Antipodal Polygons and Half-Circulant Hadamard Matrices
As known, the question on the existence of Hadamard matrices of order m = 4n, where n is an arbitrary natural number, is equivalent to the question on the possibility to inscribe a regular hypersimplex into the (4n ¡ 1)-dimensional cube. We introduced a class of Hadamard matrices of order 4n of half...
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irk-123456789-1066332016-10-02T03:02:46Z Antipodal Polygons and Half-Circulant Hadamard Matrices Medianik, A.I. As known, the question on the existence of Hadamard matrices of order m = 4n, where n is an arbitrary natural number, is equivalent to the question on the possibility to inscribe a regular hypersimplex into the (4n ¡ 1)-dimensional cube. We introduced a class of Hadamard matrices of order 4n of half-circulant type in 1997 and a class of antipodal n-gons inscribed into a regular (2n-1)-gon. In 2004 we proved that a half-circulant Hadamard ma- trix of order 4n exists if and only if there exist antipodal n-gons inscribed into a regular (2n-1)-gon. On this background there was developed the method of constructing of the Hadamard matrices of order 4n, which is universal, i.e., it can be applied to any arbitrary natural number n, including a prime number case, that usually requires the individual approach to the construction of the Hadamard matrix of corresponding order. In the paper, there are obtained the necessary and su±cient algebraic-geometric conditions for the existence of antipodal polygons allowing to justify the inductive approach to be used to the proof of existence theorems for Hadamard matrices of arbitrary order 4n, n ≥ 3. 2010 Article Antipodal Polygons and Half-Circulant Hadamard Matrices / A.I. Medianik // Журнал математической физики, анализа, геометрии. — 2010. — Т. 6, № 1. — С. 56-72. — Бібліогр.: 10 назв. — англ. 1812-9471 http://dspace.nbuv.gov.ua/handle/123456789/106633 en Журнал математической физики, анализа, геометрии Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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As known, the question on the existence of Hadamard matrices of order m = 4n, where n is an arbitrary natural number, is equivalent to the question on the possibility to inscribe a regular hypersimplex into the (4n ¡ 1)-dimensional cube. We introduced a class of Hadamard matrices of order 4n of half-circulant type in 1997 and a class of antipodal n-gons inscribed into a regular (2n-1)-gon. In 2004 we proved that a half-circulant Hadamard ma- trix of order 4n exists if and only if there exist antipodal n-gons inscribed into a regular (2n-1)-gon. On this background there was developed the method of constructing of the Hadamard matrices of order 4n, which is universal, i.e., it can be applied to any arbitrary natural number n, including a prime number case, that usually requires the individual approach to the construction of the Hadamard matrix of corresponding order. In the paper, there are obtained the necessary and su±cient algebraic-geometric conditions for the existence of antipodal polygons allowing to justify the inductive approach to be used to the proof of existence theorems for Hadamard matrices of arbitrary order 4n, n ≥ 3. |
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Medianik, A.I. Antipodal Polygons and Half-Circulant Hadamard Matrices Журнал математической физики, анализа, геометрии |
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Medianik, A.I. |
| author_sort |
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| title |
Antipodal Polygons and Half-Circulant Hadamard Matrices |
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Antipodal Polygons and Half-Circulant Hadamard Matrices |
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Antipodal Polygons and Half-Circulant Hadamard Matrices |
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Antipodal Polygons and Half-Circulant Hadamard Matrices |
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Antipodal Polygons and Half-Circulant Hadamard Matrices |
| title_sort |
antipodal polygons and half-circulant hadamard matrices |
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Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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Antipodal Polygons and Half-Circulant Hadamard Matrices / A.I. Medianik // Журнал математической физики, анализа, геометрии. — 2010. — Т. 6, № 1. — С. 56-72. — Бібліогр.: 10 назв. — англ. |
| series |
Журнал математической физики, анализа, геометрии |
| work_keys_str_mv |
AT medianikai antipodalpolygonsandhalfcirculanthadamardmatrices |
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2025-07-07T18:47:45Z |
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2025-07-07T18:47:45Z |
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Journal of Mathematical Physics, Analysis, Geometry
2010, vol. 6, No. 1, pp. 56–72
Antipodal Polygons and Half-Circulant Hadamard
Matrices
A.I. Medianik
Mathematical Division, B. Verkin Institute for Low Temperature Physics and Engineering
National Academy of Sciences of Ukraine
47 Lenin Ave., Kharkiv, 61103, Ukraine
E-mail:medianik@ilt.kharkov.ua
Received December 9, 2008
As known, the question on the existence of Hadamard matrices of order
m = 4n, where n is an arbitrary natural number, is equivalent to the ques-
tion on the possibility to inscribe a regular hypersimplex into the (4n− 1)-
dimensional cube. We introduced a class of Hadamard matrices of order 4n
of half-circulant type in 1997 and a class of antipodal n-gons inscribed into
a regular (2n-1)-gon. In 2004 we proved that a half-circulant Hadamard ma-
trix of order 4n exists if and only if there exist antipodal n-gons inscribed into
a regular (2n-1)-gon. On this background there was developed the method
of constructing of the Hadamard matrices of order 4n, which is universal,
i.e., it can be applied to any arbitrary natural number n, including a prime
number case, that usually requires the individual approach to the construc-
tion of the Hadamard matrix of corresponding order. In the paper, there are
obtained the necessary and sufficient algebraic-geometric conditions for the
existence of antipodal polygons allowing to justify the inductive approach
to be used to the proof of existence theorems for Hadamard matrices of
arbitrary order 4n, n ≥ 3.
Key words: multidimensional cube, regular hypersimplex, Hadamard
matrix, circulant matrix, antipodal polygons, necessary and sufficient
conditions, mathematical induction, existence theorem.
Mathematics Subject Classification 2000: 05B20, 52B.
