# A few more Hadamard
Partitioned Difference Families

Anamari Nakić Faculty of Electrical Engineering and Computing, University of
Zagreb, Croatia, email<EMAIL_ADDRESS>
###### Abstract

A $(G,[k_{1},\dots,k_{t}],\lambda)$ partitioned difference family (PDF) is a
partition $\cal B$ of an additive group $G$ into sets (blocks) of sizes
$k_{1}$, …, $k_{t}$, such that the list of differences of ${\cal B}$ covers
exactly $\lambda$ times every non-zero element of $G$. It is called Hadamard
(HPDF) if the order of $G$ is $2\lambda$. The study of HPDFs is motivated by
the fact that each of them gives rise, recursively, to infinitely many other
PDFs. Apart from the elementary HPDFs consisting of a Hadamard difference set
and its complement, only one HPDF was known. In this article we present three
new examples in several groups and we start a general investigation on the
possible existence of HPDFs with assigned parameters by means of simple
arguments.

Keywords: partitioned difference family, difference set, partial difference
set

##  1 Introduction

We recall that the list of differences of a subset $B$ of an additive group
$G$, denoted by $\Delta B$, is the multiset of all possible differences
between two distinct elements of $B$:

$\Delta B:=\\{x-y\,:\,x\neq y,x,y\in B\\}.$

More generally, the list of differences of a collection ${\cal
F}=\\{B_{1},\ldots,B_{t}\\}$ of subsets of $G$ is the multiset $\Delta{\cal
F}$ which is union of the lists of differences of all the $B_{i}$’s, i.e.,
$\Delta{\cal F}=\bigcup_{i=1}^{t}\Delta B_{i}$. The collection $\mathcal{F}$
is said to be a difference family (DF) if $\Delta{\cal F}$ covers every non-
zero element of $G$ a constant number $\lambda$ of times. In this case, if $K$
is the multiset of the sizes of the $B_{i}$’s, one says that $\cal F$ is a
$(G,K,\lambda)$-DF. More specifically, one writes
$(G,[k_{1},\dots,k_{t}],\lambda)$-DF if $B_{i}$ has size $k_{i}$. The
$B_{i}$’s are the base blocks of $\cal F$ and $\lambda$ is its index. One
often speaks of a $(v,K,\lambda)$-DF or $(v,[k_{1},\dots,k_{t}],\lambda)$-DF
when the group $G$ (of order $v$) is understood.

If a DF of index $\lambda$ has only one block $B$ and its size is $k$, then
one says that $B$ is a $(G,k,\lambda)$ difference set (DS). As before, one
generally speaks of a $(v,k,\lambda)$-DS if the group $G$ is understood. Among
the many classes of difference sets we have, in particular, the so called
Hadamard difference sets which are those whose parameters are
$(4u^{2},2u^{2}-u,u^{2}-u)$ for some $u$.

A $(G,[k_{1},\dots,k_{t}],\lambda)$-DF is partitioned (PDF) if its blocks
partition $G$. Here the parameters $v=o(G)$ and $\lambda$ are completely
determined by $K$. Indeed it is evident that the equalities
$k_{1}+\dots+k_{t}=v$ and $k_{1}(k_{1}-1)+...+k_{t}(k_{t}-1)=\lambda(v-1)$
hold. The notion of a PDF was introduced in [12] for the construction of
optimal constant composition codes [17, 12] which are important in various
applications (see, e.g,. [10, 14]) and it is equivalent to that of a zero
difference balanced function [11] even though, according to [7], it is far
preferable to keep the standard notation and terminology.

In a very recent article [8], PDFs have been used to prove the existence of
infinitely many resolvable linear spaces with a mandatory set of block sizes
and an automorphism group acting sharply transitively on the points and
transitively on the parallel classes.

If $B$ is $(v,k,\lambda)$-DS in $G$, then $\\{B,G\setminus B\\}$ is a
$(v,[k,v-k],v-2k+2\lambda)$-PDF. So, in particular, a Hadamard
$(4u^{2},2u^{2}-u,u^{2}-u)$-DS gives rise to a
$(4u^{2},[2u^{2}+u,2u^{2}-u],2u^{2})$-PDF. Note that here the order of the
group is twice the index. For this reason all PDFs with this property have
been called Hadamard partition difference families (HPDFs) in the paper were
they have been introduced [6]. The motivation of their study is that each HPDF
in a group $G$ gives rise to an infinite class of PDFs in suitable supergroups
of $G$.

As pointed out in [7], PDFs with (many) blocks of size $1$ are not considered
very interesting. However, in the case of HPDFs, the situation is quite
different since the PDFs arising from them by using Buratti’s construction no
longer contain blocks of size 1.

Apart from the HPDFs arising from Hadamard DSs that we call elementary, there
was only one known example of a HPDF. This makes believe that HPDFs are quite
rare. In this paper we determine three new examples in several groups.

Apart from this, the article wants to be the beginning of a general
investigation on HPDFs by means of quite simple arguments. It is structured as
follows.

In the next section we characterize the PDFs having exactly two blocks. They
necessarily consist of a difference set and its complement. In particular,
each HPDF with two blocks necessarily consists of a Hadamard difference set
and its complement.

In Section 3 we give the trivial necessary conditions for the existence of a
HPDF, we list the first admissible parameter sets and then we give some
informations on the HPDFs with exactly three blocks.

In Section 4, exploiting partial difference sets, we prove that a HPDF with
three block sizes one of which is 1 cannot exist.

In Section 5 we give one more necessary condition for the existence of a HPDF
in a group having at least one subgroup of index 2.

In Section 6 we present the three new examples mentioned earlier. We have a
$(24,[1^{3},2^{2},17],12)$-HPDF in three pairwise non-isomorphic groups, a
$(36,[3,9,24],18)$-HPDF in nine pairwise non-isomorphic groups, and a cyclic
$(40,[1,3,9,27],20)$-HPDF.

In Section 7 we determine the infinite families of descendants of our new
three examples.

Finally, in Section 8 we pose two open questions. We first ask whether there
exists a $(v,k,\lambda)$-DS with $v=2\lambda$ and $k>4$. The motivation is
that such a DS would give a $(v,[1^{v-k},k],\lambda)$-HPDF. The second
question is whether, given positive integers $q$ and $n$, there exists a PDF
whose related $K$ is $[q^{0},q^{1},q^{2},q^{3},\dots,q^{2n-1}]$. The
motivation is that such a PDF with $q=3$ would be Hadamard.

