# Does Rademacher's convergent series for p(n) define an analytic function?

Let $$p(n)$$ be the number of partitions of $$n\geq 0$$. We can let $$n$$ be any complex number in Rademacher's convergent infinite series for $$p(n)$$. (See e.g. equation (24) here.) For what $$n$$ does it converge? Does it define an analytic function for such $$n$$? If so, what are the properties of this analytic function (singularities, branch points, domain of existence, etc.)? This question asked for an analytic continuation of $$p(n)$$, so I am suggesting a possible answer.

Edit. We can write the series in the form $$p(n)=\sum A_k(z)\frac{d}{dz}f(z/k^2),$$ where $$|A_k(z)|\leq Ck^{1/2}e^{C_1(\Im z)^+},$$ where $$y^+=\max\{ y,0\},$$ and $$C_j$$ are various positive absolute contants, $$f(z)=(\sinh\sqrt{z})/\sqrt{z},\; z=C_2(n-1/24)$$. Notice that $$f$$ is an even entire function of order $$1/2$$, normal type so $$f'(0)=0$$. So $$p(n)=\sum c_k(z)f'(z/k^2),$$ where $$|c_k(z)|\leq C_3k^{-3/2}e^{C_1(\Im z)^+}$$. Since $$f'$$ is an entire function of order $$1/2$$, normal type and $$f'(0)=0$$, we have an estimate of the form $$|f'(z)|\leq C_2|z|e^{A\sqrt{|z|}},$$ and we see that the sum above must satisfy the estimate $$|p(z)|\leq C_4e^{C_1(\Im z)^++A\sqrt{|z|}}.$$

Therefore $$p(n)$$ extends to an entire function in the plane of exponential, type. An interesting question about such a function is the location of its zeros, and the related question, asymptotics for large $$|z|$$. One could try to plot $$|p(z)|$$ on a computer and look what to expect. Even a plot on the imaginary line may tell us something interesting.

Based on this argument, and pictures of Fredrik Johansson, I conjecture that zeros are asymptotic to some parabola $$y=C\sqrt{x}$$ in the right half-plane, and close to the real axis in the left half-plane (the pictures indicate infinitely many negative zeros). To prove this one has to prove the asymptotics $$p(z)=\exp\left(B(\Im z)^++A\Re\sqrt{z}+o(\sqrt{|z|})\right)$$ outside some small circles around zeros.

Remark. Unfortunately, this crude argument gives $$C_1=B=2\pi$$, and in such class of entire functions extension from integers to the complex plane is not unique.

• Where did the dependence on $e^{\pi i n}$ in $a_k$ go in this analysis? You can clearly see the $e^{\pi |z|}$ growth in the imaginary direction in the plots I made. Can this be suppressed? – Fredrik Johansson Jul 30 at 5:43
• @Fredrik Johansson: Thanks. I did not notice $e^{2\pi i n}$ in the $A_k$. Now I corrected. As a result it has exponential type (not order 1/2, as I said before), and may grow in the upper half-plane. – Alexandre Eremenko Jul 30 at 13:09

Not a direct answer to the question, but a brief numerical exploration of this function.

First, a trivial observation: we can write either $$e^{\pi i x}$$ or $$\cos(\pi x)$$ in the formula for the exponential sum $$A_k(n)$$ in the Rademacher series. This makes no difference at integer $$n$$, but gives different generalizations for noninteger $$n$$. The cosine version has the nice property of being real-valued on the real line and conjugate symmetric.

Here is a plot of the cosine-extended $$p(n)$$ on the real line:

The exponential version (real and imaginary parts):

Either version of the function seems to have simple zeros at the negative integers $$-1, -2, -3, ...$$. (This is quite nice if correct, because it matches the obvious combinatorial interpretation $$p(-n) = 0$$.) The cosine version has additional zeros on the negative real line (the first near $$-0.93$$).

At half-integers, it appears that all terms in the cosine version of the Rademacher series except the first term vanish, and so one has a trivial closed-form evaluation of $$p(k+\tfrac{1}{2})$$, $$k \in \mathbb{Z}$$. This is not the case for the exponential version.

Taking this leading term as a cue for the asymptotics on the real line, the origin is a turning point between exponential growth to the right and $$\text{oscillation} \cdot O(n^{-1})$$ behavior to the left.

Viewed in the imaginary direction, the exponential version appears to grow exponentially as $$n \to +i \infty$$ but remains small as $$n \to -i \infty$$. Plot of the real and imaginary parts of $$p(i x)$$:

The cosine version looks like the upper-half-plane exponential version:

There are additional zeros in complex plane. Since the Rademacher series converges slowly when the imaginary part of $$n$$ is large, it's a bit difficult to explore these zeros numerically. Here is a low-resolution plot of the exponential version of $$p(z)$$ on $$z \in [-4,4] + [-2,2] i$$:

And the cosine version:

The slow convergence also makes it difficult to search numerically for other potential closed forms (it is expensive to get more than ~6 digits). A faster algorithm for computing $$p(n)$$ to high precision near the origin would be very exciting.

Unoptimized Python implementation that I used to create these plots: https://gist.github.com/fredrik-johansson/7c2711887811ef9f2d7038b8451a4e63