1. Introduction
The two convex n-gons inscribed into a regular (2n − 1)-gon are said to be
antipodal if the total number of their diagonals and sides of the same length is
n for all admissible lengths [1, p. 48]. A square matrix H, whose every entry
is +1 or −1, is called the Hadamard matrix of order m if HH ′ = mI, where
c© A.I. Medianik, 2010
Antipodal Polygons and Half-Circulant Hadamard Matrices
H ′ is a transpose matrix and I is an identity one [2, p. 283]. As known, the
question on the existence of Hadamard matrices of order m = 4n, where n is
an arbitrary natural number, is equivalent to the question on the possibility to
inscribe a regular hypersimplex into the (4n − 1)-dimensional cube so that its
every vertex coincides with one of the vertices of the cube. This was established
by H. Coxeter in 1933 (see [3, p. 319]). We introduced a class of Hadamard
matrices of half-circulant type [4, p. 459] in the form
H =
(
A B
B −A
)
,
where A and B are matrices of order 2n (with bordering entries equal to 1)
containing submatrices of order 2n − 1 which are the right and, respectively,
left circulants (see [5, p. 27]). These matrices appeared to be closely connected
with the class of antipodal polygons. Namely, it is proved that a half-circulant
Hadamard matrix of order 4n exists if and only if there exist antipodal n-gons
inscribed into a regular (2n − 1)-gon (see [1, Th. 4]). Thus, the solution of the
question either on the existence of Hadamard matrices of arbitrary order 4n or
on the possibility of inscribing a regular hypersimplex into (4n− 1)-dimensional
cube depends on the solution of the question on the existence of antipodal n-gons.
Moreover, it is possible now to set the proof problem of the existence theorem
without constructing the Hadamard matrices of smallest order the existence of
which is not proved. Besides, the Hadamard matrix of order 428 was constructed
recently in [6] (twenty years later after its predecessor of order 268 [7]).
In the paper, there are obtained the necessary and sufficient algebraic-geometric
conditions for the existence of antipodal polygons (Th. 2), which, by virtue of
their universality, allow to justify the inductive approach to be used to the proof of
existence theorems for Hadamard matrices of arbitrary order 4n, n ≥ 3 (Sect. 3).
2. Existence Conditions of Antipodal Polygons
Let z = e
2πi
2n−1 , n ≥ 3. Then the vertices of a regular (2n − 1)-gon inscribed
into a unit circle with the center in the origin of coordinates are given by the
monomials zk, k = 0, 1, 2, . . . , 2n−2. Let every convex n-gon P inscribed into the
regular (2n− 1)-gon correspond to a generating polynomial p(z) =
∑2n−2
k=0 xkz
k,
where xk = 1 if the vertex with number k belongs to P , and xk = 0 otherwise.
Since P is an n-gon, we have
∑2n−2
k=0 xk = n. If dk is the number of equal diagonals
and sides of the n-gon P visible from the origin at the angle 2πk
2n−1 , then for the
polynomial p(z) the equality
|p|2 = n + 2
n−1∑
k=1
dk cos
2πk
2n− 1
,
Journal of Mathematical Physics, Analysis, Geometry, 2010, vol. 6, No. 1 57
A.I. Medianik
is valid, where
∑n−1
k=1 dk = n(n−1)/2 (see [1, Lem. 1]). The similar formula is true
for the generating polynomial p′(z) =
∑2n−2
k=0 x′kz
k of any other n-gon P ′ inscribed
into the given regular (2n− 1)-gon. If P and P ′ are antipodal, then dk + d′k = n
for every k = 1, 2, . . . , n− 1 by definition. Therefore, the generating polynomials
of antipodal n-gons satisfy the relation |p|2 + |p′|2 = n (see [1, Th. 3]). Indeed it
holds a more general equality where the rotation group of regular (2n− 1)-gon is
taken into account. In this connection, there arises a question on the existence
of an inscribed n-gon P with the number of diagonals dk, k = 1, 2, . . . , n − 1,
visible from the origin at the angle 2πk
2n−1 . It can be reduced to the solving of the
following equation system for the coefficients xi of its generating polynomial:
2n−2∑
i=0
xi = n;
2n−2∑
i=0
x2
i = n;
2n−2∑
i=0
xix|i+k| = dk,
k = 1, 2, . . . , n− 1,
(1)
where x|i+k| is the least nonnegative residue of i + k modulo 2n − 1 [1, p. 59].
The realization of the natural condition
∑n−1
k=1 dk = C2
n is also assumed. The so-
lution of system (1) was obtained in ”parametric” form (see [1, p. 60]):
x0 = 1√
2n−1
(y0 +
√
2
n−1∑
j=1
yj),
xm = 1√
2n−1
[y0 +
√
2
n−1∑
j=1
(yj cos 2πmj
2n−1 + yN−j sin 2πmj
2n−1 )],
xN−m = 1√
2n−1
[y0 +
√
2
n−1∑
j=1
(yj cos 2πmj
2n−1 − yN−j sin 2πmj
2n−1 )],
(2)
where N = 2n− 1, and
y0 =
n√
2n− 1
, y2
j + y2
N−j =
2
2n− 1
(n + 2
n−1∑
k=1
dk cos
2πkj
2n− 1
).
The similar representation is valid for the coefficients x′i of the generating poly-
nomial p′(z) of P ′. The antipodal n-gons P and P ′ generated by polynomials
p(z) =
∑2n−2
k=0 xkz
k and p′(z) =
∑2n−2
k=0 x′kz
k, for which the coefficients are given
by (2) and by the similar formula with y = {y0, y1, . . . , y2n−2} substituted by
y′ = {y′0, y′1, . . . , y′2n−2}, exist if and only if the conditions
y2
j + y2
N−j + y′j
2 + y′ 2N−j =
2n
2n− 1
(3)
are true for all j = 1, 2, . . . , n− 1 (see [1, Lem. 3]).
58 Journal of Mathematical Physics, Analysis, Geometry, 2010, vol. 6, No. 1
Antipodal Polygons and Half-Circulant Hadamard Matrices
Let w̄0, w̄m and w̄N−m denote the right sides of equations (2), and let w̄ =
w̄3
0 +
∑n−1
m=1(w̄
3
m + w̄3
N−m). Consider now a system of equations with respect to
the coordinates of vector y
∂w̄
∂yi
= 3yi, i = 0, 1, 2, . . . , 2n− 2. (4)
Thus, valid is the following statement: antipodal n-gons P and P ′ and, conse-
quently, a corresponding to them Hadamard matrix of order 4n exist if and only if
system (4) has two solutions y = {y0, y1, . . . , y2n−2} and y′ = {y′0, y′1, . . . , y′2n−2}
such that y0 = y′0 = n√
2n−1
, and the rest of the coordinates of vectors y and y′
satisfy antipodal conditions (3) (see [1, Th. 5]). Moreover, the derivatives of (4)
have the form
∂w̄
∂y0
=
3√
2n− 1
2n−2∑
i=0
y2
i ,
∂w̄
∂yk
=
3
√
2√
2n− 1
[
√
2y0yk +
n−k−1∑
j=1
(yjyj+k + yN−jyN−j−k)
+
1
2
k−1∑
r=1−k
(y[(k+r)/2]y[(k−r)/2] − yN−[(k+r)/2]yN−[(k−r)/2])], (5)
∂w̄
∂yN−k
=
3
√
2√
2n− 1
[(
√
2− 1)y0yN−k +
2n−2∑
s=n
(y|s+k|′ − y|s−k|′)ys],
where k = 1, 2, . . . , n− 1, |s± k|′ = min(|s± k|, N − |s± k|), and [(k± r)/2]
equals k±r
2 if it is integer, or N−(k±r)
2 otherwise.
