On the density of the set of known Hadamard orders

Abstract

Let S(x) be the number of n ≤ x for which a Hadamard matrix of order n exists. Hadamard’s conjecture states that S(x) is about x/4. From Paley’s constructions of Hadamard matrices, we have that

$$ S(x) = \Omega\left( \frac{x}{\log x} \right). $$

In a recent paper, the first author suggested that counting the products of orders of Paley matrices would result in a greater density. In this paper we use results of Kevin Ford to show that it does:

$$S(x) \geq \frac{x}{\log x} \exp\left((C+o(1))(\log \log \log x)^2 \right)\,, $$

where C = 0.8178....

This bound is surprisingly hard to improve upon. We show that taking into account all the other major known construction methods for Hadamard matrices does not shift the bound. Our arguments use the notion of a (multiplicative) monoid of natural numbers. We prove some initial results concerning these objects. Our techniques may be useful when assessing the status of other existence questions in design theory.

Appendices

Appendix

A Monoids and sets of natural numbers

In this appendix, we prove two theorems showing that taking products of sets does not greatly increase asymptotic density. We give two sets of proofs; one elementary and self-contained, and the other shorter but depending on results on generating functions.

In this paper, we use some standard notation to discuss the growth of the counting function of a set: Let f:$\mathbb{N}$→$\mathbb{R}$ be a function. Then

• “A(x) = O(f(x))” means that there is a constant C > 0 and x ₀ ∈ $\mathbb{N}$ such that A(x) < C f(x) for all x ≥ x ₀,

• “A(x) = Ω(f(x))” means that there is a constant C > 0 and x ₀ ∈ $\mathbb{N}$ such that A(x) > C f(x) for all x ≥ x ₀.

• “A(x) = Θ(f(x))” means that there are constants c ₁ > c ₂ > 0 and x ₀ ∈ $\mathbb{N}$ such that c ₁ f(x) ≥ A(x) ≥ c ₂ f(x) for all x ≥ x ₀.

• “A(x) = o(f(x))” means that for any constant C > 0 there is some x ₀ ∈ $\mathbb{N}$ such that A(x) < C f(x) for all x ≥ x ₀.

2.1 A.1 Elementary proofs

For any subset $\mbox{$\cal{A}$}$ of $\mathbb{N}$ and any x ∈ $\mathbb{N}$, let

$$ a(x)=|\mbox{$\cal{A}$}\cap(x/2,x]|\qquad\mbox{and}\qquad\bar{a}(x)=|\mbox{$\cal{A}$}\cap[x/2,x]|\,. $$

Lemma A.1

Let $\mbox{$\cal{A}$},\mbox{$\cal{B}$}$ and $\mbox{$\cal{C}$}=\mbox{$\cal{A}$}\mbox{$\cal{B}$}$ be subsets of $\mathbb{N}$ which are monoids. Then, for all x ∈ $\mathbb{N}$,

$$\label{ProductDensityBound} \frac{c(x)}{b(x)} \leq \sum\limits_{k=1}^{\lceil\log_2x\rceil} \frac{(b(x/2^{k-1})+b(x/2^k))\bar{a}(2^k)}{b(x)}\,. $$

(6)

Moreover, if the righthand side is bounded by a constant c₁, say, for all x ∈ $\mathbb{N}$, then C(x) = Θ(B(x)).

Proof

Since every element of $\mbox{$\cal{C}$}\cap(x/2,x]$ can be written in the form ab, where, for some k ∈ {1,2,...,⌈log₂ x⌉}, $a\in \mbox{$\cal{A}$}\cap[2^{k-1},2^k]$ and $b\in \mbox{$\cal{B}$}\cap(x/2^{k+1},x/2^{k-1}]$, we have

$$ c(x) \leq \sum\limits_{k=1}^{\lceil\log_2x\rceil} (b(x/2^{k-1})+b(x/2^k))\bar{a}(2^k)\,. $$

Dividing through by b(x) then gives (6). We now prove the second part of the lemma. By hypothesis, we have

$$ c(x) \leq b(x)\left\{ \sum\limits_{k=1}^{\lceil\log_2x\rceil} \frac{(b(x/2^{k-1})+b(x/2^k))\bar{a}(2^k)}{b(x)} \right \} \leq c_1 b(x)\,. $$