##  2 PDFs and HPDFs with two blocks

The following fact is extremely well-known (see, e.g., [1]).

######  Proposition 2.1.

If $B$ is a $(v,k,\lambda)$-DS in $G$, then the complement $\overline{B}$ of
$B$ in $G$ is a $(v,v-k,v-2k+\lambda)$-DS.

The above explains why, as said in the introduction, a $(v,k,\lambda)$-DS and
its complement give a $(v,[k,v-k],v-2k+2\lambda)$-PDF.

Although the following is a rather straightforward generalization of
Proposition 2.1, we are not aware whether it was ever stated explicitly
before.

######  Proposition 2.2.

Let $B$ be a $k$-subset of an additive group $G$ of order $v$ and let
$\overline{B}$ be the complement of $B$ in $G$. Let $g$ be a non-zero element
of $G$ and let $\lambda(g)$ be its multiplicity in $\Delta B$. Then the
multiplicity $\overline{\lambda}(g)$ of $g$ in $\Delta\overline{B}$ is
$v-2k+\lambda(g)$.

###### Proof.

Given subsets $X$ and $Y$ of $G$, set

$\Lambda_{X\times Y}(g)=\\{(x,y)\in X\times Y\ |\ x-y=g\\}.$

We have $|\Lambda_{B\times B}(g)|=\lambda$ and
$|\Lambda_{\overline{B}\times\overline{B}}(g)|=\overline{\lambda}$ by
assumption. Let $(x_{1},y_{1})$, …, $(x_{\lambda},y_{\lambda})$ be the
$\lambda$ pairs of $\Lambda_{B\times B}(g)$ and set

$B^{\prime}=B\setminus\\{x_{1},\dots,x_{\lambda}\\}\quad{\rm and}\quad
B^{\prime\prime}=B\setminus\\{y_{1},\dots,y_{\lambda}\\}.$

It is easy to see that

$\Lambda_{{B}\times\overline{B}}(g)=\\{(b,-g+b)\ |\ b\in
B^{\prime}\\}\quad{\rm and}\quad\Lambda_{\overline{B}\times{B}}(g)=\\{(g+b,b)\
|\ b\in B^{\prime\prime}\\}$

so that $|\Lambda_{{B}\times\overline{B}}(g)|=|B^{\prime}|=k-\lambda$ and
$|\Lambda_{\overline{B}\times{B}}(g)|=|B^{\prime\prime}|=k-\lambda$. Finally,
it is evident that $\Lambda_{G\times G}(g)=\\{(g+y,y)\ |\ y\in G\\}$ so that
$|\Lambda_{G\times G}(g)|=|G|=v$. Considering that the four sets $B\times B$,
$\overline{B}\times\overline{B}$, ${B}\times\overline{B}$, and
$\overline{B}\times{B}$ partition $G\times G$, we can write

$|\Lambda_{{B}\times{B}}(g)|+|\Lambda_{\overline{B}\times\overline{B}}(g)|+|\Lambda_{{B}\times\overline{B}}(g)|+|\Lambda_{\overline{B}\times{B}}(g)|=|\Lambda_{G\times
G}(g)|$

and hence, from what seen above,
$\lambda+\overline{\lambda}+(k-\lambda)+(k-\lambda)=v$ which immediately gives
the assertion. ∎

######  Proposition 2.3.

A PDF with only two blocks necessarily consists of a difference set and its
complement. More specifically, a HPDF with only two blocks necessarily
consists of a Hadamard difference set and its complement.

###### Proof.

If $\cal F$ is a $(v,[k,v-k],\lambda)$-PDF, then ${\cal
F}=\\{B,\overline{B}\\}$ with $B$ a $k$-subset of an additive group $G$ of
order $v$ and $\overline{B}=G\setminus B$. Given any element $g\in
G\setminus\\{0\\}$ denote by $\mu(g)$ and $\overline{\mu}(g)$ the
multiplicities of $g$ in $\Delta B$ and $\Delta\overline{B}$, respectively. By
definition of a PDF we have $\mu(g)+\overline{\mu}(g)=\lambda$ for every $g\in
G$. On the other hand we have $\overline{\mu}(g)=v-2k+\mu(g)$ by Proposition
2.2 and then $v-2k+2\mu(g)=\lambda$, i.e., $\mu(g)={2k-v+\lambda\over 2}$ is a
constant. This means that $B$ is a difference set and the first assertion
follows.

Now assume that $\\{B,\overline{B}\\}$ is a $(v,[k,v-k],\lambda)$-HPDF so that
we have $v=2\lambda$. Without loss of generality we can assume that $k\leq
v-k$, i.e., $k\leq{v\over 2}$. Then we deduce that $B$ is a $(v,k,\mu)$-DS
with $\mu={2k-\lambda\over 2}$. Considering that $v$ is even, $k-\mu$ must be
a square by the Bruck-Ryser-Chowla theorem. It follows that ${\lambda\over
2}=u^{2}$, hence $\mu=k-u^{2}$ for some integer $u$. Then, from the trivial
identity $\mu(v-1)=k(k-1)$, we get $(k-u^{2})(4u^{2}-1)=k(k-1)$. Hence $k$ is
a solution of the quadratic equation $x^{2}-4u^{2}x+4u^{4}-u^{2}=0$ so that
either $k=2u^{2}+u$ or $k=2u^{2}-u$. On the other hand we have assumed
$k\leq{v\over 2}$, hence $k=2u^{2}-u$ and $\mu=u^{2}-u$. We conclude that $B$
is a $(4u^{2},2u^{2}-u,u^{2}-u)$-DS, i.e., $B$ is a Hadamard difference set.
The assertion follows. ∎

##  3 Some necessary conditions

The following proposition is straightforward.

######  Proposition 3.1.

Let $\mathcal{F}$ be a $(G,[k_{1},\ldots,k_{t}],\lambda)$-HPDF. Then

  1. (i)

$k_{1}+\cdots+k_{t}=2\lambda$;

  2. (ii)

$k_{1}^{2}+\cdots+k_{t}^{2}=\lambda(2\lambda+1)$;

  3. (iii)

$\lambda$ is even, hence $o(G)\equiv 0$ $\pmod{4}$.

