Determining the Lambert series for $xq+x^2q^4+x^3q^9+...+x^nq^{n^2}+...$

Question

I am trying to determine the polynomials $P_n(x)$ from $$ xq+x^2q^4+x^3q^9+...+\ x^nq^{n^2}+...=\sum_{n\geqslant1}\frac{P_n(x)q^n}{1-xq^n}; $$ that is, $$ \sum_{d|n}x^{\frac nd-1}P_d(x)=\begin{cases}x^k,&n=k^2,\\0&\text{otherwise.}\end{cases} $$

Or if you prefer, the Dirichlet generating function is $$ \sum_{n\geqslant1}\frac{P_n(x)}{n^s}=\frac{x\operatorname{Li}_{2s}(x)}{\operatorname{Li}_s(x)}. $$ In particular ($p$, $p_1$, $p_2$, $p_3$ are primes), \begin{align*} P_1(x)&=x\\ P_p(x)&=-x^p\\ P_{p^2}(x)&=-x^{p^2}+x^{2p-1}+x^p\\ P_{p_1p_2}(x)&=-x^{p_1p_2}+2x^{p_1+p_2-1}\\ P_{p^3}(x)&=-x^{p^3}+2x^{p^2+p-1}-x^{3p-2}-x^{2p-1}\\ P_{p_1p_2^2}(x)&=-x^{p_1p_2^2}+2x^{p_1+p_2^2-1}+2x^{p_1p_2+p_2-1}-3x^{p_1+2p_2-2}-x^{p_1+p_2-1}\\ P_{p_1p_2p_3}(x)&=-x^{p_1p_2p_3}+2x^{p_1p_2+p_3-1}+2x^{p_1p_3+p_2-1}+2x^{p_2p_3+p_1-1}-6x^{p_1+p_2+p_3-2} \end{align*} but how to handle the whole thing?

In fact ultimately I only need an explicit (well...) expression for $P_n'(1)$. Note that $P_n(1)$ is the Liouville function.

@Alexey Yes, my motivation is to understand better a particular formal group described in a question before — მამუკა ჯიბლაძე, Sep 06 '15 at 04:51
In what sense do you only need an explicit expression for $P'_n(1)$? — Kevin Smith, Sep 06 '15 at 12:27
I have some complicated expression for $P_n'(1)$, something like $\sum_{d_1d_2|n}\mu(d_1)\mu(d_2)\text{something}(\frac n{d_1d_2})$ and I would like to have a more transparent formula, that's all I need for now. — მამუკა ჯიბლაძე, Sep 06 '15 at 12:38

score 1 · Answer 1 · edited Sep 09 '15 at 09:03

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This problem crucially involves the analogue of the Mobius function $\mu(n)$ for polylogarithms. However, the "polylogarithmic" analogues $\mu_n(x)$, say with $\mu(n)=\mu_n(1)$, are generally not multiplicative, so one cannot use the Fundamental Theorem of Arithmetic to determine them (this is one of the properties Lambert series work nicely with). I'm pretty sure no non-recursive (see below) formula is known for $\mu_n(x)$ and, therefore, neither for the polynomials $P_n(x)$.

Let $\mu_n(x)$, $n\in\mathbb{N}$, be the unique sequence of polynomials in $x$ of degree $n$ with the property that $$\sum_{d|n}\mu_{n/d}(x)x^{d-1}=\delta(n)$$ so $\mu_n(x)$ is the Dirichlet inverse of $x^{n-1}$, in other words, $$x{Li}^{-1}_s(x)=\sum_1^{\infty}\frac{\mu_n(x)}{n^s}.$$

Forming the Dirichlet product of the Dirichlet product in the question (your second displayed equation) with $\mu_{n}$ gives an expression for $P_n$ explicitly in terms of $\mu_n$, that is $$P_n(x)=\sum_{d|n: d=k^2}\mu_{n/d}(x)x^{\sqrt{d}}.$$

Now let $$f*g(n)=\sum_{d|n}f\left(\frac{n}{d}\right)g(d)$$ denote the Dirichlet convolution of $f$ and $g$, and $$f^{* j}(n)=\underbrace{f* f* \cdots *f}_{j\text{ copies of }f}(n)$$ denote Dirichlet self-convolution to order $j$, that is, the Dirichlet convolution of $f$ with itself $j-1$ times. The reason I claim that no non-recursive solution to your problem is known is simply that, when $f(n)={n\choose m}$, I believe a closed form for the Dirichlet self convolution to order $j$ is not known, i.e, one must actually carry out $j$ nested summations, and this is related to the fact that the binomial coefficient is not multiplicative (of course this could be incorrect so if anyone knows otherwise please comment on the solution here).

