Generating functions finding $ a_n $ formula

Question

When I have generating function in form

$$ A(x) = \frac{11x-1}{(1-3x)(1-7x)} $$

I know that the one way to find formula for $ a_n $ is to find partial fraction from this formula and then change this to power series.

But when I have to deal with generating function of integer partition I don't know how I should start.

I have function:

$$ \sum_{n=0}^\infty a_nx^n = (1+x+x^2) \prod_{i=1}^\infty \frac{1}{1-x^{2i+1} }$$

And I have to check if it is true for this function that:

$a_{100}>20 $ or $a_7 = 5$ or $a_n \neq 0$ for $n = 0,1,2...$

I know that $(1+x+x^2) \prod_{i=1}^\infty \frac{1}{1-x^{2i+1} }$ is partition where every element is odd which I can write as: $ (1+x+x^2)*(1+x^3+x^6+x^9+...)*(1+x^5+x^{10}+....)*... $

which I can write as

$ \frac{1-x^3}{1-x}*(\frac{1}{1-x^3}\frac{1}{1-x^5}*...) $

and I don't know what to do later. I think it's wrong thinking, or maybe is here any trick to check it fast.

Answer 1

First, note that all the terms in $$ \prod_1^\infty \frac1{1-x^{2i+1}} $$ are of the form $\sum_{k=0}^\infty x^{ki}$ which has only non-negative coefficients and has a leading term of $1$. $(1+x+x^2)$ is also of that form. Therefore, if we compare the expansion of $$ (1+x+x^2) \prod_1^m \frac1{1-x^{2i+1}} = \sum b_n x^n $$ to the expansion of $$ (1+x+x^2) \prod_1^{m+1} \frac1{1-x^{2i+1}} = \sum c_n x^n $$ we see that each of the $c_n$ is the sum of $b_n$ arising from the leading $1$ in $\frac1{1-x^{2m+3}}$ and other non-negative terms; thus for all $n$, we know $c_n \geq b_n$. So if a coefficient $a_n$ in a partial product is positive, then the coefficient in the full product will be at least as large.

Now look at the first case: $$ (1+x+x^2) \frac1{1-x^3} = \frac1{1-x} \sum x^n $$ which has all its $a_n = 1 > 0$. So all the $a_n$ in the full product are > 0$.

Now consider that $$(1+x+x^2+x^3+\cdots)(1+x^5+x^{10}+\cdots) = \sum (1+\left\lfloor\frac{n}{5}\right\rfloor) x^n $$ and then when we include the $1+x^7+ x^14\cdots)$ factor we have $$ \sum (1+\left\lfloor\frac{n}{5}\right\rfloor + \left\lfloor\frac{n}{7}\right\rfloor ) x^n $$ So $a_7 = 3 \neq 5$.

Finally, $$ a_{100} = \left(1+\left\lfloor\frac{100}{5}\right\rfloor + \left\lfloor\frac{100}{7}\right\rfloor + \left\lfloor\frac{100}{9}\right\rfloor +\cdots + \left\lfloor\frac{100}{9}\right\rfloor \right) \\= 1 + 20 + 14 + 11 + 9 + 7 + 6 + 2\cdot 5 +3\cdot 4 + 5\cdot 3 + 8\cdot 2 + 50\cdot 1 = 171>20 $$ I suppose you could tell this just from the first two terms...

Answer 2

The expression $(1+x+x^2) \prod_{i=1}^{\infty} \frac{1}{1-x^{2i+1}}$ counts partitions into $1,5,7,9,...$

To answer your three questions:

1) The number of partitions of $100$ into $1$ and $5$ is already exactly $20$, so the number of partitions of $100$ into $1,5,7,9,11,...$ is strictly bigger than $20$.

2) The number of partitions of $7$ into $1,5,7$ is $3$, not $5$.

3) The number of partitions into $1,5,7,9,11,...$ is never $0$ because any $n$ can be written in the form $n = \underbrace{1 + ... + 1}_{n\, \mathrm{times}}$.