Now we have the following partition

$$ \mbox{$\cal{C}$}\cap[1,x]=\bigcup\limits_{k=1}^{\lceil\log_2 x\rceil}\mbox{$\cal{C}$}\cap(x/2^k,x/2^{k-1}] $$

for $\mbox{$\cal{C}$}\cap[1,x]$ and a similar partition for $\mbox{$\cal{B}$}\cap[1,x]$. So

$$ C(x)= \sum\limits_{k=1}^{\lceil\log_2x\rceil} c(x/2^{k-1}) \leq c_1\sum\limits_{k=1}^{\lceil\log_2x\rceil} b(x/2^{k-1}) =c_1B(x)\,. $$

Since $\mbox{$\cal{B}$}\subset\mbox{$\cal{C}$}$, we then have B(x) ≤ C(x) ≤ c ₁ B(x). This completes the proof of the second part of the lemma.□

Proof of Theorem 3.2

For some constants c ₁,c ₂ > 0,

$$ b(x) = B(x)-B(\lfloor{x/2}\rfloor) \geq c_1 x^{\beta}-c_2 \left(\frac{x}{2}\right)^{\beta} = \left(\frac{x}{2}\right)^{\beta}(2^{\beta}c_1-c_2) $$

Now

$$ \begin{array}{lll} \sum\limits_{k=1}^{\lceil\log_2x\rceil} \frac{(b(x/2^{k-1})+b(x/2^k))\bar{a}(2^k)}{b(x)} &\leq \sum\limits_{k=1}^{\lceil\log_2x\rceil}\frac{B(x/2^{k-1})A(2^k)}{b(x)}\\ &\leq c_3 \sum\limits_{k=1}^{\lceil\log_2x\rceil} \left(\frac{x}{2^{k-1}}\right)^{\beta}2^{k\alpha} \left(\frac{2}{x}\right)^{\beta}\\ &\leq c_4 \sum\limits_{k=1}^{\lceil\log_2x\rceil} 2^{k(\alpha-\beta)}\,, \end{array} $$

which is bounded since α < β. So Lemma A.1 implies that C(x) = Θ(B(x)).□

We now prove Theorem 3.3, that the size of a monoid is at most slightly bigger than its generating set:

Proof of Theorem 3.3

Fix ε > 0. We prove M(x) = O(x ^α + ε). Put α ₀ = α + ε/2. Let x ₀ be such that $G(x)\leq \tfrac{1}{2}x^{\alpha_{0}}$ for all x ≥ x ₀. Let $\mbox{$\cal{G}$}_{0}=\mbox{$\cal{G}$}\cap [1,x_0)$, and let $\mbox{$\cal{G}$}_{1}=\mbox{$\cal{G}$} \cap [x_0,\infty)$. Let $\mbox{$\cal{M}$}_{0}$ be the monoid generated by $\mbox{$\cal{G}$}_{0}$, and let $\mbox{$\cal{M}$}_{1}$ be the monoid generated by $\mbox{$\cal{G}$}_{1}$. Then the following statements hold:

• (A) $G_1(x) \leq \tfrac{1}{2}(x^{\alpha_0})$,

• (B) $M_{0}(x)=O((\log x)^{|\mbox{$\cal{G}$}_0|})$,

• (C) $\mbox{$\cal{M}$} = \mbox{$\cal{M}$}_{0}\mbox{$\cal{M}$}_{1}$,

• (D) $M(x)\leq M_0(x)M_1(x)=O((\log x)^{|\mbox{$\cal{G}$}_0|}M_1(x))$.

So, noting item (D), in order to prove that M(x) = O(x ^α + ε), it is sufficient to prove that $M_1(x)=O(x^{\alpha_1})$, for all α ₁ ∈ (α ₀,α + ε).

Fix α ₁ ∈ (α ₀,α + ε). We prove $M_1(x)=O(x^{\alpha_1})$. Let n = ⌈log₂ x⌉. Any element y of $\mbox{$\cal{M}$}_{1} \cap [1,x]$ corresponds to a partition of n as follows: Suppose y = y ₁ y ₂ ...y _r where y ₁ ≤ y ₂ ≤ ... ≤ y _r are elements of $\mbox{$\cal{G}$}_1$. Put $a_{i}= \lfloor\log_{2}y_{i}\rfloor$. Then a ₁ + a ₂ + ...a _r  = m ≤ n, and 0 ≤ a ₁ ≤ a ₂ ≤ ... ≤ a _r . Thus replacing a _r with a _r ′ = a _r  + n − m, we see that any product $y=y_1y_2\dots y_r\in\mbox{$\cal{M}$}_1\cap[1,x]$ of r elements y _i of $\mbox{$\cal{G}$}_1$ maps to a partition of n into at most r pieces. The number of such y sequences y ₁,y ₂,...,y _r with $\lfloor\log_2 y_i\rfloor = a_i$ is at most

$$ G_1(2^{a_1+1}) G_1(2^{a_2+1}) \dots G_1(2^{a_r+1}) \leq 2^{a_1\alpha_0}2^{a_2\alpha_0}\dots2^{a_r\alpha_0} \leq 2^{n\alpha_0}\,. $$