Anyway, if $$\frac{\mu_n(x)}{x}=\sum_0^{n}c_k(n)(x-1)^k,$$ then $$c_k(n)=\sum_{j_1+2j_2+\cdots=k}\frac{(-1)^{j_1+j_2+\cdots}}{j_1!j_2!\cdots}\prod_{p|.}{j_1+j_2+\cdots\choose \alpha_p}*{.\choose 1}^{* j_1}*{.\choose 2}^{* j_2}*\cdots(n)$$ where $\alpha_p$ is the exponent of $p$ in the factorisation of $n$ and the summation extends over all partitions of $k$. This is derived using Faa di Bruno's formula for derivatives of composite functions applied to the reciprocal of $Li_s(x)$. From here you can easily get a formula for $P_n(x)$.

edited Sep 09 '15 at 09:03

მამუკა ჯიბლაძე

17,468

answered Sep 05 '15 at 12:58

Kevin Smith

2,470

Very interesting. Do you know any investigations in the nature of $\mu_n(x)$? I also tried to work with available recursive expressions for $P_n$, but unfortunately could not obtain anything sufficiently explicit for my purposes. – მამუკა ჯიბლაძე Sep 05 '15 at 16:56
Or could you extract more information about just $P_n'(1)$ with your approach? – მამუკა ჯიბლაძე Sep 05 '15 at 16:58
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On the nature of $\mu_n(x)$, one would expect something to have been written about reciprocal polylogarithms, but it's not a particularly google-friendly search I think. – Kevin Smith Sep 05 '15 at 18:15
I don't even figure out what an efficient keyword would be for searching... – მამუკა ჯიბლაძე Sep 05 '15 at 18:18
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I have edited the answer to include a formula for $\mu_n(x)$. – Kevin Smith Sep 08 '15 at 12:04
Could you please explain what is $\star_j$? – მამუკა ჯიბლაძე Sep 09 '15 at 06:24
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I have changed the notation slightly and included clearer definitions. This formula really is elementary - if you differentiate successively $Li^{-1}_s(x)$ w.r.t. $x$, then look at the coefficient of $n^{-s}$, you'll find this is the result. I don't know about mathematica but the convolutions will be easy to implement in Sage. – Kevin Smith Sep 09 '15 at 08:52
Oh I see, so this $_j$ was just numbering, I thought these were some different kinds of convolutions. I've added underbrace there for clarity, if it's not OK just remove it. – მამუკა ჯიბლაძე Sep 09 '15 at 09:01
I see. Feel free to edit it the way you find most appropriate. – მამუკა ჯიბლაძე Sep 09 '15 at 09:05
Like I said, you can just grind this out using Faa di Bruno, using the fact that the coefficients of $\zeta^{-m}(s)$ are the $$\prod_{p|n}{m\choose \alpha_p}.$$ – Kevin Smith Sep 09 '15 at 09:09
Let us continue this discussion in chat. – Kevin Smith Sep 09 '15 at 09:16

მამუკა ჯიბლაძე · Answer 2 · 2015-09-09T06:26:19.697

Decided to add an expression for $P_n'(1)$ as I finally got it simplified to certain extent. I decided still to leave no answers approved as (1) I would be grateful for any possible further alternative expressions for $P_n'(1)$ and (2) the question itself is broader and seems to be interesting to others in this generality. Besides, this expression for $P_n'(1)$ somehow gives hopes for (and maybe hints at) further simplifying $P_n(x)$ along the lines of the answer given by Kevin Smith: after all, knowing $n$ values like $P_n(k)$, $P_n'(l)$, $P_n''(m)$, ... one could reconstruct $P_n(x)$, right?

So what I obtained is $$ P_n'(1)=\lambda(n)\surd(n)-\sum_{1<d|n}\lambda\left(\frac nd\right)\varphi(d) $$ where $\lambda$ is the Liouville function, $\varphi$ is the Euler totient function and $\surd(n)$ is the square root of the largest square dividing $n$.

This may be seen from \begin{align*} \sum_{n\geqslant1}\frac{P_n'(1)}{n^s}=\left.\frac d{dx}\left(\frac{x\text{Li}_{2s}(x)}{\text{Li}_s(x)}\right)\right|_{1\leftarrow x}&=\left.\frac{\text{Li}_{2s-1}(x)+\text{Li}_{2s}(x)}{\text{Li}_s(x)} -\text{Li}_{2s}(x)\frac{\text{Li}_{s-1}(x)}{\text{Li}_s(x)^2}\right|_{1\leftarrow x}\\ &=\frac{\zeta(2s-1)}{\zeta(s)}+\frac{\zeta(2s)}{\zeta(s)}-\frac{\zeta(2s)}{\zeta(s)}\frac{\zeta(s-1)}{\zeta(s)}. \end{align*}

Determining the Lambert series for $xq+x^2q^4+x^3q^9+...+x^nq^{n^2}+...$

2 Answers2