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\markright{\sc the electronic journal of combinatorics 5
(1997), \#R1\hfill}

\begin{document}
\title{New symmetric designs from regular Hadamard matrices}
\author{Yury J. Ionin\thanks{The author acknowledges with thanks the Central Michigan
University
Research Professor award.}\\
Department of Mathematics\\
Central Michigan University \\
Mt. Pleasant, MI 48859, USA\\
{\small\texttt{yury.ionin@cmich.edu}}}

\maketitle
\begin{abstract}
For every positive integer $m$, we construct a symmetric
$(v,k,\lambda )$-design with parameters $v=\frac{h((2h-1)^{2m}-1)}{h-1}$,
$k=h(2h-1)^{2m-1}$, and $\lambda =h(h-1)(2h-1)^{2m-2}$, where
$h=\pm 3\cdot 2^d$ and $|2h-1|$ is a prime power.  For $m\geq 2$ and
$d\geq 1$, these parameter values were previously undecided.  The tools
used in the construction are balanced generalized weighing matrices
and regular Hadamard matrices of order $9\cdot 4^d$.
\end{abstract}
\smallskip
\centerline{Submitted: October 30, 1997; Accepted: November 17, 1997}
\vspace{1cm}

MR Subject Number: {05B05}\hfill

Keywords: Symmetric design, regular Hadamard matrix, balanced generalized
weighing matrix\hfill



\bigskip


\begin{large}
\section{Introduction}

Let $v>k>\lambda\geq 0$ be integers.  A {\em symmetric}
$(v,k,\lambda )${\em -design} is an incidence structure $(P,{\mathcal B})$,
where $P$ is a set of cardinality $v$ (the point-set) and ${\mathcal B}$ is a
family of $v$ $k$-subsets (blocks) of $P$ such that any two distinct points
are contained in exactly $\lambda$ blocks.  If $P=\{ p_1,...,p_v\}$ and
${\mathcal B}=\{ B_1,...,B_v\}$, then the $(0,1)$-matrix $M=[m_{ij}]$ of
order $v$,
where $m_{ij}=1$ if and only if $p_j\in B_i$, is the {\em incidence matrix}
of the design.  A $(0,1)$-matrix $X$ of order $v$ is the
incidence matrix of a symmetric
$(v,k,\lambda )$-design if and only if it satisfies the equation
$XX^T=(k-\lambda )I+\lambda J$, where $I$ is the identity matrix and $J$ is
the
all-one matrix of order $v$.  For references, see \cite{BJL} or
\cite[Chapter 5]{CRC}.

A {\em Hadamard matrix} of order $n$ is an $n$ by $n$ matrix $H$ with entries
equal to $\pm 1$ satisfying $HH^T=nI$.  A Hadamard matrix is {\em
regular} if its row and column sums are constant.  This sum is always
even and if we denote it $2h$, then the order of the matrix is
equal to $4h^2$.  Replacing $-1$s in a regular
Hadamard matrix of order $4h^2$ by $0$s yields the incidence matrix
of a symmetric $(4h^2,2h^2-h,h^2-h)$-design usually called a {\em
Menon design}.  Conversely, replacing
$0$s by $-1$s in the incidence matrix
of a symmetric $(4h^2,2h^2-h,h^2-h)$-design yields a regular Hadamard matrix
of order $4h^2$.  For references, see \cite{Seb}.  In this paper,
we will be interested in regular
Hadamard matrices of order $9\cdot 4^{d}$, where $d$ is a positive
integer.  If $H$ is such a matrix, then the Kronecker product of a
regular Hadamard matrix of order 4 and $H$ is a regular Hadamard matrix
of order $9\cdot 4^{d+1}$.  Therefore, one can obtain a family of
regular Hadamard matrices of order $9\cdot 4^{d}$, starting with a
regular Hadamard matrix of order $36$.

A {\em balanced generalized weighing matrix} $\text{BGW}(v,k,\lambda )$
over a (multiplicatively written) group $G$ is a matrix $W=[\omega _{ij}]$
of \
order $v$ with entries from the set $G\cup\{ 0\}$ such that (i) each row
and each column of $W$ contain exactly $k$ non-zero entries and (ii)
for any distinct rows $i$ and $h$, the multiset
$$\{\omega _{hj}^{-1}\omega _{ij}\colon 1\leq j\leq v,\omega _{ij}\neq 0,
\omega _{hj}\neq 0\}$$
contains exactly $\lambda /|G|$ copies of every element of $G$.