Since w̄ is a homogeneous polynomial of third degree by definition, then
∂w̄
y0
, ∂w̄
yk
, ∂w̄
yN−k
are homogeneous polynomials of second degree, moreover, the last
ones are homogeneous polynomials of first degree relatively to the variables
y1, y2, . . . , yn−1 and yn, yn+1, . . . , y2n−2. As for the expressions for ∂w̄
yk
, they con-
tain no summands with yjys, 0 < j < n, s ≥ n, but they contain quadratic
differences 1
2(y2
[ k
2
]
− y2
N−[ k
2
]
) for r = 0.
By analogy with the homogeneous polynomial w̄ = w̄(y), we introduce a poly-
nomial w̄′ = w̄′30 +
∑n−1
m=1(w̄
′3
m + w̄′3N−m), where w̄′ are right sides of equality (2)
after substitution of the coordinates of vector y = {y0, y1, . . . , y2n−2} by the cor-
responding coordinates of vector y′ = {y′0, y′1, . . . , y′2n−2}. Then the equations,
valid for the coordinates of vectors y and y′ such that antipodal n-gons
as well as a half-circulant Hadamard matrix of order 4n exist, can be represented
Journal of Mathematical Physics, Analysis, Geometry, 2010, vol. 6, No. 1 59
A.I. Medianik
in the form
W̄i = ∂w̄
∂yi
− 3yi = 0, i = 0, 1, 2, . . . , 2n− 2,
W̄ ′
i = ∂w̄′
∂y′i
− 3y′i = 0, i = 0, 1, 2, . . . , 2n− 2,
W̄2n−1 = y0 − n√
2n−1
= 0, W̄ ′
2n−1 = y′0 − n√
2n−1
= 0,
Yj = y2
j + y2
N−j + y′ 2j + y′ 2N−j − 2n
2n−1 = 0,
j = 1, 2, . . . , n− 1.
(6)
Thus, by Theorem 5 [8] for n ≥ 3 there directly follows
Theorem 1. Antipodal convex n-gons and a half-circulant Hadamard matrix
of order 4n exist if and only if there do not exist the polynomials Ai, A′i, A2n−1,
A′2n−1, Bj depending on the variables y0, y1, . . . , y2n−2, y
′
0, . . . , y
′
2n−2 such that the
left sides of equations (6) satisfy the identity
2n−2∑
i=0
(AiW̄i + A′iW̄ ′
i ) + A2n−1W̄2n−1 + A′2n−1W̄
′
2n−1 +
n−1∑
j=1
BjYj ≡ 1.
The obtained results submitted in this section, as shown further, will be used
as a background for applying the inductive approach to the proof of the existence
theorems for antipodal polygons.
3. Basis of Inductive Approach
First, we have to simplify the system of equations (6). For this we find an ex-
pression for the homogeneous polynomial of third degree w̄(y0, y1, . . . , y2n−2).
By definition, w̄(y) equals the sum of right sides of (2) raised to the third power
for all m = 1, 2, . . . , n − 1, where cos 2πmj
2n−1 and sin 2πmj
2n−1 , j = 1, 2, . . . , n − 1,
are used as coefficients. Let us transform its expression by applying the formulas
cosα cosβ = 1
2 [cos(α−β)+cos(α+β)] and sinα sinβ = 1
2 [cos(α−β)−cos(α+β)],
where α and β are expressions of the form 2πmj
2n−1 , and also an identity
1
2
+
n−1∑
j=1
cos
2πmj
2n− 1
≡ 0,
which is valid for all m 6≡ 0 (mod 2n − 1) (it follows directly from the more
general identity (see problem N16 in [9, p. 88])). Substituting twice the products
60 Journal of Mathematical Physics, Analysis, Geometry, 2010, vol. 6, No. 1
Antipodal Polygons and Half-Circulant Hadamard Matrices
of cosines and sines for the sum and the difference, respectively, and changing
the summation order, finally we obtain
w̄ =
1√
2n− 1
[y3
0 + 3y0
2n−2∑
i=1
y2
i +
1√
2
∑
3j=N
(y3
j − 3yjy
2
N−j)
+
3√
2
s6=j∑
s=|2j|′
(y2
j − y2
N−j)ys + 3
√
2
n−1∑
j 6=N/3
(−1)|2j|′yjyN−jyN−|2j|′
+ 3
√
2
s<N−j−s∑
j<s
((yjys − yN−jyN−s)y|j+s|′
+ (−1)j+s+|j+s|′(yjyN−s + ysyN−j)yN−|j+s|′)],
where the value of index j varies from 1 to n − 1. Moreover, N = 2n − 1 and
the sign of absolute value, in particular |2j|′, denotes min(2j,N − 2j). Let us
represent this polynomial as a polynomial of third degree with respect to y0
w̄ =
1√
2n− 1
(y3
0 + 3y0
2n−2∑
i=1
y2
i + 3
√
2w),
where w is a kernel of w̄ not containing variable y0 and having the form
w =
1
2
∑
3j=N
(y3
j /3− yjy
2
N−j) +
n−1∑
j 6=N/3
[
1
2
(y2
j − y2
N−j)y|2j|′
+ (−1)|2j|′yjyN−jyN−|2j|′ ] +
s<N−j−s∑
j<s
[(yjys − yN−jyN−s)y|j+s|′
+ (−1)j+s+|j+s|′(yjyN−s + ysyN−j)yN−|j+s|′ ]. (7)
Notice that if N can not be divided by 3, then the first sum in (7) is absent and
therefore the condition j 6= N/3 in the second sum drops out. It should be added
that if j = s in the last sum, then it practically coincides with the second sum,
and if j = s = |j + s|′, then it coincides also with the first sum.
As follows from (5), the first equation in (6) under i = 0 has the form
3√
2n− 1
2n−2∑
j=0
y2
j − 3y0 = 0.