###### Proof.

Let $\mathcal{F}$ be a $(G,[k_{1},\ldots,k_{t}],\lambda)$-HPDF. By definition
of a HPDF, we have $k_{1}+\cdots+k_{t}=|G|=2\lambda$. From
$\Delta\mathcal{F}=\lambda\,(G\setminus\\{0\\})$ we obtain

$|\Delta\mathcal{F}|=k_{1}(k_{1}-1)+\cdots+k_{t}(k_{t}-1)=\lambda(2\lambda-1)$

and then

$k_{1}^{2}+\cdots k_{t}^{2}=\lambda(2\lambda-1)+k_{1}+\cdots+k_{t}.$

Then (ii) follows taking into consideration (i).

We obviously have $k_{1}+\cdots+k_{t}\equiv k_{1}^{2}+\cdots+k_{t}^{2}$ (mod
2). Thus, using (i) and (ii), we get $2\lambda\equiv\lambda(2\lambda+1)$ (mod
2) and (iii) follows. ∎

As a consequence of the above proposition, every HPDF in a group $G$ has
$o(G)$ necessarily doubly even. However, it is worth to observe that this can
be deduced in another way as follows. Assuming that $\cal F$ is a HPDF in $G$
and $o(G)\equiv 2$ (mod 4), then $\lambda={o(G)\over 2}$ would be odd and this
is absurd since the multiplicity in $\Delta{\cal F}$ of any involution $i$ of
$G$ is necessarily even: indeed, if $x-y=i$ is a representation of $i$ as a
difference from $\cal F$, then $y-x=i$ is a representation of $i$ as a
difference from $\cal F$ as well.

In the following table we list the very first parameter sets satisfying the
necessary conditions of Proposition 3.1 disregarding those were $K$ has size 2
in view of Proposition 2.3.

$\begin{array}[]{cc}\begin{array}[]{c|c|c}v&K&\lambda\\\
\hline\cr{\color[rgb]{1,0,0}{20}}&{\color[rgb]{1,0,0}{[1,2,3,14]}}&{\color[rgb]{1,0,0}10}\\\
{\color[rgb]{0,.5,.5}24}&{\color[rgb]{0,.5,.5}[1^{3},2^{2},17]}&{\color[rgb]{0,.5,.5}12}\\\
\color[rgb]{1,0,0}28&{\color[rgb]{1,0,0}{[1,9,18]}}&\color[rgb]{1,0,0}14\\\
\color[rgb]{1,0,0}28&{\color[rgb]{1,0,0}{[3,6,19]}}&\color[rgb]{1,0,0}14\\\
{\color[rgb]{0,0,1}32}&{\color[rgb]{0,0,1}[2^{2},6,22]}&{\color[rgb]{0,0,1}16}\\\
{\color[rgb]{0,.5,.5}36}&{\color[rgb]{0,.5,.5}[3,9,24]}&{\color[rgb]{0,.5,.5}18}\\\
\color[rgb]{1,0,0}36&{\color[rgb]{1,0,0}{[3,4^{2},25]}}&\color[rgb]{1,0,0}18\\\
\color[rgb]{1,0,0}36&{\color[rgb]{1,0,0}{[1^{5},6,25]}}&\color[rgb]{1,0,0}18\\\
{\color[rgb]{0,.5,.5}40}&{\color[rgb]{0,.5,.5}[1,3,9,27]}&{\color[rgb]{0,.5,.5}20}\\\
40&[3^{4},28]&20\\\ 40&[1^{2},3^{2},4,28]&20\\\
\color[rgb]{1,0,0}40&{\color[rgb]{1,0,0}{[1^{4},4^{2},28]}}&\color[rgb]{1,0,0}20\\\
40&[1^{3},2^{2},5,28]&20\\\ \end{array}\end{array}$

An HPDF with “blue parameter set” has been found by Buratti [6] and it is the
only non-elementary HPDF known at this moment.

In Section 6 we will determine an HPDF with “green parameter set” in several
groups.

By exhaustive computer search we have checked that an HPDF with “red parameter
set” does not exist.

The existence of an HPDF with “uncolored parameter set” is still in doubt.

The HPDFs with only two blocks have been already characterized in Proposition
2.3. Let us see what we can say about HPDFs with three blocks.

######  Proposition 3.2.

In a $(v,[k_{1},k_{2},k_{3}],\lambda)$-HPDF we necessarily have

$k_{1,2}={2\lambda-k_{3}\pm\sqrt{2\lambda(2k_{3}+1)-3k_{3}^{2}}\over 2}$

###### Proof.

By Proposition 3.1 we have $k_{1}+k_{2}+k_{3}=2\lambda$ and
$k_{1}^{2}+k_{2}^{2}+k_{3}^{2}=\lambda(2\lambda+1)$. The second identity can
be rewritten as $(k_{1}+k_{2})^{2}-2k_{1}k_{2}+k_{3}^{2}=2\lambda^{2}+\lambda$
and then, in view of the first identity, we have $(2\lambda-
k_{3})^{2}-2k_{1}k_{2}+k_{3}^{2}=2\lambda^{2}+\lambda$ which gives

$k_{1}\cdot k_{2}={2\lambda^{2}-(4k_{3}+1)\lambda+2k_{3}^{2}\over 2}$

Considering that $k_{1}+k_{2}=2\lambda-k_{3}$ and recalling that two numbers
having sum $s$ and product $p$ are the solutions of the quadratic equation
$x^{2}-sx+p=0$, after trivial computations we get the assertion. ∎

The possible block-sizes of a $(v,[k_{1},k_{2},k_{3}],\lambda)$-HPDF are
strongly limited by the above proposition. Indeed we have the following.

######  Corollary 3.3.

The existence of a $(v,[k_{1},k_{2},k_{3}],\lambda)$-HPDF necessarily implies
that no prime divisor of $(2k_{1}+1)(2k_{2}+1)(2k_{3}+1)$ is congruent to $5$
$($mod $6)$.