In this paper, we will use a balanced generalized weighing matrix
$\text{BGW}(q^m+q^{m-1}+\dots +q+1,q^m,q^m-q^{m-1})$ over a cyclic
group $G$ of
order $t$, where $q$ is a prime power, $m$ is a positive integer, and $t$ is a
divisor of $q-1$.  Such matrices are known to exist \cite[IV.4.22]{CRC} and
have been applied to constructing symmetric designs by Rajkundlia
\cite{Raj}, Brouwer \cite{Bro}, Fanning \cite{Fan}, and the author
\cite{Ion1,Ion2}.  If $\mathcal{M}$ is a set of $m$ by $n$ matrices,
$G$ is a group of bijections $\mathcal{M}\to\mathcal{M}$, and $W$ is
a balanced generalized weighing matrix over $G$, then, for any
$P\in\mathcal{M}$, $W\otimes P$ denotes the matrix obtained by
replacing every entry $\sigma$ in $W$ by the matrix $\sigma P$.
In Section 2 (Lemma \ref{main}), we will prove the following modification
of a result from \cite{Ion2}:

Let $\mathcal{M}$ be a set of matrices of order $v$ containing the
incidence matrix $M$ of a symmetric $(v,k,\lambda )$-design with
$q=\frac{k^2}{k-\lambda}$ a prime power.  Let $G$ be a finite
cyclic group of bijections $\mathcal{M}\rightarrow\mathcal{M}$
such that (i) $(\sigma P)(\sigma Q)^T=PQ^T$ for any
$P,Q\in\mathcal{M}$ and $\sigma\in G$, (ii)
$\sum_{\sigma\in G}\sigma M=\frac{k|G|}{v}J$,
and (iii) $|G|$ divides $q-1$.
If $W$ is a balanced generalized weighing matrix
$\text{BGW}(q^m+\dots +q+1,q^m,q^m-q^{m-1})$ over $G$,
then $W\otimes M$ is the incidence
matrix of a symmetric $(v(q^m+q^{m-1}+\dots +q+1),kq^m,\lambda q^m)$-design.

In order to apply this lemma, we need a symmetric $(v,k,\lambda )$-design
to start with.  In the paper \cite{Ion2}, we have shown that the designs
corresponding to certain McFarland and Spence difference
sets (or their complements) serve as such starters.  In Section 3
of this paper, we show that for $h=\pm 3\cdot 2^d$, if $|2h-1|$ is a prime
power, then there is a symmetric $(4h^2,2h^2-h,h^2-h)$-design, which
can also serve as a starter.  As a result, we show that for any
positive integers $m$ and $d$, if $h=\pm 3\cdot 2^d$ and $|2h-1|$ is a prime
power, then there exists a symmetric $(v,k,\lambda )$-design with
$$v=\frac{h((2h-1)^{2m}-1)}{h-1}, k=h(2h-1)^{2m-1}, \lambda
=h(h-1)(2h-1)^{2m-2}.$$
These parameters are new, except $m=1$ (Menon designs) and $d=0$
(constructed by the author in \cite{Ion2}).

\section{Preliminaries}

Throughout this paper, we will denote identity,
zero, and all-one matrices of suitable orders by $I$, $O$, and $J$,
respectively.

If $W$ is a balanced generalized weighing matrix of order $w$ over a group
$G$ of
bijections on a set $\mathcal{M}$ of matrices of order $n$, then, for
any $P\in\mathcal{M}$, we will denote by $W\otimes P$ the matrix of
order $nw$ obtained by replacing every nonzero entry $\sigma$ in $W$
by the matrix $\sigma P$ and every zero entry in $W$ by the zero matrix of
order $n$.

The following lemma represents a slight modification of a result
proven in \cite{Ion2}.  Since it is crucial
for this paper and the proof is short, we will repeat it here.

\begin{lem}\label{main}
Let $v>k>\lambda\geq 0$ be integers.  Let $\mathcal{M}$ be a
set of matrices of order $v$ and $G$ a finite
group of bijections $\mathcal{M}\rightarrow\mathcal{M}$ satisfying the
following conditions:

(i) $\mathcal{M}$ contains the incidence matrix $M$ of a symmetric
$(v,k,\lambda )$-design;

(ii) for any $P,Q\in\mathcal{M}$ and $\sigma\in G$,
$$(\sigma P)(\sigma Q)^T=PQ^T;$$

(iii) $\sum_{\sigma\in G}\sigma M=\frac{k|G|}{v}J;$

(iv) $q=\frac{k^2}{k-\lambda}$ is a prime power;

(v) $G$ is cyclic and $|G|$ divides $q-1$.