The first equation of the second group of (6) has a similar form for
i = 0. Substituting into them the known value y0 = y′0 = n√
2n−1
(one of the
Journal of Mathematical Physics, Analysis, Geometry, 2010, vol. 6, No. 1 61
A.I. Medianik
necessary conditions for the existence of antipodal n-gons), we obtain
2n−2∑
i=1
y2
i −
n(n− 1)
2n− 1
= 0,
2n−2∑
i=1
y′2i −
n(n− 1)
2n− 1
= 0.
It is obvious that these two equations and the last n− 1 equations of system (6)
are linearly dependent, since
n−1∑
j=1
(y2
j + y2
N−j + y′j
2 + y′ 2N−j −
2n
2n− 1
) =
2n−2∑
i=1
(y2
i + y′2i )− 2n(n− 1)
2n− 1
.
Thus, one of the first two equations may not be included into transformed system
of equations (6). Without loss of generality, we eliminate the second of them,
containing y′i, as well as y0 = n√
2n−1
and y′0 = n√
2n−1
from the rest of the equations
W̄i, W̄ ′
i , i = 1, 2, . . . , 2n− 2, by using the expressions of (5) for the derivatives of
polynomial w̄. Then (6) takes the form
Y0 = −n(n−1)
2n−1 +
2n−2∑
i=1
y2
i = 0;
Yj = − 2n
2n−1 + y2
j + y2
N−j + y′2j + y′2N−j = 0,
j = 1, 2, . . . , n− 1;
Wi = yi√
4n−2
+ ∂w
∂yi
= 0, i = 1, 2, . . . , 2n− 2;
W ′
i = y′i√
4n−2
+ ∂w′
∂y′i
= 0, i = 1, 2, . . . , 2n− 2,
(8)
where w′ is obtained from w by substituting y′j , y
′
N−j for yj , yN−j and so on.
Evidently that the existence of solution of the system of algebraic equations
(6) leads to the existence of solution of system (8), and conversely, adding the
relations y0 = y′0 = n√
2n−1
to the solution of system (8), we obtain the solution
of system (6). Thereby, the following theorem is valid.
Theorem 2. Antipodal convex n-gons exist if and only if there do
not exist polynomials Bj(y, y′), j = 0, 1, 2, . . . , n − 1, Ci(y, y′) and C ′i(y, y′),
i = 1, 2, . . . , 2n − 2, such that for the left sides of non-homogeneous algebraic
equations (8) the following identity is valid:
L =
n−1∑
j=0
BjYj +
2n−2∑
i=1
(CiWi + C ′iW ′
i ) ≡ 1. (9)
This relation means that after a similar reduction in polynomial L all coeffi-
cients, except a constant term, turn into 0. Thus, a question on the proof of the
62 Journal of Mathematical Physics, Analysis, Geometry, 2010, vol. 6, No. 1
Antipodal Polygons and Half-Circulant Hadamard Matrices
existence of antipodal polygons is reduced to a question on the nonexistence of
such polynomials of any finite degree m, for which L(y, y′) satisfies identity (9).
In this connection, we denote a polynomial in the left side of (9) by Lm, where m
points out its degree. Since the degree of Yj(y, y′), Wi(y),W ′
i (y
′) is two, then the
degree of polynomials Bj(y, y′), Ci(y, y′) and C ′i(y, y′) is m − 2. Let us denote
the coefficients of polynomial Lm(y, y′): K0 is a constant term, Ki is a coefficient
at yi, i = 1, 2, . . . , 2n− 2, Ki2j′ is a coefficient at y2
i y
′
j , etc. By Lw we denote the
following linear combination of the coefficients Lm(y, y′):
Lw =
∑
3j=N
(Kj3 −Kj(N−j)2) +
n−1∑
j 6=N/3
[Kj2|2j|′ −K(N−j)2|2j|′
+ (−1)|2j|′Kj(N−j)(N−|2j|′)] +
s<N−j−s∑
j<s
[Kjs|j+s|′ −K(N−j)(N−s)|j+s|′
+ (−1)j+s+|j+s|′(Kj(N−s)(N−|j+s|′) + Ks(N−j)(N−|j+s|′)]. (10)
By Lw′ we denote the corresponding linear combination of the coefficients with
primes. It is easy to see that Lw includes only coefficients of polynomial Lm
found at its terms coinciding with the monomials of polynomial w and, moreover,
having the same signs as w. Since it is a homogeneous polynomial of third degree,
then, by Euler rule,
∑2n−2
i=1 yj
∂w
∂yi
= 3w. Hence, in a homogeneous polynomial of
second degree ∂w
∂yi
the coefficient sign of the monomial ypyq coincides with the
one of ypyqyi in polynomial w, and therefore, the coefficient sign of ypyqyi (some
of these numbers can be equal each other) of polynomial Lm, m ≥ 3, included
in the combination Lw, coincides with the coefficient sign of ypyq in polynomial
∂w
∂yi
as well as with the coefficient sign of yqyi in polynomial ∂w
∂yp
and the one of
ypyi in polynomial ∂w
∂yq
. The same can said about the linear combination Lw′ .
Let us denote by Lm
w the expression Lw of (10) written via the coefficients of
polynomial Lm.
Lemma 1. If in relation (9) the degree of polynomial L is three, then L3
w
is expressed via the coefficients of polynomials Ci, i = 1, 2, . . . , 2n − 2, in the
following way:
L3
w = (2n− 3)
2n−2∑
i=1
Ci
i , (11)
where Ci
i is a coefficient of yi in Ci.
P r o o f. Since in all equations of system (8) the degree is two, then the
required expression for L3
w a priori may contain only the coefficients of polyno-
mials Bj , j = 0, 1, . . . , n − 1, and Ci, i = 1, 2, . . . , 2n − 2, with the first degree
Journal of Mathematical Physics, Analysis, Geometry, 2010, vol. 6, No. 1 63
A.I. Medianik
terms but actually it contains only the last coefficients. Non of the coefficients
of polynomial Bj with the first degree terms is contained in L3
w, as the term y2
j
is contained in the equality Yj together with the term y(N−j)2 and by (10), the
summands Kj2|2j|′ and K(N−j)2|2j|′ are contained in Lw with opposite signs.
As follows from (7), the derivative ∂w
∂yi
for i = |2j|′ contains the semidifference
y2
j−y2
N−j
2 and the terms of the form ±yαyβ (α 6= β). If the degree of L is three, then
from polynomial Ci, i = 1, 2, . . . , 2n − 2, the summands Ci
i = 1
2sign2(Kj2i)Ci
i +
1
2sign2(K(N−j)2i)Ci
i and Ci
i = sign2(Kαβi)Ci
i (for each admissible pair of indices
α, β) enter into L3
w. Since in the derivative ∂w
∂yi
under i < n there are 2n − 4
summands with coefficients equal to 1 (as seen from (5)) for ∂w̄
∂yk
: 2(n− k− 1) +
2(k − 1) = 2n − 4, one summand with the coefficient +1
2 and one with the
coefficient −1
2 , then Ci
i is contained in L3
w with the coefficient 2n−4+2· 12 = 2n−3.
