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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.

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    $\begingroup$ The Wikipedia article actually points to en.wikipedia.org/wiki/Euler_function, where it says "not to be confused with Euler totient function"). $\endgroup$
    – user700480
    Commented Nov 20, 2019 at 13:19
  • $\begingroup$ 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... $\endgroup$ Commented Nov 20, 2019 at 13:23
  • $\begingroup$ By the way, you're very close to modular forms - keep going; they're wonderful ;) $\endgroup$ Commented Nov 20, 2019 at 13:29

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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.

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  • $\begingroup$ Awesome, thank you! $\endgroup$ Commented Nov 20, 2019 at 14:25

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