Then, for any positive integer $m$, there exists
a symmetric $(vw,kq^m,\lambda q^m)$-design, where
$w=\frac{q^{m+1}-1}{q-1}$.
\end{lem}

\noindent\textbf{Proof.}
Let $W=[\omega _{ij}]$, $i,j=1,2,\dots ,w$ be a balanced generalized
weighing matrix
\newline$\text{BGW}(w,q^m,q^m-q^{m-1})$ over $G$.  We claim that
$W\otimes M$ is the incidence matrix of a symmetric $(vw,kq^m,\lambda
q^m)$-design.  It suffices to show that, for $i,h=1,2,\dots ,w$,
\begin{equation*}
\sum_{j=1}^w(\omega _{ij}M)(\omega _{hj}M)^T=
\begin{cases}
(k-\lambda )q^mI+\lambda q^mJ\text{ if }i=h,\\
\lambda q^mJ\text{ if }i\neq h.
\end{cases}
\end{equation*}

If $i=h$, we have for some $\sigma _j\in G$,
$$\sum_{j=1}^w(\omega _{ij}M)(\omega _{hj}M)^T=
\sum_{j=1}^{q^m}(\sigma _jM)(\sigma _jM)^T=
\sum_{j=1}^{q^m}MM^T=(k-\lambda )q^mI+\lambda q^mJ.$$

If $i\neq h$, we have for some $\sigma _j,\tau _j\in G$,
$$\sum_{j=1}^w(\omega _{ij}M)(\omega _{hj}M)^T=
\sum_{j=1}^{q^m-q^{m-1}}(\sigma _jM)(\tau _jM)^T
=\sum_{j=1}^{q^m-q^{m-1}}(\tau _j^{-1}\sigma _jM)M^T$$
$$=\frac{q^m-q^{m-1}}{|G|}\big(\sum_{\sigma\in G}\sigma M\big) M^T
=\frac{k(q^m-q^{m-1})}{v}JM^T=\frac{k^2(q^m-q^{m-1})}{v}J=\lambda q^mJ.$$
 $\Box$

\begin{dfn}
Let $v>k>\lambda >0$ be integers.  A $(v,k,\lambda )$-difference set
is a $k$-subset of an (additively written) group $\Gamma$ of order $v$
such that the multiset $\{ x-y\colon x,y\in\Gamma\}$ contains
exactly $\lambda$ copies of each nonzero element of $\Gamma$.
\end{dfn}

Several infinite families of difference sets are known (see
\cite{CRC} or \cite{Jung} for references).  We will mention the McFarland
family
having parameters $(p^{d+1}(r+1),p^dr,p^{d-1}(r-1))$, where $p$ is a
prime power, $d$ is a positive integer, and
$r=\frac{p^{d+1}-1}{p-1}$, and the Spence family having parameters
$(3^{d+1}(3^{d+1}-1)/2,3^{d}(3^{d+1}+1)/2,3^{d}(3^{d}+1)/2)$,
where $d$ is a positive integer.

If $\Delta$ is a $(v,k,\lambda )$-difference set in a group $\Gamma$
and $\mathcal{B}=\{\Delta +x\colon x\in\Gamma\}$,
then $\text{dev}(\Delta )=(\Gamma ,\mathcal{B})$ is a symmetric
$(v,k,\lambda )$-design.

In order to apply Lemma \ref{main}, we need a symmetric
$(v,k,\lambda )$-design with $q=\frac{k^2}{k-\lambda}$ a prime
power, a set $\mathcal{M}$ of matrices of order $v$ containing the
incidence matrix this design, and a cyclic group $G$ satisfying conditions
(ii), (iii), and (v) of Lemma \ref{main}.
In the paper \cite{Ion2}, we have shown that
$(v,k,\lambda )$ can be the parameters of any McFarland or Spence
difference set or their complement with $q=\frac{k^2}{k-\lambda }$ a prime
power.
In this paper, we will use the Spence $(36,15,6)$-difference set in
$\Gamma =\mathbb{Z}_3\oplus\mathbb{Z}_3\oplus\mathbb{Z}_4$ and the
complementary $(36,21,12)$-difference set.  In the
next section, we will reproduce the construction of the corresponding
$\mathcal{M}$ and $G$ given in \cite{Ion2}

\section{$(36,15,6)$- and $(36,21, 12)$-difference sets}

We start with a brief description of the Spence $(36,15,6)$-difference set in
$\Gamma =\mathbb{Z}_3\oplus\mathbb{Z}_3\oplus\mathbb{Z}_4$.