The same coefficient is for i = N −|2j|′ ≥ n. In this case, as follows from (5), ∂w̄
∂yi
contains 2n−3 summands with coefficients equal to ±1). Moreover, when j runs
the values from 1 to n− 1, then i = |2j|′ runs the same values but in other order,
and i = N − |2j|′ runs all values from n to 2n− 2. Thus, L3
w contains every Ci
i ,
i = 1, 2, . . . , 2n− 2, with the coefficient 2n− 3, as shown in (11).
The conclusion of the lemma will be proved completely if L3
w, after
substituting the coefficients of polynomials Ci from (9), does not contain any
other terms of the form Ct
i , t 6= i. Complement terms can appear only when in
some equation Wt from (8) there is at least one summand ypyq with the coefficient
sign(Kw
pqt), where Kw
pqt is a coefficient at ypyqyt in the polynomial w, which occurs
in another equation Wi with the coefficient sign(Kw
pqi), where Kw
pqi is a coefficient
at ypyqyi in w. Furthermore, p 6= q and p + q 6= N . Otherwise, the term ypyq
could not be contained in two different equations of system (8) and, as follows
from the form of polynomial w of (7), both equations Wt and Wi have the term
yN−pyN−q but with the sign sign(Kw
(N−p)(N−q)i) in the equation Wi and with the
sign sign(Kw
(N−p)(N−q)t) in Wt.
In the last sum of the polynomial w of (7) replace index s by s − j, and
|j + s|′ by |j + (s − j)|′ = s. Taking into account that j + (s − j) =
s < N − s, we may conclude that this sum contains also the summands (yjys−j−
yN−jyN−(s−j))ys + (yjyN−(s−j) + yN−jys−j)yN−s = (yjys + yN−jyN−s)ys−j +
(yjyN−s − yN−jyN−(s−j))yN−(s−j). Hence, the signs of the monomials ypyq and
yN−pyN−q are identical in one of the equations Wt or Wi, and they are opposite in
another one. Thus, in L3
w the term Ct
i (t 6= i) as well as Ci
t is with the coefficient
equal to sign(Kw
pqt) · sign(Kw
pqi) + sign(Kw
(N−p)(N−q)t) · sign(Kw
(N−p)(N−q)i) = 0,
i.e., L3
w actually does not contain Ct
i . If a product of the first two co-factors is 1,
then, by virtue of the proved above, a product of the last two coefficients is −1,
and conversely. Lemma 1 is proved.
64 Journal of Mathematical Physics, Analysis, Geometry, 2010, vol. 6, No. 1
Antipodal Polygons and Half-Circulant Hadamard Matrices
Notice that due to the symmetry of the equations Wi and W ′
i in (8), with
respect to their variables, a similar relation is valid for L3
w′ :
L3
w′ = (2n− 3)
2n−2∑
i=1
C ′i
i′ . (12)
Lemma 2. A constant term K0 of the polynomial L3(y, y′) is a linear com-
bination of its other coefficients, namely,
K0 = − n
4n− 2
2n−2∑
i=1
(Ki2 + Ki′2) +
n(Lw + Lw′)
(2n− 3)(4n− 2)3/2
. (13)
P r o o f. Since the left side of relation (9) is a polynomial of third degree, and
the left sides of equations (8) are polynomials of second degree by the condition
of the lemma, then the polynomials Bj , j = 0, 1, 2, . . . , n− 1, and Ci, C ′i,
i = 1, 2, . . . , 2n− 2, are of first degree. Therefore,
Bj = Bj
0 +
2n−2∑
k=1
(Bj
kyk + Bj
k′y
′
k),
Ci = Ci
0 +
2n−2∑
s=1
(Ci
sys + Ci
s′y
′
s), (14)
C ′i = C ′i
0 +
2n−2∑
s=1
(C ′i
s ys + C ′i
s′y
′
s).
Substituting these expressions into (9) and taking into account expressions
for Yj , j = 0, 1, . . . , n− 1, Wi and W ′
i , i = 1, 2, . . . , 2n− 2, from (8), we can find
K0 = −n(n− 1)
2n− 1
B0
0 −
2n
2n− 1
n−1∑
j=1
Bj
0. (15)
Calculate now the coefficients of the polynomial L3 of the second powers y2
i
and y′2i . Let j = 1, 2, . . . , n− 1. Then
Kj2 = B0
0 + Bj
0 + 1
2C
|2j|′
0 +
Cj
j√
4n−2
,
K(N−j)2 = B0
0 + Bj
0 − 1
2C
|2j|′
0 +
CN−j
N−j√
4n−2
,
Kj′2 = Bj
0 + 1
2C
′|2j|′
0 +
C′j
j′√
4n−2
,
K(N−j)′2 = Bj
0 − 1
2C
′|2j|′
0 +
C′N−j
(N−j)′√
4n−2
.
Journal of Mathematical Physics, Analysis, Geometry, 2010, vol. 6, No. 1 65
A.I. Medianik
By summing up these equations termwise, we obtain
2n−2∑
i=1
(Ki2 + Ki′2) = (2n− 2)B0
0 + 4
n−1∑
j=1
Bj
0 +
2n−2∑
i=1
(Ci
i + C ′i
i′ )
√
4n− 2
.
Eliminating B0
0 and Bj
0 from (15) and applying the last equality, we find
K0 = − n
4n− 2
2n−2∑
i=1
(Ki2 + Ki′2) +
n
(4n− 2)3/2
2n−2∑
i=1
(Ci
i + C ′i
i′ ).
Using Lemma 1, for K0 we obtain the required linear expression via the
coefficients of polynomial L3(y, y′). Lemma 2 is proved.
From the lemma above it follows that in the case when the polynomial L in
relation (9) is of third degree, then its constant term is a linear combination of
its other coefficients turned into 0 by identity (9), and therefore it must turn into
0 too, what contradicts to the same identity (9). To prove that K0 equals null
at any degree of polynomial L, it is possible to use the mathematical induction
principle. For this it is sufficient to verify a possibility of the inductive passage
at least in the case of the third degree polynomial L when its degree becomes
increased by one.
