# Partition function and "Euler function" - what does it mean?

Denote the partition function by $$p_k(n)$$, and define it as a count of the number of possible sequences of positive integers $$a+b+c+...=n$$ where the $$a,b,c,...$$ are not necessarily distinct (so that, for example, $$1+3+4$$ is not counted as distinct from $$1+4+3$$).

I know Wikipedia can be unreliable. But I read here that "The multiplicative inverse of its [the partition function's] generating function is the Euler function." Does this mean Euler's totient function $$\phi(n)$$? It's impossible to tell from the context or the links. I've searched around and I have come out none the wiser.

The Wikipedia link asserts that the "Euler function" is given by

$$\phi(q)=\prod_{k=1}^\infty (1-q^k)$$

But it goes into no further detail apart from cryptic references to "$$q$$-series. And when you follow links to that, you end up with no answers at all.

Could someone please explain what the quote above means - ideally in algebraic form? I understand what a generating function is (though the article doesn't say which sort it is referring to), but I would really like to see that this under-explained statement as a formula.

• The Wikipedia article actually points to en.wikipedia.org/wiki/Euler_function, where it says "not to be confused with Euler totient function").
– user700480
Commented Nov 20, 2019 at 13:19
• Hi @Stinking Bishop (a fine cheese, BTW!). Yes, I got to that belatedly. So, is the answer that the two functions are self-referential - they are both defined in terms of the other, but there's no wider context?> I feel like I'm missing something basic here... Commented Nov 20, 2019 at 13:23
• By the way, you're very close to modular forms - keep going; they're wonderful ;) Commented Nov 20, 2019 at 13:29

This is what it means: if you write a formal power series:

$$P(x)=\sum_{n=0}^{\infty}p(n)x^n$$

and you write the Euler function as a formal product (which then can be unwound as a formal power series):

$$\phi(x)=\prod_{n=1}^{\infty}(1-x^n)$$

then, at least formally, $$P(x)\phi(x)=1$$. ("Formally" means that you can do the additions and multiplications to get all the coefficients in the product, without regard to convergence, and then all the coefficients of $$x^n$$ in $$P(x)\phi(x)$$, except for the constant coefficient $$1$$, turn out to be $$0$$!)

The identity is also valid for the actual power series whenever both sides exist and converge.

• Awesome, thank you! Commented Nov 20, 2019 at 14:25