We consider $\Gamma$ as the set of triples $(x_1,x_2,x_3)$, where
$x_1,x_2\in\{ 0,1,2\}$ and $x_3\in\{ 0,1,2,3\}$ with the mod 3 and the
mod 4 addition, respectively.  Consider
$\mathbb{Z}_3\oplus\mathbb{Z}_3$ as a 2-dimensional vector space over
the field GF(3).  Let $L_1,L_2,L_3,L_4$ be its 1-dimensional
subspaces.  Put $D_1=\{ (x_1,x_2,0)\in\Gamma\colon
(x_1,x_2)\not\in L_1\}$ and, for $i=2,3,4$, $D_i=\{
(x_1,x_2,i-1)\in\Gamma\colon (x_1,x_2)\in L_i\}$.  Then $D=D_1\cup D_2\cup
D_3\cup D_4$ is a $(36,15,6)$-difference set in $\Gamma$ \cite[Theorem
11.2]{Jung}.

In order to obtain the incidence matrix of the corresponding symmetric
design, we have to select an order on $\Gamma$.  We will assume
that $(x_1,x_2,x_3)$ precedes $(y_1,y_2,y_3)$ in $\Gamma$ if and only
if there is $i$ such that $x_i<y_i$ and $x_j=y_j$ whenever $j>i$.
Let $M$ be the $(0,1)$-matrix of order $36$ whose rows and columns are
indexed by elements of $\Gamma$ in this order and $(x,y)$-entry is
equal to $1$ if and only if $y-x\in D$.  Then $M$ is the incidence
matrix of a symmetric $(36,15,6)$-design.  In order to describe the
structural properties of $M$ which will be important in the sequel,
we introduce the following operation $\rho$ on the set
of 3 by 3 block-matrices.

\begin{dfn}
Let $P=[P_{ij}]$ be a $3$ by $3$ block-matrix with square blocks (in
particular, $P$ can be a $3$ by $3$ matrix).  Denote by $\rho P$ the
matrix obtained by applying
the cyclic permutation $\rho =(123)$ of degree $3$ to the set of columns of
$P$, i.e.,
\begin{equation*}
\rho
\begin{bmatrix}
P_{11} & P_{12} & P_{13}\\
P_{21} & P_{22} & P_{23}\\
P_{31} & P_{32} & P_{33}
\end{bmatrix}
=
\begin{bmatrix}
P_{13} & P_{11} & P_{12}\\
P_{23} & P_{21} & P_{22}\\
P_{33} & P_{31} & P_{32}
\end{bmatrix}
.
\end{equation*}
\end{dfn}

The above incidence matrix $M$ of a symmetric $(36,15,6)$-design
can be represented as a 4 by 4 block-matrix
\begin{equation*}
M=
\begin{bmatrix}
M_1 & M_2 & M_3 & M_4\\
M_4 & M_1 & M_2 & M_3\\
M_3 & M_4 & M_1 & M_2\\
M_2 & M_3 & M_4 & M_1
\end{bmatrix},
\end{equation*}
where each $M_i$ is a 9 by 9 matrix.  Further, each $M_i$ can be
represented as a 3 by 3 block-matrix
\begin{equation*}
M_i=
\begin{bmatrix}
M_{i1} & M_{i2} & M_{i3}\\
M_{i3} & M_{i1} & M_{i2}\\
M_{i2} & M_{i3} & M_{i1}
\end{bmatrix},
\end{equation*}
where each $M_{ij}$ is a matrix of order 3, $M_{11}=O$,
$M_{12}=M_{13}=J$, $M_{21}=M_{22}=M_{23}=M_{31}=M_{41}=I$,
$M_{32}=M_{43}=\rho I$, and $M_{33}=M_{42}=\rho ^2I$.