###### Proof.

Assume that a $(v,[k_{1},k_{2},k_{3}],\lambda)$-HPDF exists and let $p$ be a
prime factor of $(2k_{1}+1)(2k_{2}+1)(2k_{3}+1)$. Up to a reordering of the
indices we can assume that $p$ is a divisor of $2k_{3}+1$ and hence $p$ cannot
divide $k_{3}$. By Proposition 3.2 it is clear that
$2\lambda(2k_{3}+1)-3k_{3}^{2}$ must be a perfect square. So, in particular,
it must be a square modulo $p$. Considering that $p$ is a divisor of
$2k_{3}+1$ we have that $2k_{3}+1\equiv 0$ (mod $p$). Also, $k_{3}^{2}$ is a
non-zero square of $\mathbb{Z}_{p}$ since $p$ does not divide $k_{3}$. We
conclude that $-3$ must be a square of $\mathbb{Z}_{p}$ and then, by the
Quadratic Law of Reciprocity, we have either $p=3$ or $p\equiv 1$ (mod 6). ∎

As a consequence of Corollary 3.3, in a $(v,[k_{1},k_{2},k_{3}],\lambda)$-HPDF
we cannot have, for instance, blocks of size 2, 5, 7, 8, 11, 12, 14, 16, 17, …

##  4 Exploiting partial difference sets

Note that Corollary 3.3 does not forbid the existence of a

$(v,[k_{1},k_{2},k_{3}],\lambda)\mbox{-HPDF}$

with a block of size 1 and indeed, using Proposition 3.2, we can see, for
instance, that for every $n\geq 2$

$\displaystyle\biggl{(}{3^{2n-1}+1\over 2},\ \biggl{[}{3^{2n-1}+3^{n}\over
2},\ {3^{2n-1}-3^{n}\over 2},\ 1\biggl{]},\ {3^{2n-1}+1\over 4}\biggl{)}$

is an admissible parameter set for a HPDF. On the other hand we are going to
see that a $(v,[k_{1},k_{2},1],\lambda)$-HPDF cannot exist. To prove this we
have to exploit a result on partial difference sets.

A $(v,k,\alpha,\beta)$ partial difference set (PDS) in a group $G$ is a
$k$-subset $B$ of $G$ such that $\Delta B$ covers $\alpha$ times every non-
zero element of $B$ and $\beta$ times every non-zero element of $G\setminus
B$.

Among the various necessary conditions for the existence of a non-trivial PDS
we will need the following (see Proposition 3(d) in [18]).

######  Lemma 4.1.

If there exists a $(v,k,\alpha,\beta)$-PDS and
$\gamma:=(\alpha-\beta)^{2}+4(k-\beta)$ is not a square, then we have
$(v,k,\alpha,\beta)=(4t+1,2t,t-1,t)$ for a suitable integer $t$.

In the next proposition we will see that if a PDF in a group $G$ has exactly
three blocks one of which has size 1, then each of the other two blocks is a
PDS or a DS in $G$.

######  Proposition 4.2.

Let ${\cal F}=\\{B_{1},B_{2},\\{0\\}\\}$ be a
$(v,[k_{1},k_{2},1],\lambda)$-PDF in $G$. Then, for $i=1,2$ we have:

  1. (i)

$B_{i}$ is a $(v,{v-1\over 2},{v-3\over 4})$-DS if $v-\lambda$ is odd;

  2. (ii)

$B_{i}$ is a $(G,k_{i},\alpha_{i},\alpha_{i}+1)$-PDS with
$\alpha_{i}=k_{i}+{\lambda-v\over 2}$ if $v-\lambda$ is even.

###### Proof.

For $i=1,2$ and for $g\in G$, let $\lambda_{i}(g)$ be the multiplicity of $g$
in $\Delta B_{i}$. The assumption that $\cal F$ is a
$(v,[k_{1},k_{2},1],\lambda)$-PDF means that we have

$\lambda_{1}(g)+\lambda_{2}(g)=\lambda\quad\forall\ g\in G\setminus\\{0\\}$
(1)

Let $\overline{B}_{1}$ be the complement of $B_{1}$ in $G$ and let
$\overline{\lambda_{1}}(g)$ be the multiplicity of $g$ in
$\Delta\overline{B_{1}}$. We have $\overline{B_{1}}=B_{2}\cup\\{0\\}$ so that
the possible representations of $g\in G$ as a difference from
$\overline{B_{1}}$ which are not representations of $g$ as a difference from
$B_{2}$ are $g=g-0$ if $g\in B_{2}$, and $g=0-(-g)$ if $-g\in B_{2}$. Thus we
can write:

$\lambda_{2}(g)=\begin{cases}\overline{\lambda_{1}}(g)&{\rm if}\
\\{g,-g\\}\subset B_{1};\cr\overline{\lambda_{1}}(g)-2&{\rm if}\
\\{g,-g\\}\subset B_{2};\cr\overline{\lambda_{1}}(g)-1&{\rm if}\
|\\{g,-g\\}\cap B_{i}|=1\ {\rm for}\ i=1,2\cr\end{cases}$ (2)

In the following, we distinguish two cases according to the parity of
$v-\lambda$ and keep in mind that we have

$\overline{\lambda_{1}}(g)=v-2k_{1}+\lambda_{1}(g)$ (3)

in view of Proposition 2.2.

Case 1: $v-\lambda$ is odd.

Observe that for any given $g\in G\setminus\\{0\\}$ we cannot have
$\\{g,-g\\}\subset B_{1}$ otherwise (1), (2) and (3) would give
$\lambda=v-2k_{1}+2\lambda_{1}(g)$ contradicting the assumption that
$v-\lambda$ is odd.

Analogously, we cannot have $\\{g,-g\\}\subset B_{2}$ otherwise (1), (2) and
(3) would give $\lambda=v-2k_{1}+2\lambda_{1}(g)-2$ contradicting again that
$v-\lambda$ is odd.

We conclude that $g$ and $-g$ lie in different $B_{i}$’s for every $g\in
G\setminus\\{0\\}$. It follows, in particular, that $G$ is involution-free: if
$g$ is an involution, we cannot have $g\in B_{1}$ and $-g\in B_{2}$ since $g$
and $-g$ are the same. Thus $G$ has odd order and we have $B_{2}=-B_{1}$ so
that $k_{1}=k_{2}={v-1\over 2}$. Considering that ${\cal F}$ is a
$(v,[k_{1},k_{2},1],\lambda)$-PDF we must have
$\lambda(v-1)=k_{1}(k_{1}-1)+k_{2}(k_{2}-1)$ and hence
$\lambda(v-1)=2\cdot{v-1\over 2}\cdot{v-3\over 2}$ which gives
$\lambda={v-3\over 2}$. Finally, (2) and (3) give
$\lambda_{2}(g)=v-2\cdot{v-1\over 2}+\lambda_{1}(g)-1=\lambda_{1}(g)$. It
follows, by (1), that $\lambda_{1}(g)=\lambda_{2}(g)={\lambda\over
2}={v-3\over 4}$. It is now clear that $B_{1}$ and $B_{2}$ are $(v,{v-1\over
2},{v-3\over 4})$-DSs.