Let the degree of polynomial L of (9) be four. Then the polynomials Bj , Ci
and C ′i of (14) are added the summands of second degree denoted Bj
{2}, C
i
{2}
and Ci
{2}, respectively. Consequently, the right side of equality (13) obtains the
difference ∆2
K0
induced by the new summands
∆2K0 = − n
4n− 2
2n−2∑
i=1
(∆2Ki2 + ∆2Ki′2) +
n(∆2Lw + ∆2Lw′)
(2n− 3)(4n− 2)3/2
. (16)
Substituting new expressions for Bj , Ci, C ′i of (9), we get
∆2Ki2 = −n(n−1)
2n−1 B0
i2 − 2n
2n−1
n−1∑
j=1
Bj
i2
,
∆2Ki′2 = −n(n−1)
2n−1 B0
i′2 − 2n
2n−1
n−1∑
j=1
Bj
i′2 ,
(17)
where B0
i2 and Bj
i2
are coefficients at y2
i in the polynomials B0 and Bj ,
j = 1, 2, . . . , n−1, whereas B0
i′2 , B
j
i′2 are analogous coefficients. To find ∆2Lw we
use (10) for Lw expressed via the coefficients of polynomial L
∆2Lw = {
2n−2∑
i=1
yiC
i
{2}√
4n− 2
}Lw =
2n−2∑
i=1
I⊂Lw∑
i∈I
sign(Kw
I )Ci
I/i
√
4n− 2
=
Cw√
4n− 2
, (18)
66 Journal of Mathematical Physics, Analysis, Geometry, 2010, vol. 6, No. 1
Antipodal Polygons and Half-Circulant Hadamard Matrices
where the braces mean that in product yiC
i
{2} from all possible terms KI of third
degree we can take the coefficients only of the terms that are in the polynomial w,
and with the same sign as in (10). Notice that the subscript in Ci
I/i is two-valued
obtained by eliminating index i from I. Thus, a linear combination Cw in ∆2Lw
is obtained from Lw by raising one by one the subscripts of every summand and
substituting ”K” by ”C”(if in any summand of Lw the index is repeated, then
it can be raised only once). An expression similar to (18) can be obtained for
∆2Lw′ .
Substituting expressions (17) and (18) into (16), setting
B =
n− 1
2
2n−2∑
i=1
(B0
i2 + B0
i′2) +
2n−2∑
i=1
n−1∑
j=1
(Bj
i2
+ Bj
i′2),
we find
∆2K0 =
n2B
(2n− 1)2
+
n(Cw + Cw′)
(2n− 3)(4n− 2)2
, (19)
where Cw′ is determined by analogy with Cw (see equality (18)).
Representation (13) for the constant term K0 given by (15), obtained under
supposition that the degree m of the polynomial L(y, y′) is three, when passing
to m = 4 has the following form:
K0 = − n
4n− 2
2n−2∑
i=1
(Ki2 + Ki′2) +
n(Lw + Lw′)
(2n− 3)(4n− 2)3/2
−∆2K0, (20)
where the first two groups of summands coincide with those of (13) in form,
but now they belong to the polynomial L4(y, y′), and ∆2K0 (19) is expressed
via the coefficients of additional summands of the second degree of polynomials
Bj , Ci, C ′i of (14). To find the representation ∆2K0 via the coefficients of the
polynomial L4(y, y′) we have to prove the following lemma.
Lemma 3. If the degree of the polynomial L in relation (9) is four, then for
its coefficients there are valid the following equalities:
[[2
2n−2∑
i=1
Ki4 +
n−1∑
s=1
(Ks4 + K(N−s)4 + Ks2(N−s)2)]]
−2
2n−2∑
i=1
(Ki2i′2 + Ki2(N−i)′2) +
4
n− 1
2n−2∑
p,q=1
Kp2q′2 =
8B
n− 1
+ C, (21)
[[2
2n−2∑
i=1
Ki4 +
∑
p≤q
Kp2q2 ]]−
2n−2∑
i=1
(Ki2i′2 + Ki2(N−i)′2)
Journal of Mathematical Physics, Analysis, Geometry, 2010, vol. 6, No. 1 67
A.I. Medianik
+
n + 1
n− 1
2n−2∑
p,q=1
Kp2q′2 =
4nB
n− 1
+ Cw + Cw′ , (22)
where B = n−1
2
2n−2∑
i=1
(B0
i2 +B0
i′2)+
2n−2∑
i=1
n−1∑
j=1
(Bj
i2
+Bj
i′2), C = [[
n−1∑
s=1
(C |2s|′
s2 −C
|2s|′
(N−s)2
+
(−1)|2s|′CN−|2s|′
s(N−s) ]], Cw =
I⊂Lw∑
i∈I
sign(Kw
I )Ci
I/i, Cw′ =
I⊂Lw∑
i∈I
sign(Kw
I )C ′i
(I/i)′ , and
the expression enclosed within double square brackets should be added the same
expression with the subscripts i′, s′, (N − s)′, p′, q′.
P r o o f. All coefficients of L in relations (21) and (22) are coefficients at the
products of monomials of second degree from Bj
{2}, C
i
{2}, C
′i
{2} and monomials of
second degree in the left sides of the equations of system (8): {B0
{2}
∑2n−2
i=1 y2
i +∑n−1
j=1 Bj
{2}(y
2
j + y2
N−j + y′2j + y′2N−j) +
∑2n−2
i=1 (Ci
{2}
∂w
∂yi
+ C ′i
{2})
∂w′
∂y′i
. Notice that
not all these products are considered but only those the first co-factors of which
have the coefficients Bj
i2
(Bj
i′2), or Ci
αβ(C ′i
α′β′), where α and β are such that there
is a monomial of the form yαyβyi in w as well as in w′. We have for s < n:
Ks4 =B0
s2 + Bs
s2 +
1
2
C
|2s|′
s2 ,
K(N−s)4 =B0
(N−s)2 + Bs
(N−s)2 −
1
2
C
|2s|′
(N−s)2
,
Ks2(N−s)2 =B0
s2 + B0
(N−s)2 + Bs
s2 + Bs
(N−s)2 (23)
− 1
2
(C |2s|′
s2 − C
|2s|′
(N−s)2
) + (−1)|2s|′CN−|2s|′
s(N−s) .
By summing up these equalities termwise over s from 1 to n− 1, we have
n−1∑
s=1
(Ks4 + K(N−s)4 + Ks2(N−s)2)
= 2
2n−2∑
i=1
B0
i2 + 2
n−1∑
s=1
(Bs
s2 + Bs
(N−s)2) +
n−1∑
s=1
(−1)|2s|′CN−|2s|′
s(N−s) .