Let $\mathcal{M}$ be the set of block-matrices $P=[P_{ij}]$,
$i,j=1,2,3,4$, where each $P_{ij}$ is a block-matrix
$P_{ij}=[P_{ijkl}]$, $k,l=1,2,3$, satisfying the following conditions:

(i) each $P_{ijkl}$ is a $(0,1)$-matrix of order 3;

(ii) for $i=1,2,3,4$, there is a unique $h_i=h_i(P)\in\{ 1,2,3,4\}$
such that
$$(P_{ijk1},P_{ijk2},P_{ijk3})\text{ is a permutation of
}(O,J,J)\text{ for }j=h_i\text{ and all }k$$
and
$$P_{ijkl}\in\{ I,\rho I,\rho ^2I\}\text{ for }j\ne h_i\text{ and all
}k,l.$$

Clearly, the above matrix $M$ is an element of $\mathcal{M}$.

Define a bijection $\sigma\colon\mathcal{M}\to\mathcal{M}$ by $\sigma
P=P^\prime$, where

(i) for $i=1,2,3,4$ and $j=2,3,4$, $P_{ij}^\prime =P_{i,j-1}$;

(ii) for $i=1,2,3,4$, if $h_i=4$, then $P_{i1}^\prime =\rho P_{i4}$;

(iii) for $i=1,2,3,4$, if $h_i\ne 4$, then $P_{i1kl}^\prime =\rho
P_{i4kl}$ for all $k,l$.

Let $G$ be the cyclic group generated by $\sigma$.  Then $|G|=12$.
\medskip

{\em Claim.}  For any $P,Q\in\mathcal{M}$, $(\sigma P)(\sigma
Q)^T=PQ^T$.

\noindent\textbf{Proof.}
Let $P,Q\in\mathcal{M}$ and let $P^\prime =\sigma P$ and $Q^\prime =\sigma
Q$.  It suffices to show that, for $i=1,2,3,4$,
\begin{equation}\label{1}
P_{i1}^\prime Q_{i1}^{\prime T}=P_{i4}Q_{i4}^T.
\end{equation}
If $h_i(P)=h_i(Q)=4$ or $h_i(P)\ne 4$ and $h_i(Q)\ne 4$, then
$P_{i1}^\prime$ is obtained from $P_{i4}$ by the same permutation of
columns as $Q_{i1}^\prime$ from $Q_{i4}$, so \eqref{1} is clear.
Suppose $h_i(P)=4$ and $h_i(Q)\ne 4$.  Then
$(P_{i4k1},P_{i4k2},P_{i4k3})$ is a permutation of $(O,J,J)$ and
matrices $Q_{i4k1},Q_{i4k2},Q_{i4k3}$ have the same row sum (equal to
1).  Therefore
$$\sum_{l=1}^3P_{i1kl}^\prime Q_{i1kl}^{\prime
T}=\sum_{l=1}^3P_{i4kl}Q_{i4kl}^T=2J,$$
and \eqref{1} follows.
 $\Box$

It is readily verified that
\begin{equation}\label{2}
\sum_{n=0}^{11}\sigma ^nM=5J.
\end{equation}

Thus, the set $\mathcal{M}$, the matrix $M$, and the group $G$
satisfy Lemma \ref{main} for $(v,k,\lambda )=(36,15,6)$ with
$|G|=12$.  Note that the sum of the entries of any row of any
matrix $P\in\mathcal{M}$ is equal to 15.

Let $\overline{M}=J-M$ and $\overline{\mathcal{M}}=\{ J-P\colon
P\in\mathcal{M}\}$.  Without changing $G$, we obtain that
$\overline{\mathcal{M}}$, $\overline{M}$, and $G$
satisfy Lemma \ref{main} for $(v,k,\lambda )=(36,21,12)$.  The
sum of the entries of any row of any
matrix $P\in\overline{\mathcal{M}}$ is equal to 21.

Note that the described $(36,15,6)$-design and $(36,21,12)$-design are
symmetric $(4h^2,2h^2-h,h^2-h)$-designs with $h=3$ and $h=-3$,
respectively.