Case 2: $v-\lambda$ is even.

Here, there is no $g\in G$ such that $g\in B_{1}$ and $-g\in B_{2}$. Indeed,
in the opposite case, (1), (2) and (3) would give
$\lambda=v-2k_{1}+2\lambda_{1}(g)-1$ contradicting the assumption that
$v-\lambda$ is even. Thus, by (2), we have

$\lambda_{2}(g)=\begin{cases}v-2k_{1}+\lambda_{1}(g)&{\rm if}\ g\in B_{1};\cr
v-2k_{1}+\lambda_{1}(g)-2&{\rm if}\ g\in B_{2}\end{cases}$

Using again (1) and (3) we get $\lambda=v-2k_{1}+2\lambda_{1}(g)$ or
$v-2k_{1}+2\lambda_{1}(g)-2$ according to whether $g\in B_{1}$ or $g\in
B_{2}$, respectively. Solving these identities with respect to
$\lambda_{1}(g)$ we finally get

$\lambda_{1}(g)=\begin{cases}\alpha_{1}&{\rm if}\ g\in
B_{1};\cr\alpha_{1}+1&{\rm if}\ g\in B_{2}\end{cases}$

This precisely means that $B_{1}$ is a
$(G,k_{1},\alpha_{1},\alpha_{1}+1)$-PDS. In the same way, exchanging the roles
of $B_{1}$ and $B_{2}$ one gets that $B_{2}$ is a
$(G,k_{2},\alpha_{2},\alpha_{2}+1)$-PDS.

∎

######  Corollary 4.3.

A $(v,[k_{1},k_{2},1],\lambda)$-HPDF cannot exist.

###### Proof.

Assume that there exists a $(v,[k_{1},k_{2},1],\lambda)$-HPDF so that
$v-\lambda=\lambda$ is even. Up to a translation we can assume that the block
of size 1 is $\\{0\\}$. Then, by Proposition 4.2, the block of size $k_{1}$ is
a $(2\lambda,k_{1},\alpha,\alpha+1)$-PDS with $\alpha=k_{1}-{\lambda\over 2}$.
Here the parameter $\gamma$ mentioned in Lemma 4.1 is
$1+4(k_{1}-k_{1}+{\lambda\over 2})=2\lambda+1$. Thus, in view of the same
lemma, $2\lambda+1$ is a perfect square, say $2\lambda+1=\mu^{2}$, otherwise
we should have $v=4t+1$ which is absurd. By Proposition 3.2 it is also
necessary that $6\lambda-3$ is a perfect square. Now note that we have
$6\lambda-3=3(2\lambda-1)=3(\mu^{2}-2)$ so that $3$ should divide $\mu^{2}-2$.
This is absurd since 2 is not a square modulo 3. ∎

##  5 Exploiting subgroups of index 2

The following proposition exploits the possible existence of a subgroup of
index 2.

######  Proposition 5.1.

Let ${\cal F}=\\{B_{1},\dots,B_{t}\\}$ be a
$(G,[k_{1},\ldots,k_{t}],\lambda)$-HPDF, assume that $G$ has a subgroup $H$ of
index $2$, and set $|B_{i}\cap H|=s_{i}$ for $i=1,\dots,t$. Then the following
identities hold:

$s_{1}+\ ...\ +s_{t}=\lambda\quad\quad{\rm and}\quad\quad
2s_{1}(k_{1}-s_{1})+\ ...\ +2s_{t}(k_{t}-s_{t})=\lambda^{2}$

###### Proof.

For $i=1,\dots,t$, set $B^{\prime}_{i}=B_{i}\cap H$ and
$B^{\prime\prime}_{i}=B_{i}\setminus H$ so that $|B^{\prime}_{i}|=s_{i}$ and
$|B^{\prime\prime}_{i}|=k_{i}-s_{i}$. The first identity follows from the fact
that the $B_{i}$’s partition $H$ which has order $\lambda$. Now note that we
have

$\Delta B_{i}=\Delta B^{\prime}_{i}\ \cup\ \Delta B^{\prime\prime}_{i}\ \cup\
(B^{\prime}_{i}-B^{\prime\prime}_{i})\ \cup\ (B^{\prime\prime}_{i}-B_{i}).$

Also note that $\Delta B^{\prime}_{i}\ \cup\ \Delta B^{\prime\prime}_{i}$ is a
multisubset of $H$ and that both $B^{\prime}_{i}-B^{\prime\prime}_{i}$ and
$B^{\prime\prime}_{i}-B^{\prime}_{i}$ are multisubsets of $G\setminus H$ of
size $|B^{\prime}_{i}|\cdot|B^{\prime\prime}_{i}|=s_{i}(k_{i}-s_{i})$. Thus we
can say that $\Delta B_{i}$ has exactly $2s_{i}(k_{i}-s_{i})$ elements in
$G\setminus H$. Then, considering that $\Delta{\cal F}$ covers every element
of $G\setminus H$ exactly $\lambda$ times, we conclude that we have
$\sum_{i=1}^{t}2s_{i}(k_{i}-s_{i})=\lambda\cdot|G\setminus H|=\lambda^{2}$,
i.e., the second identity holds. ∎

As a consequence we have the following.

######  Corollary 5.2.

If there exists a $(G,[k_{1},\ldots,k_{t}],\lambda)$-HPDF and $G$ has a
subgroup of index $2$, then the diophantine system

$\left\\{\begin{array}[]{rcrcl}x_{1}&+\ ...\ +&x_{t}&=&\lambda\\\
2x_{1}(k_{1}-x_{1})&+\ ...\
+&2x_{t}(k_{t}-x_{t})&=&\lambda^{2}\end{array}\right.$

has a solution $(s_{1},\dots,s_{t})$ with $0\leq s_{i}\leq k_{i}$ for each
$i$.