(24)
Since there are no summands of the form y′2i in the left side of the first
equation of system (8), by analogy we obtain
n−1∑
s=1
(Ks′4 + K(N−s)′4 + Ks′2(N−s)′2)
= 2
n−1∑
s=1
(Bs
s′2 + Bs
(N−s)′2) +
n−1∑
s=1
(−1)|2s|′C ′N−|2s|′
s′(N−s)′ .
(25)
68 Journal of Mathematical Physics, Analysis, Geometry, 2010, vol. 6, No. 1
Antipodal Polygons and Half-Circulant Hadamard Matrices
Further we have
Ks2s′2 = B0
s′2 + Bs
s′2 + Bs
s2 +
1
2
(C |2s|′
s′2 + C
′|2s|′
s2 ),
K(N−s)2(N−s)′2 = B0
(N−s)′2 + Bs
(N−s)′2 + Bs
(N−s)2 −
1
2
(C |2s|′
(N−s)′2 + C
′|2s|′
(N−s)2
),
Ks2(N−s)′2 = B0
(N−s)′2 + Bs
(N−s)′2 + Bs
s2 +
1
2
(C |2s|′
(N−s)′2 − C
′|2s|′
s2 ),
K(N−s)2s′2 = B0
s′2 + Bs
s′2 + Bs
(N−s)2 −
1
2
(C |2s|′
s′2 − C
′|2s|′
(N−s)2
).
By summing up these equalities termwise over s from 1 to n− 1, we get
2n−2∑
i=1
(Ki2i′2 + Ki2(N−i)′2) = 2
2n−2∑
i=1
B0
i′2 + 2
n−1∑
s=1
(Bs
s2 + Bs
s′2 + Bs
(N−s)2 + Bs
(N−s)′2).
(26)
Substituting the values Ks4 and K(N−s)4 from (23) into the first summand of
equation (21) with regard to relations (24–26), we obtain
[[2
2n−2∑
i=1
Ki4 +
n−1∑
s=1
(Ks4 + K(N−s)4 + Ks2(N−s)2)]]
−2
2n−2∑
i=1
(Ki2i′2 + Ki2(N−i)′2) = 4
2n−2∑
i=1
(B0
i2 −B0
i′2) + C. (27)
To establish the first conclusion of the lemma we have to find the sum∑
p<q
Kp2q′2 , taking into account that under 1 ≤ p < q < N
Kp2q′2 = B0
q′2 + B
|p|′
q′2 + B
|p|′
p2 +
1
2
[(−1)p+|p|′C |2p|′
q′2 + (−1)q+|q|′C |2q|′
p2 ],
where |p|′ = min(p,N − p), |2p|′ = min(|2p|, N − |2p|), and |2p| is the smallest
positive residue by modulus 2n− 1
∑
p<q
Kp2q′2 = (2n− 2)
2n−2∑
i=1
B0
i′2 + 2
2n−2∑
i=1
n−1∑
j=1
(Bj
i2
+ Bj
i′2). (28)
Multiplying this equality by 4
n−1 and adding termwise to equality (27), we
obtain relation (21) of the lemma.
To prove the second conclusion of the lemma, first we have to find
p+q 6=N∑
p<q
Kp2q2 ,
where 1 ≤ p < q < N (the cases of p = q and p+ q = N are in (23)). If p < q < n
or n ≤ p < q, then
Kp2q2 = B0
p2 + B0
q2 + B
|p|′
q2 + B
|q|′
p2 +
1
2
[(−1)p+|p|′C |2p|′
q2
Journal of Mathematical Physics, Analysis, Geometry, 2010, vol. 6, No. 1 69
A.I. Medianik
+(−1)q+|q|′C |2q|′
p2 ] + sign(Kw
pq|q−p|′)C
|q−p|′
pq + sign(Kw
pq|q+p|′)C
|q+p|′
pq . (29)
Thus, if p < n ≤ q, then
Kp2q2 = B0
p2 + B0
q2 + B
|p|′
q2 + B
|q|′
p2 +
1
2
[(−1)p+|p|′C |2p|′
q2 + (−1)q+|q|′C |2q|′
p2 ]
+sign(Kw
pq(N−|q−p|′))C
N−|q−p|′
pq + sign(Kw
pq(N−|q+p|′))C
N−|q+p|′
pq . (30)
There are similar representations for Kp′2q′2 . Using equalities (23, 26, 29, 30),
we obtain
[[2
2n−2∑
i=1
Ki4 +
2n−2∑
p≤q
Kp2q2 ]]−
2n−2∑
i=1
(Ki2i′2 + Ki2(N−i)′2)
= 2n
2n−2∑
i=1
B0
i2 − 2
2n−2∑
i=1
B0
i′2 + 2
2n−2∑
i=1
n−1∑
j=1
(Bj
i2
+ Bj
i′2) + Cw + Cw′ . (31)
Multiplying equation (28) by n−1
n+1 and adding to (31), we obtain the required
relation (22). Lemma 3 is proved.
As one can see, to find an expression for ∆2K0 (19) via the polynomial L4,
relations (21) and (22) are insufficient. There should be found a relation
between the sum of the differences Cw + Cw′ and C, for which we will define
a new linear combination L2w of the coefficients of polynomial L4 constructed
by the linear combination Lw. Namely, if Lw contains the coefficients Kαβi
and Kiγδ, then L2w contains the coefficient Kαβγδ with the sign: sign(Kαβγδ) =
sign(Kαβi) · sign(Kiγδ). Moreover, by definition, L2w contains only one term with
given indices αβγδ and no one of the form Kα2γ2 or Kαβ(N−α)(N−β). The sign of
Kαβγδ does not depend on permutation of indices α, β, γ, δ, what can be shown
by means of arguments used above (mainly in the proof of Lemma 1). A linear
combination L2w′ is defined in a similar way.
Lemma 4. For the linear combinations L2w and L2w′ of the coefficients of
polynomial L4(y, y′) there is true the equality
L2w + L2w′ = (2n− 5)(Cw + Cw′) + C, (32)
where Cw, Cw′ and C have the same values as in Lemma 3.
P r o o f. To prove this lemma, it is not necessary first to find the linear
combination Lw and then to define the above combination L2w. It is sufficient to
use the system of equations (8) having the derivatives ∂w
∂yi
, i = 1, 2, . . . , 2n − 2.