\section{Using the Kronecker product}

The next lemma will allow us to double the parameter $h$ in a family of
symmetric $(4h^2,2h^2-h,h^2-h)$-designs satisfying Lemma \ref{main}.

\begin{lem}\label{A}
Let an integer $h\ne 0$, a set $\mathcal{M}$ of
matrices of order $4h^2$, and a finite cyclic group
$G=\langle\sigma\rangle$ of bijections
$\mathcal{M}\to\mathcal{M}$ satisfy the following conditions:

(i) $\mathcal{M}$ contains the incidence matrix $M$ of a symmetric
$(4h^{2},2h^2-h,h^2-h)$-design;

(ii) for any $P,Q\in\mathcal{M}$, $(\sigma P)(\sigma Q)^{T}=PQ^{T}$;

(iii) $\sum_{n=0}^{|G|-1}\sigma ^nM=\frac{(2h-1)|G|}{4h}J.$

(iv) the sum of the entries of any row of any matrix
$P\in\mathcal{M}$ is equal to $2h^2-h$.

Then there exists a set $\mathcal{M}_1$ of
matrices of order $16h^2$ and a cyclic group $G_1=\langle\tau\rangle$ of
bijections
$\mathcal{M}_1\to\mathcal{M}_1$ satisfying the following conditions:

(a) $\mathcal{M}_1$ contains the incidence matrix $M_1$ of a symmetric
$(16h^{2},8h^2-2h,4h^2-2h)$-design;

(b) for any $R,S\in\mathcal{M}_1$, $(\tau R)(\tau S)^{T}=RS^{T}$;

(c) $\sum_{n=0}^{|G_1|-1}\tau ^nM_1=\frac{(4h-1)|G_1|}{8h}J;$

(d) the sum of the entries of any row of any matrix
$R\in\mathcal{M}_1$ is equal to $8h^2-2h$;

(e) $|G_1|=2|G|$.
\end{lem}

\noindent\textbf{Proof.}
For any $P\in\mathcal{M}$, define
\begin{equation*}
R_{P}=
\begin{bmatrix}
J-P & P & P & P\\
P & J-P & P & P\\
P & P & J-P & P\\
P & P & P & J-P
\end{bmatrix}
.
\end{equation*}

It is well known and readily verified that $M_1=R_{M}$ is the incidence
matrix of a
symmetric $(16h^{2},8h^2-2h,4h^2-2h)$-design.

Let $\mathcal{M}_1=\{ R_{P}\colon P\in\mathcal{M}\}$.  Then
$M_1\in\mathcal{M}_1$, so $\mathcal{M}_1$ satisfies (a).  Condition (d)
is implied by (iv).  Any
matrix $R\in\mathcal{M}_1$ can be divided into eight $4h^{2}$ by
$8h^{2}$ cells $R_{ij}$, $1\leq i\leq 4$, $1\leq j\leq 2$.  Observe that each
$R_{ij}$ is of one of the two following types:

(type 1) $R_{ij}=[P\quad J-P]$ or $R_{ij}=[J-P\quad P]$, $P\in\mathcal{M}$;

(type 2) $R_{ij}=[P\quad P]$, $P\in\mathcal{M}$.

Observe also that $R_{i1}$ and $R_{i2}$ are not of the same type.

For any $R\in\mathcal{M}_1$, denote by
$\tau R$ a $(0,1)$-matrix of order $16h^{2}$ divided into eight $4h^{2}$ by
$8h^{2}$ cells $\tau R_{ij}$, $1\leq i\leq 4$, $1\leq j\leq 2$,
where
\begin{equation*}
\tau R_{i2}=R_{i1}
\end{equation*}
and
\begin{equation*}
\tau R_{i1}=
\begin{cases}
J-R_{i2} & \text{if }R_{i2}\text{ is of type 1,}\\
\left[\sigma P\quad\sigma P\right] & \text{if }R_{i2}=\left[ P\quad P\right] .
\end{cases}
\end{equation*}

In order to verify (b), it suffices to show that, for $i=1,2,3,4$, $(\tau
R_{i1})(\tau S_{i1})^{T}=R_{i2}S_{i2}^{T}$.

If $R_{i2}$ and $S_{i2}$ are of type (1), then $(\tau R_{i1})(\tau
S_{i1})^{T}=(J-R_{i2})(J-S_{i2})^T=8h^2J-R_{i2}J^T-JS_{i2}^T+R_{i2}S_{i2}^{T}=
R_{i2}S_{i2}^{T}$ for the row sum of any matrix of type 1 is equal to
$4h^2$.  If $R_{i2}=[P\quad P]$ and $S_{i2}=[Q\quad
Q]$, where $P,Q\in\mathcal{M}$, then $(\tau R_{i1})(\tau
S_{i1})^{T}=2(\sigma P)(\sigma Q)^{T}=2PQ^{T}=R_{i2}S_{i2}^{T}$.  If
$R_{i2}=[P\quad P]$ and $S_{i2}$ is of type 1, then
$(\tau R_{i1})(\tau S_{i1})^{T}=(\sigma P)J=(2h^2-h)J=R_{i2}S_{i2}^{T}$.