As application of the above corollary one can see that none of these $K$,
though admissible, can be the multiset of block-sizes of a HPDF:

$[50,20,5,1];\quad[52,23,2,1,1,1];\quad[73,38,3,2];$
$\quad[77,28,8,3];\quad[79,31,7,3];\quad[81,21,16,1,1];\quad[104,35,14,3].$

As a matter of fact we have to admit that the admissible $K$ which are ruled
out by Corollary 5.2 do not appear so many. On the other hand Proposition 5.1
has been useful to limit our computer search for HPDFs of small orders.

##  6 New HPDFs

In this section we present HPDFs with the three new parameter sets

$(24,[1^{3},2^{2},17],12),\quad\quad(36,[3,9,24],18)\quad{\rm
and}\quad(40,[1,3,9,27],20).$

We will use, in particular, some dihedral groups and dicyclic groups of small
orders. We recall that the dihedral group of order $2n$, denoted $D_{2n}$, is
the group with defining relations

$\langle x,y\ |\ x^{n}=1;\ y^{2}=1;\ yx^{i}=x^{-i}y\rangle$

We also recall that the dicyclic group of order $4n$, denoted $Q_{4n}$, is the
group with defining relations

$\langle x,y\ |\ x^{2n}=1;\ y^{2}=x^{n};\ yx^{i}=x^{-i}y\rangle$

###  6.1 $(24,[1^{3},2^{2},17],12)$-HPDFs

We found an example of a $(24,[1^{3},2^{2},17],12)$-HPDF in each of the
following groups: $C_{3}\rtimes C_{8}$, $SL(2,3)$ and $C_{3}\times D_{8}$.

$G=C_{3}\rtimes C_{8}$

This is the semidirect product of $C_{3}$ by $C_{8}$ with defining relations

$C_{3}\rtimes C_{8}=\langle a,b\,|\,a^{8}=b^{3}=1,ab^{-1}=ba\rangle$

Thus the elements of $G$ are of the form $a^{i}b^{j}$ with $0\leq i\leq 7$ and
$0\leq j\leq 2$. The difference (even though we should say “ratio” since we
are in multiplicative notation) between two elements $a^{i_{1}}b^{j_{1}}$ and
$a^{i_{2}}b^{j_{2}}$ is given by

$(a^{i_{1}}b^{j_{1}})(a^{i_{2}}b^{j_{2}})^{-1}=a^{i_{1}-i_{2}}b^{(-1)^{i_{2}}(j_{1}-j_{2})}$
(4)

Let ${\cal F}=\\{B_{1},B_{2},B_{3},B_{4},B_{5},B_{6}\\}$ be the partition of
$G$ defined as follows:

$B_{1}=\\{1,a,a^{2},a^{3},a^{4},a^{6},a^{7},b,ab,a^{3}b,a^{4}b,a^{5}b,a^{6}b,b^{2},ab^{2},a^{2}b^{2},a^{4}b^{2}\\};$
$B_{2}=\\{a^{3}b^{2}\\};\quad B_{3}=\\{a^{5}b^{2}\\};\quad
B_{4}=\\{a^{7}b^{2}\\};$ $B_{5}=\\{a^{5},\ a^{2}b\\};\quad B_{6}=\\{a^{7}b,\
a^{6}b^{2}\\}.$

Using (4) it is straightforward to check that $\cal F$ is a
$(G,[1^{3},2^{2},17],12)$-HPDF.

$G=SL(2,3)$

This is the 2-dimensional special linear group over $\mathbb{Z}_{3}$. Its
elements are the $2\times 2$ matrices with elements in $\mathbb{Z}_{3}$ and
determinant equal to 1. Let ${\cal
F}=\\{B_{1},B_{2},B_{3},B_{4},B_{5},B_{6}\\}$ be the partition of $G$ defined
as follows:

$B_{1}=\biggl{\\{}\begin{pmatrix}2&1\cr 0&2\end{pmatrix}\biggl{\\}};\quad\quad
B_{2}=\biggl{\\{}\begin{pmatrix}1&2\cr 0&1\end{pmatrix}\biggl{\\}};\quad\quad
B_{3}=\biggl{\\{}\begin{pmatrix}0&2\cr 1&2\end{pmatrix}\biggl{\\}};$
$B_{4}=\biggl{\\{}\begin{pmatrix}0&2\cr 1&0\end{pmatrix},\
\begin{pmatrix}1&1\cr 1&2\end{pmatrix}\biggl{\\}};\quad\quad\quad
B_{5}=\biggl{\\{}\begin{pmatrix}2&1\cr 2&0\end{pmatrix},\
\begin{pmatrix}2&2\cr 0&2\end{pmatrix}\biggl{\\}};$ $B_{6}=G\ \setminus\
(B_{1}\ \cup\ B_{2}\ \cup\ B_{3}\ \cup\ B_{4}\ \cup\ B_{5}).$

It is straightforward to check that $\cal F$ is a
$(G,[1^{3},2^{2},17],12)$-HPDF.

$G=\mathbb{Z}_{3}\times D_{8}$

The reader can easily recognize that the partition of $G$ into the blocks
listed below is a $(G,[1^{3},2^{2},17],12)$-HPDF.

$B_{1}=\\{(0,x^{2})\\};\quad B_{2}=\\{(2,xy)\\};\quad B_{3}=\\{(2,x^{3}y)\\};$
$B_{4}=\\{(1,x^{3}),\ (2,x^{3})\\};\quad\quad B_{5}=\\{(1,y),\ (2,x^{2}y)\\};$
$B_{6}=G\ \setminus\ (B_{1}\ \cup\ B_{2}\ \cup\ B_{3}\ \cup\ B_{4}\ \cup\
B_{5}).$

###  6.2 $(36,[3,9,24],18)$-HPDFs

We found an example ${\cal F}=\\{A,B,C\\}$ of a $(36,[3,9,24],18)$-HPDF in the
groups

$\mathbb{Z}_{6}\times\mathbb{Z}_{6},\quad\quad\mathbb{Z}_{3}\times\mathbb{Z}_{12},\quad\quad\mathbb{Z}_{3}\times\mbox{Q}_{12},\quad\quad
D_{6}\times D_{6},\quad\quad\mathbb{Z}_{6}\times D_{6},$ $\mathbb{Z}_{3}\times
A_{4},\quad\quad\quad\mathbb{Z}_{3}\rtimes\mbox{Q}_{12},\quad\quad\quad\mathbb{Z}_{3}^{2}\rtimes\mathbb{Z}_{4},\quad\quad\quad\mathbb{Z}_{2}\times\mathbb{Z}_{3}\rtimes
D_{6}.$

We present our example for each of the first five.