Now Hardy and Ramanujan showed that the number p(n) of partitions of n is asymptotic to

$$ \exp(\pi\sqrt{2n/3})/4n\sqrt{3}=O(x^{\delta})\,, $$

for all δ > 0. So, choosing δ = α ₁ − α ₀, we have

$$ M_1(x)\leq p(n) 2^{\lceil\log_2 x\rceil\alpha_0} = O(x^{\alpha_1-\alpha_0+\alpha_0})=O(x^{\alpha_1})\,. $$

□

2.2 A.2 Proofs using generating functions

We will use generating functions to show that these constructions do not increase the density of known Hadamard orders. Since we are interested in the properties of products of sets $\mbox{$\cal{C}$} = \mbox{$\cal{A}$} \mbox{$\cal{B}$}$, functions of the form

$$ \sum\limits_{n \in \mbox{$\cal{A}$}} z^{\log_2 n} $$

are useful, since multiplying elements corresponds to adding the powers in terms of the series. In the context of smooth numbers, Bernstein [2] estimated such functions by looking at

$$ a(z) = \sum\limits_{k \geq 0} a_k z^k := \sum\limits_{n \in \mbox{$\cal{A}$}} z^{\lfloor \log_2 n\rfloor} . $$

These series have many fewer terms, and so are easier to analyze. Note that $a_k = A(2^k) - A(2^{k-1})$ is the number of k-bit elements of $\mbox{$\cal{A}$}$. We will prove results about a _k , i.e. A(x) for x a power of two, but since A(x) is monotone increasing, and all the coefficients of the generating function are nonnegative, this will suffice.

Lemma A.2

Let $\mbox{$\cal{C}$}$ be the set of products of elements of sets $\mbox{$\cal{A}$}$ and $\mbox{$\cal{B}$}$ with series a(z) and b(z). Then

$$ c(z) \leq a(z) \frac{b(z)}{1-z}. $$

Proof

The coefficient of z ⁿ in $a(z) \frac{b(z)}{1-z} = a(z) b(z) (1 + z + z^2 \cdots)$ is

$$ \sum\limits_{k=0}^n a_k B(2^{n-k}). $$

Any n-bit element of $\mbox{$\cal{C}$}$ can be written as a product of a k-bit element of $\mbox{$\cal{A}$}$ and an element of $\mbox{$\cal{B}$}$ of at most n − k bits. □

We may use the analytic properties of series like this to bound the size of the corresponding counting function. Flajolet and Sedgewick [7] give a wealth of such results. Their Theorem IV.7 relates the growth rate of power series coefficients to singularities of the corresponding function:

Theorem A.3

If $f(z)= \sum f_n z^n$ has positive coefficients and is analytic at 0 and

$$ R = \sup \{ r \geq 0 | f {\rm \ is \ analytic \ at \ all \ points \ of\ } 0 \leq z < r\} $$

then $\lim \sup |f_n|^{1/n} = (1/R)$.

From Corollary 2.2 we have $s_k = \Theta(2^{(1-\epsilon) k})$ for any ε > 0, so by the ratio test the radius of convergence of s(z) is 1/2. The coefficients of generating functions for the other monoids have smaller growth, and so a larger radius of convergence. Theorem VI.12 of [7] shows that the size of the product set $\mbox{$\cal{A}$} \mbox{$\cal{B}$} = \{ ab | a \in \mbox{$\cal{A}$}, b \in \mbox{$\cal{B}$} \}$ of two sets with different growth rates is a constant times the size of the larger set, proving Theorem 3.2:

Theorem A.4

Suppose $a(z) = \sum a_n z^n$ and $b(z) = \sum b_n z^n$ are power series with radii of convergence α > β ≥ 0, respectively. Suppose b_n − 1/b _n approaches a limit b as $n \longrightarrow \infty$. If a(b) ≠ 0, then c _n ~a(b) b _n , where $\sum c_n z^n = a(z) b(z)$.

Finally consider the monoid generated by a set $\mbox{$\cal{A}$}$. The generating function for the monoid will be

$$ {\rm Exp}(a(z)) := \exp \left( a(z) + \frac{1}{2} a(z^2) + \frac{1}{3} a(z^3) + \cdots \right). $$

This function has the same radius of convergence as a(z) (see Section IV.4 of [7]), giving Theorem 3.3.