Let $G_1$ be the group of bijections $\mathcal{M}_1\to\mathcal{M}_1$ generated
by $\tau$.  Then (e) is satisfied, and we have to verify (c).  For
$n=1,2,\dots ,2|G|-1$, let $A_n$ be the $(i,j)$-block of the 4 by 4
block-matrix $\tau ^nM_1$.  Then there is $P\in\mathcal{M}$ such that the
multiset $\{ A_n\colon 0\leq n\leq 2|G|-1\}$ is the union of
$\{ \sigma ^nP\colon 0\leq n\leq |G|-1\}$ and the multiset consisting
of $\frac{|G|}{2}$ copies of $P$ and $\frac{|G|}{2}$ copies of $J-P$.
Therefore,
$$\sum_{n=0}^{2|G|-1}A_n=\sum_{n=0}^{|G|-1}\sigma ^nP+\frac{|G|}{2}J
=\frac{(2h-1)|G|}{4h}J+\frac{|G|}{2}J=\frac{(4h-1)|G_1|}{8h}J.$$
 $\Box$

The following theorem is now immediate by induction.

\begin{thm}\label{corol}
Let an integer $h\ne 0$, a set $\mathcal{M}$ of
matrices of order $4h^2$, and a finite cyclic group
$G$ of bijections
$\mathcal{M}\to\mathcal{M}$ satisfy conditions (i)--(iv) of Lemma
\ref{A}.  Then, for any positive integer $d$, there exists a non-empty set
$\mathcal{M}_{d}$ of matrices of order $4^{d+1}h^{2}$ and a cyclic
group $G_{d}$ of bijections $\mathcal{M}_{d}\to\mathcal{M}_{d}$
satisfying the following conditions:

(a) $\mathcal{M}_{d}$ contains the incidence matrix $M_d$ of a symmetric
design with parameters
$$(4^{d+1}h^{2},2^{2d+1}h^2-2^{d}h,2^{2d}h^2-2^{d}h);$$

(b) for any $P,Q\in\mathcal{M}_{d}$ and $\tau\in G_{d}$, $(\tau P)(\tau
Q)^{T}=PQ^{T}$;

(c) $\sum_{\tau\in G_{d}}\tau M_d=\frac{(2^{d+1}h-1)|G_d|}{2^{d+2}h}J;$

(d) the sum of the entries of any row of any matrix
$R\in\mathcal{M}_d$ is equal to $2^{2d+1}h^2-2^{d}h$;

(e) $|G_{d}|=2^{d}|G|$.
\end{thm}

We combine Theorem
\ref{corol} and Lemma \ref{main} and obtain the main result of this paper.

\begin{thm}
If $h=\pm 3\cdot 2^d$, where $d$ is a positive integer and
$|2h-1|$ is a prime power, then, for any positive integer $m$, there
exists a symmetric
$(\frac{h((2h-1)^{2m}-1)}{h-1},h(2h-1)^{2m-1},h(h-1)(2h-1)^{2m-2})$-design.
\end{thm}

\noindent\textbf{Proof.}
We start with the set $\mathcal{M}$ or $\overline{\mathcal{M}}$
described in Section 3 and apply Theorem \ref{corol} to this set to
obtain the set of matrices $\mathcal{M}_d$ or
$\overline{\mathcal{M}}_d$ and the group $G_d$.
Then we apply Lemma \ref{main}.  Properties (ii) and (iii) required in Lemma
\ref{main} are implied by (b) and (c) of Theorem \ref{corol}.  The
parameter $q$ of Lemma \ref{main} is equal to $(2h_d-1)^2$, where $h_d=\pm
3\cdot 2^d$, so $q$ is a prime power.
Since $|G|=12$, we have $|G_d|=3\cdot 2^{d+2}=4|h_d|$, so $|G_d|$ divides
$q-1$.
 $\Box$

\begin{rem}
These parameters are new, except $m=1$ (Menon designs).
\end{rem}
\end{large}

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\end{document}