$G=\mathbb{Z}_{6}\times\mathbb{Z}_{6}$

$A=\\{(1,1),(1,3),(1,5)\\};$

$B=\\{\\{0,2),(0,3),(1,4),(2,0),(2,5),(3,4),(4,1),(4,4),(5,4)\\};$

$C=G\setminus(A\ \cup\ B).$

$G=\mathbb{Z}_{3}\times\mathbb{Z}_{12}$

$A=\\{(1,1),(1,5),(1,9)\\};$

$B=\\{\\{0,2),(0,3),(0,4),(1,2),(1,8),(1,11),(2,0),(2,2),(2,7)\\};$

$C=G\setminus(A\ \cup\ B).$

$G=\mathbb{Z}_{3}\times\mbox{Q}_{12}$

$A=\\{(0,xy),(1,xy),(2,xy)\\};$

$B=\\{\\{1,1),(0,x^{3}),(0,x^{2}),(2,y),(1,x^{5}),(2,x^{4}y),(2,x^{4}),(2,x^{2}y),(2,x)\\};$

$C=G\setminus(A\ \cup\ B).$

$G=D_{6}\times D_{6}$

$A=\\{(y,xy),(xy,xy),(x^{2}y,xy)\\};$

$B=\\{(1,x^{2}y),(x,y),(x^{2},1),(x^{2},x),(x^{2},x^{2}),(x^{2},xy),(y,1),(xy,x^{2}),(x^{2}y,x)\\};$

$C=G\setminus(A\ \cup\ B).$

$G=\mathbb{Z}_{6}\times D_{6}$

$A=\\{(1,xy),(3,xy),(5,xy)\\};$

$B=\\{\\{0,x),(1,x^{2}),(2,1),(3,1),(4,x^{2}),(4,y),(4,xy),(4,x^{2}y),(5,x)\\};$

$C=G\setminus(A\ \cup\ B).$

###  6.3 A cyclic $(40,[1,3,9,27],20)$-HPDF

This is our unique example of a HPDF in a cyclic group. Also, it has maximum
order among the few non-elementary HPDFs that are known at this moment. Thus
it is surprising that this is also the unique example that we have been able
to get by hand without any use of the computer. The idea was the following.
Start from any cyclic $(40,13,4)$-DS, that is a Singer difference set. Such a
difference set is available in the literature. One is, for instance, the
following (see [13], page 427):

$D=\\{1,2,3,5,6,9,14,15,18,20,25,27,35\\}$

By Corollary 2.1, $\overline{D}:=\mathbb{Z}_{40}\setminus D$ is a
$(40,27,18)$-DS. Then, if we are able to partition $D$ into three subsets $A$,
$B$, $C$ of sizes $1$, $3$, $9$ in such a way that $\Delta A\ \cup\ \Delta B\
\cup\ \Delta C$ is all $\mathbb{Z}_{40}\setminus\\{0\\}$ twice, it is obvious
that

${\cal F}=\\{A,B,C,\overline{D}\\}$

would be a $(40,[1,3,9,27],20)$-HPDF. Well, the desired partition of $D$ has
$A$, $B$, $C$ as follows:

$A=\\{13\\};\quad B=\\{5,15,25\\};\quad C=\\{1,2,3,6,9,14,18,20,27\\}.$

This is readily seen from the difference tables of $B$ and $C$ below (of
course $\Delta A$ is empty).

| 5 | 15 | 25
---|---|---|---
$5$ | $-$ | 30 | 20
$15$ | $\bf 10$ | $-$ | 30
$25$ | $\bf 20$ | 10 | $-$
| 1 | 2 | 3 | 6 | 9 | 14 | 18 | 20 | 27
---|---|---|---|---|---|---|---|---|---
$1$ | $-$ | 39 | 38 | 35 | 32 | 27 | 23 | 21 | 14
$2$ | 1 | $-$ | 39 | 36 | 33 | 28 | 24 | 22 | 15
$3$ | 2 | 1 | $-$ | 37 | 34 | 29 | 25 | 23 | 16
$6$ | 5 | 4 | 3 | $-$ | 37 | 32 | 28 | 26 | 19
$9$ | $\bf 8$ | 7 | 6 | 3 | $-$ | 35 | 31 | 29 | 22
$14$ | $\bf 13$ | 12 | 11 | 8 | 5 | $-$ | 36 | 34 | 27
$18$ | $\bf 17$ | 16 | 15 | 12 | 9 | 4 | $-$ | 38 | 31
$20$ | $\bf 19$ | 18 | 17 | 14 | 11 | 6 | 2 | $-$ | 33
$27$ | $\bf 26$ | 25 | 24 | 21 | 18 | 13 | 9 | 7 | $-$

It is worth observing that $C$ is a $(10,4,9,2)$ relative difference set. It
means that $\Delta C$ is precisely twice $\mathbb{Z}_{10\cdot 4}\setminus N$
where $N$ is the subgroup of $\mathbb{Z}_{10\cdot 4}$ of order 4 (see [13]).

##  7 New infinite families of PDFs

Most composition constructions for PDFs make use of difference matrices and
lead to PDFs whose block sizes belong, almost all, to the set of block sizes
of the component PDFs (see [9, 15]). Buratti [6] motivated the introduction of
HPDFs showing that any single $(v,K,\lambda)$-HPDF is the ancestor of an
infinite series of PDFs where, apart from one special block of size
$2\lambda$, the size of every other block is the double of some $k\in K$. We
are going to recall the main application of his construction and then we
determine the descendants of our new examples.

Following [4], the maximal prime power factors of a given integer $v$ will be
called components of $v$, and $\mathbb{F}_{v}$ will denote the ring which is
the direct product of the fields whose orders are the components of $v$. Thus,
for instance, $\mathbb{F}_{63}=\mathbb{F}_{9}\times\mathbb{F}_{7}$.

######  Theorem 7.1.

[6] If there exists a $(G,[k_{1},\ldots,k_{t}],\lambda)$-HPDF and all the
components of $2n+1$ are greater than $2\cdot\max\\{k_{1},\dots,k_{t}\\}$,
then there exists a
$(2\lambda(2n+1),[(2k_{1})^{n},\ldots,(2k_{t})^{n},2\lambda],2\lambda)$-PDF in
$G\times\mathbb{F}_{2n+1}$.

