Number of d-Calabi-Yau partitions

Question

This problem arises from algebraic geometry/representation theory, see https://arxiv.org/pdf/1409.0668.pdf (chapter 2).

We call a partition $p=[p_1,...,p_n]$ with $2 \leq p_1 \leq p_2 \leq ... \leq p_n$ d-Calabi-Yau (for $d \geq 1$) when $n-d-1=\sum\limits_{i=1}^{n}{\frac{1}{p_i}}$.

For example for $d=1$ there are 4 1-Calabi-Yau partitions: [2,2,2,2],[3,3,3],[2,4,4] and [2,3,6]. For $d=2$ there are 18, see example 2.15. (b) in https://arxiv.org/pdf/1409.0668.pdf .

Question 1: Is there a nice formula for the $d$-Calabi-Yau partitions for a fixed $d$?

It seems hard to calculate the number for $d \geq 3$ even with a computer since a term $p_i$ can be pretty big.

Question 2: For a given $d$, what is the maximal value a term $p_i$ can have?

For $d=1$ it is 6 and for $d=2$ it is 42.

Answer 1

When $d=n-2$ (which constitutes the bulk of the cases) these partitions are known as Egyptian fractions, and their number is tabulated at http://oeis.org/A002966. No closed formula seems to exist.

For arbitrary $n$ and $d$ there is http://oeis.org/A156869, but one has to be careful because the value 1 is allowed there. So whilst there is no nice formula, the number of $d$-CY partitions should be given by http://oeis.org/A156871 (with the caveat that 1 is allowed).

For your second question, one has to list all Egyptian fractions (but that is not a straightforward thing to do) and read off the answer, although there might be an algorithm to write down the "biggest" partition somehow. In any case, the maximum for $d=3$ seems to be 1806 if I'm not mistaken (see also https://oeis.org/A007018).

Answer 2

Following up on @pbelmans's answer and researching an article mentioned in OEIS A007018, I believe your Question 2 was answered just under 100 years ago by David Curtiss (On Kellogg's Diophantine problem, Amer. Math. Monthly 29 (1922) 380-387).*

Curtiss confirms Kellogg's conjecture that the maximum $x_i$ in any $$\frac{1}{x_1} + \frac{1}{x_2} + \cdots + \frac{1}{x_n} = 1$$ is given by the sequence $u_1 = 1$ and $u_{k+1} = u_k(u_k+1)$ which begins 1, 2, 6, 42, 1806, 3263442. On p386 he explains, "But the value $u_n$ is actually attained by giving to the $x$'s the values $u_k+1$, so that $u_n$ is the maximum of $f_{n-1}(x)$." (That last expression is something he defined to simplify the proof.) Indeed, \begin{gather*} \frac{1}{1+1} + \frac{1}{2+1} + \frac{1}{6} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = 1, \\ \frac{1}{1+1} + \frac{1}{2+1} + \frac{1}{6+1} + \frac{1}{42} = \frac{1}{2} + \frac{1}{3} + \frac{1}{7} + \frac{1}{42} = 1, \\ \frac{1}{1+1} + \frac{1}{2+1} + \frac{1}{6+1} + \frac{1}{42+1} + \frac{1}{1806} = \frac{1}{2} + \frac{1}{3} + \frac{1}{7} + \frac{1}{43} + \frac{1}{1806} = 1, \dots \end{gather*}

* There's also a sci.math.research thread from 1996 where a claim that Curtiss's proof is wrong is retracted, but the proof does seem to be difficult to follow. Gerry Myerson provided a reference to a simpler 1995 proof by Izhboldin and Kurliandchik.

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I'm including a comment to @katago here so that answers to both questions are in one place. Looking through MathSciNet, the most recent upper bound may be Browning & Elsholtz 2011, something like $$(1.264085...)^{(5/12)^{2^{n−1}}}$$ where apparently the 1.26 is a known constant that arises in these problems. There are also lower bounds in the literature.

So the short answer to Q1 is no, at least not yet.