12}}$.

If either $12$ is chosen, clearly we have $1$ remaining choice left for the product, so this gives us $\color{red}2$ cases.

If $10$ is chosen, the remaining $2$ must come from the first polynomial, so this gives us $\color{red}1$ case.

If $8$ is chosen, the $4$ can come from either the first or second polynomials, so this gives us $\color{red}2$ cases.

If the $6$ is chosen, we can either get the $6$ from the first polynomial, or get $2$ from it and $4$ from the second, giving us $\color{red}2$ cases.

In total, we have $\color{red}7$ cases from this case.

Case $2$: The terms from the last $4$ polynomials are $1$. We can essentially ignore them. In this case, the choice of the term from the second polynomial fully determines which term from the first polynomial we need to choose to get a term of coefficient $12$. So, there are $\color{red}4$ cases.

So, our final total is $$7+4=\color{red}{11}$$Wolfram Alpha confirms the answer here (below is the full polynomial)