Applying the above theorem using the new examples of HPDFs obtained in Section
6 we obtain the following results.

######  Corollary 7.2.

If all the components of $2n+1$ are greater than $34$, then there exists a
$(48n+24,[34^{n},4^{2n},2^{3n},24],24)$-PDF in $G\times\mathbb{F}_{2n+1}$ for
each of the three groups $G$ considered in 6.1.

The first $n$ for which the above corollary can be applied is 18. In this way
one gets a $(984,[34^{18},4^{36},2^{54},24],24)$-PDF in
$G\times\mathbb{F}_{37}$.

######  Corollary 7.3.

If all the components of $2n+1$ are greater than $48$, then there exists a
$(72n+36,[6^{n},18^{n},48^{n},36],36)$-PDF in $G\times\mathbb{F}_{2n+1}$ for
each of the nine groups $G$ considered in 6.2.

The first $n$ for which the above corollary can be applied is 24. In this way
one gets a $(1764,$ $[6^{24},18^{24},48^{24},36],36],36)$-PDF.

######  Corollary 7.4.

If all the components of $2n+1$ are greater than $54$, then there exists a
$(80n+40,[2^{n},6^{n},18^{n},54^{n},40],40)$-PDF in
$\mathbb{Z}_{40}\times\mathbb{F}_{2n+1}$.

The first $n$ for which the above corollary can be applied is 29. In this way
one gets a $(2360,[2^{29},6^{29},18^{29},54^{29},40],40)$-PDF.

##  8 A pair of open questions

It is obvious that any $(v,k,\lambda)$ difference set $B$ with $v=2\lambda$
gives rise to a $(v,[1^{v-k},k],\lambda)$-HPDF consisting of $B$ and all
possible singletons $\\{g\\}$ with $g\in G\setminus B$. Thus, for our
purposes, it is worth to look for difference sets with this property. Let us
call them Pell difference sets. The existence of a Pell $(v,k,\lambda)$-DS
obviously requires that $k(k-1)=\lambda(2\lambda-1)$ and, by the Bruck-Ryser-
Chowla theorem, that $k-\lambda$ is a square. On the other hand the latter
condition is redundant. Indeed, as shown in the following, the first condition
implies the second.

######  Proposition 8.1.

If $k$ and $\lambda$ are integers such that $k(k-1)=\lambda(2\lambda-1)$, then
$k-\lambda$ is a square.

###### Proof.

Set $d=\gcd(k,\lambda)$. Thus we have $k=de$ and $\lambda=df$ with $e$ and $f$
coprime integers. From the given equality we get $e(de-1)=f(2df-1)$ and then,
considering that $\gcd(e,f)=1$, we necessarily have $de-1=fg$ and $2df-1=eg$
for some integer $g$. Subtracting the second equality from the first one we
get $d(e-2f)=g(f-e)$. We clearly have $\gcd(d,g)=1$ and hence $d=f-e$ and
$g=e-2f$. We conclude that $k-\lambda=d(e-f)=d^{2}$. ∎

Apart from the trivial $(4,3,2)$-DS, no other Pell difference set is known.

######  Question 8.2.

Does there exist a Pell difference set of order $v>4$?

Here are the first possible triples $(v,k,\lambda)$ for which a non-trivial
Pell $(v,k,\lambda)$ difference set may exist:

$\begin{array}[]{l}(120,\ 85,\ 60)\\\ (4060,\ 2871,\ 2030)\\\ (137904,\
97513,\ 68952)\\\ (4684660,\ 3312555,\ 2342330)\\\ (159140520,\ 112529341,\
79570260)\\\ \end{array}$

Let us consider the first one. A putative $(120,85,60)$ is the complement of a
$(120,35,10)$-DS. Even though several authors investigated the possible
existence of a DS with these parameters, they have been only able to rule out
some groups (see [2]).

In view of these consideration and the fact that the admissible values of $v$
grow up very rapidly, we are afraid that Question 8.2 is really hard.

We come now to another question which is probably easier.

The parameter set of the last HPDF constructed in Section 6 can be written as

$\left({3^{4}-1\over 2},[3^{0},3^{1},3^{2},3^{3}],{3^{4}-1\over 4}\right).$

Inspired by this, we have noticed that

$\left({q^{2n}-1\over
q-1},[q^{0},q^{1},q^{2},q^{3},\dots,q^{2n-1}],{q^{2n}-1\over q+1}\right)$

is an admissible parameter set of a PDF for every positive integer $q$ (not
necessarily a prime power!). Indeed, if we set $v={q^{2n}-1\over q-1}$,
$k_{i+1}=q^{i}$ for $0\leq i\leq 2n-1$, and $\lambda={q^{2n}-1\over q+1}$, we
see that we have:

  * 1)

$\displaystyle k_{1}+k_{2}+\dots+k_{2n}=\sum_{i=0}^{2n-1}q^{i}={q^{2n}-1\over
q-1}=v$;

  * 2)

$\displaystyle k_{1}(k_{1}-1)+k_{2}(k_{2}-1)+\dots+k_{2n}(k_{2n}-1)=$

$\begin{array}[]{l}=(k_{1}^{2}+k_{2}^{2}+\dots+k_{2n}^{2})-(k_{1}+k_{2}+\dots+k_{2n})=\\\
\\\ =\displaystyle\sum_{i=0}^{2n-1}q^{2i}-{q^{2n}-1\over q-1}={q^{4n}-1\over
q^{2}-1}-{q^{2n}-1\over q-1}=\\\ \\\
=\displaystyle{(q^{2n}-1)(q^{2n-1}-q)\over q^{2}-1}=\lambda(v-1).\\\
\end{array}$

Thus the following question naturally arises.

######  Question 8.3.

Given positive integers $q$ and $n$, does there exist a PDF whose $K$ is
$[q^{0},q^{1},q^{2},q^{3},\dots,q^{2n-1}]$?

For now, we know that the answer is positive for $q=3$ and $n=1,2$. A positive
answer for $q=3$ and any $n$ would give the first infinite family of HPDFs.

## Acknowledgements

The author is supported in part by the Croatian Science Foundation under the
project 9752.

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