In fact, any monomial ypyq contained in the derivative ∂w
∂yi
has sign(Kw
ipq) which
coincides with the sign of the corresponding monomial in the polynomial w (7)
70 Journal of Mathematical Physics, Analysis, Geometry, 2010, vol. 6, No. 1
Antipodal Polygons and Half-Circulant Hadamard Matrices
(if w does not contain the monomial yiypyq, we suppose sign(Kw
ipq) = 0). Hence,
the form of L2w expressed via Cw and C can be determined by the coefficients of
the fourth degree polynomial
∑2n−2
i=1 Ci
{2}
∂w
∂yi
, where Ci
{2} is a quadratic part of
the polynomial Ci of (9).
Let i = |2j|′, j < n. Then, in the i-equation of system (8) the deriva-
tive ∂w
∂yi
contains 2n − 4 summands with coefficients equal to ±1 and the semi-
difference 1
2(y2
j − y2
N−j). Moreover, there is a monomial yN−pyN−q along with
ypyq, p < q < n or n ≤ p < q, and by definition there is neither summand
Kpq(N−p)(N−q) nor Kj2(N−j)2 in L2w. Thus, if L2w contains the summand Kpqrs,
then it contains Ci
pq with the sign sign(Kw
irs) ·sign(Kpqrs) = sign(Kw
ipq), as by defi-
nition sign(Kpqrs) = sign(Kw
ipq) · sign(Kw
irs), and the coefficient at sign(Kw
ipq)C
i
pq
is (2n− 4)− 2 + 2 · 1
2 = 2n− 5. Besides, L2w contains sign(Kw
ij2)Ci
j2 = C
|2j|′
j2 and
sign(Kw
i(N−j)2)C
i
(N−j)2 = −C
|2j|′
(N−j)2
(see the first two summands of the second
sum in (7)) with the coefficient 2n− 4 = (2n− 5) + 1. Thus, when i = |2j|′ < n,
L2w contains
(2n− 5)[
n−1∑
i=1
∑
p<q
sign(Kw
ipq)C
i
pq +
n−1∑
j=1
(C |2j|′
j2 − C
|2j|′
(N−j)2
)] +
n−1∑
j=1
(C |2j|′
j2 − C
|2j|′
(N−j)2
)
as summands. When i = N − |2j|′ ≥ n, the derivative ∂w
∂yi
contains 2n − 3
summands with the coefficients ±1. What is more, it contains the summand ypyq,
p < n ≤ q, when p + q 6= N , as well as the summand yN−pyN−q and the term
yjyN−j . Therefore, in L2w there is a summand sign(Kw
ipq)C
i
pq with the coefficient
(2n−4)−2+1 = 2n−5 and a summand sign(Kw
ij(N−j))C
i
j(N−j) = (−1)|2j|′C |2j|′
j(N−j)
(see the last summand of the second sum in (7)) with the coefficient 2n − 4 =
(2n− 5) + 1 if sign(Kw
ij(N−j)) 6= 0. Thus, when i ≥ n, L2w contains the following
summands:
(2n− 5)[
2n−2∑
i=n
∑
p<q
sign(Kw
ipq)C
i
pq +
n−1∑
j=1
(−1)|2j|′C |2j|′
j(N−j))] +
n−1∑
j=1
(−1)|2j|′C |2j|′
j(N−j)).
Summing up the obtained relations termwise and taking into account that
L2w′ satisfies similar equalities, we arrive to equality (32). Lemma 4 is proved.
As we can see, by Lemmas 3 and 4, the expressions B,Cw Cw′ , contained in the
difference ∆2K0 (19), are combinations of the coefficients L4(y, y′) at monomials
of fourth degree. Thus, by (20), the constant term K0 of the polynomial L4 is
a linear combination of its other coefficients equal to null by (9), and K0 must
turn into null too, what contradicts to (9). Hence, a degree of the polynomial
L(y, y′) of (9) should be greater than four.
Journal of Mathematical Physics, Analysis, Geometry, 2010, vol. 6, No. 1 71
A.I. Medianik
Our method of constructing of half-circulant Hadamard matrices of order 4n
(see Basic Lemma in [4]) is universal, i.e., it can be applied to any integer n,
and, consequently, universal is the initial equation system (6) for the proposed
inductive approach (notice that the well-known method used by J. Williamson
[10] works only for odd n). Hence, the obtained above results give an opportunity
to use the inductive approach to the proof of the existence of antipodal n-gons
for any integer n ≥ 3 as well as of half-circulant Hadamard matrices of order 4n.
Thus, there is valid the following theorem.
Therem 3. If for any natural m ≥ 3 the assumption that a constant term
of the polynomial Lm(y, y′) defined by (9) is a linear combination of its other
coefficients involves the same assertion for the polynomial Lm+1, then there exists
a half-circulant Hadamard matrix of any order 4n, n ≥ 3.
Notice that (by analogy with the considered above passage from degree m = 3
to degree m + 1 = 4 of polynomial L) to prove the specified in Theorem 3
properties of the polynomial L of any finite degree m + 1, it is sufficient to show
that the difference ∆m−1K0 obtained by replacing the degree m by m + 1 of L,
can be expressed linearly via the coefficients of Lm+1 at terms of degree (m + 1).
References
[1] A.I. Medianik, Antipodal n-Gons Inscribed into a Regular (2n − 1)-Gon and the
Half-Circulant Hadamard Matrices of Order 4n. — Mat. Fiz., Anal., Geom. 11
(2004), 45–66. (Russian)
[2] M. Hall, Combinatorial Theory. Mir, Moscow, 1970. (Russian)
[3] W. Ball and H. Coxeter, Mathematical Recreations and Essays. Mir, Moscow,
1986. (Russian)
[4] A.I. Medianik, Regular Simplex Inscribed into a Cube and Hadamard Matrices of
Half-Circulant Type. — Mat. Fiz., Anal., Geom. 4 (1997), 458–471. (Russian)
[5] R. Bellman, Introduction to Matrix Analysis. Nauka, Moscow, 1969. (Russian)
[6] H. Khraghani and B. Tayfeh-Rezaie, A Hadamard Matrix of Order 428. —
J. Combin. Designs 13 (2005), 435–440.
[7] K. Sawade, A Hadamard Matrix of Order 268. — Graphs Combin. 1 (1985), No. 2,
185–187.
[8] A.I. Medianik, On a Regular Hypersimplex Insrribed into the Multidimensional
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[9] G. Polya and G. Szego, Aufgaben und Lehrsatze aus der Analysis. Band II. Nauka,
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