Ann.
Acad.
Rom.
Sci.
Ser.
Math.
Appl.
ISSN
2066-6594
Vol.
18,
No.
4/2026
INVERTIBLE
CHEBYSHEV–BINOMIAL
TRANSFORMATIONS
∗
Mircea
Merca
†
Dedicated
to
Professor
Constantin
Fetecau
on
the
occasion
of
his
80th
anniversary
DOI 10.56082/annalsarscimath.2026.4.149
Abstract
We
show
that
explicit
expansions
of
the
Chebyshev
polynomials
of
the
first
kind
naturally
induce
several
invertible
triangular
transfor-
mations
involving
classical
binomial
arrays
and
closely
related
ballot-
number
arrays.
Using
a
unified
coefficient-extraction
method,
we
de-
rive
direct
and
inverse
formulas
relating
the
coefficients
2
n
n
−
k
and
the
ballot-number
differences
2
n
n
+
k
−
2
n
n
+
k
+1
to
the
sequences
n
k
2
n
−
k
and
n
k
4
n
−
k
.
The
resulting
identities
define
explicit
mutually
inverse
triangular
transformations.
The
closed
forms
appearing
on
their
right-
hand
sides
admit
natural
combinatorial
interpretations
in
terms
of
dis-
joint
subsets
and
block
decompositions.
We
also
introduce
a
signed
refinement
of
the
classical
enumeration
of
pairs
of
disjoint
subsets,
which
yields
an
additional
Chebyshev–binomial
transformation
and
its
explicit
inverse.
Keywords:
Chebyshev
polynomials,
symmetric
Laurent
polynomials,
binomial
transformations,
combinatorial
interpretations.
MSC:
055A19,
11B65,
33C45,
05A15.
∗
Accepted
for
publication
on
July
17,
2026
†
mircea.merca@upb.ro
,
Department
of
Mathematical
Models
and
Methods,
Funda-
mental
Sciences
Applied
in
Engineering
Research
Center,
National
University
of
Science
and
Technology
Politehnica
Bucharest,
060042
Bucharest,
Romania
and
Academy
of
Ro-
manian
Scientists,
050044
Bucharest,
Romania
149
Invertible
Chebyshev–binomial
transformations
150
1
Introduction
Let
n
and
k
be
nonnegative
integers
with
n
≥
k
.
The
quantity
n
k
2
n
−
k
admits
several
well-known
combinatorial
interpretations.
A
classical
interpretation
is
that
it
counts
the
pairs
(
A,
B
)
of
disjoint
subsets
of
an
n
-set
X
satisfying
|
A
|
=
k
.
Indeed,
one
first
chooses
the
subset
A
in
n
k
ways,
and
then
each
of
the
remaining
n
−
k
elements
may
either
belong
to
B
or
remain
outside
A
∪
B
,
giving
2
n
−
k
possibilities.
A
second
interpretation
arises
from
perfect
matchings
[4].
Let
X
=
X
1
X
2
·
·
·
X
n
be
a
partition
of
a
2
n
-set
into
2-element
blocks.
Then
n
k
2
n
−
k
is
precisely
the
number
of
(
n
+
k
)-subsets
of
X
intersecting
every
block
X
i
.
Indeed,
such
a
subset
must
contain
at
least
one
element
from
each
block,
and
exactly
k
blocks
contribute
both
of
their
elements.
The
same
quantity
also
appears
in
the
classical
identity
n
i
=
k
i
k

n
i
=
n
k
2
n
−
k
,
which
follows
immediately
from
the
elementary
relation
i
k

n
i
=
n
k

n
−
k
i
−
k
and
the
binomial
theorem.
The
binomial
coefficients
2
n
n
−
k
count
walks
of
length
2
n
on
the
integer
line,
starting
at
the
origin,
with
steps
+1
and
−
1,
and
ending
at
2
k
.
Consequently,
z
−
n
(1
+
z
)
2
n
=
n
k
=
−
n
2
n
n
−
k
z
k
M.
Merca
151
is
the
corresponding
generating
function.
The
main
idea
of
this
paper
is
to
work
in
the
space
of
symmetric
Laurent
polynomials
A
sym
:=
{
f
(
z
)
∈
C
[
z,
z
−
1
]
:
f
(
z
)
=
f
(
z
−
1
)
}
.
Every
element
of
this
space
can
be
written
uniquely
as
c
0
+
N
d
=1
c
d
z
d
+
z
−
d
(cf.
[5,
p.
2]).
Hence,
{
1
}
∪
{
z
d
+
z
−
d
}
d
≥
1
is
its
natural
monomial
basis.
Upon
setting
u
=
z
+
z
−
1
,
we
have
z
d
+
z
−
d
=
2
T
d
u
2
,
d
≥
1
.
Thus,
the
transition
from
the
monomial
symmetric
Laurent
basis
to
the
ordinary
monomial
basis
in
u
is
governed
by
the
coefficients
of
the
scaled
Chebyshev
polynomials
2
T
d
(
u/
2).
Combined
with
coefficient
extraction,
this
transition
induces
an
invertible
triangular
transformation
between
2
n
n
−
k
and
n
k
2
n
−
k
.
The
direct
transformation
is
given
in
Theorem
1,
while
its
inverse
appears
in
Corollary
2.
Explicit
representations
of
Chebyshev
polynomials
and
the
correspond-
ing
connection
coefficients
between
the
Chebyshev
and
monomial
bases
are
classical;
see,
for
example,
[6,
7].
These
connections
have
been
used
in
sev-
eral
contexts
involving
binomial
sums.
In
particular,
Cigler
[3]
studied
the
relation
between
the
moments
of
Chebyshev
polynomials
and
central
bino-
mial
coefficients,
while
Bacher
[2]
obtained
a
Pascal-type
triangular
array
from
identities
involving
Chebyshev
polynomials.
More
recently,
Adegoke,
Frontczak
and
Goy
[1]
used
explicit
Chebyshev
representations
to
derive
binomial
sums
involving
Fibonacci
and
Lucas
numbers.
The
present
work
follows
a
different
direction.
Starting
from
the
substi-
tutions
u
=
z
+
z
−
1
and
v
=
z
−
z
−
1
,
Invertible
Chebyshev–binomial
transformations
152
we
use
the
resulting
Chebyshev
identities
as
explicit
connection
formulas
that
rewrite
the
symmetric
and
antisymmetric
Laurent
expansions
arising
below
as
polynomial
expansions
in
u
or
v
.
We
then
combine
these
formu-
las
systematically
with
coefficient
extraction.
This
procedure
yields
three
explicit
pairs
of
mutually
inverse
triangular
transformations.
They
relate,
respectively,
2
n
n
−
k
to
n
k
2
n
−
k
,
2
n
n
−
k
to
n
k
4
n
−
k
,
and
2
n
n
−
k
−
2
n
n
−
k
−
1
to
n
k
4
n
−
k
.
To
the
best
of
our
knowledge,
these
particular
pairs
of
mutually
inverse
triangular
transformations
have
not
been
recorded
explicitly
in
the
forms
presented
here.
Thus,
our
contribution
lies
not
in
the
underlying
Chebyshev
expansions,
which
are
classical,
but
in
using
a
common
Laurent
coefficient-
extraction
framework
to
derive
three
pairs
of
mutually
inverse
triangular
transformations
and
to
connect
their
closed
forms
with
natural
combinato-
rial
interpretations.
We
further
show
that
the
same
mechanism
extends
to
a
signed
refinement
of
the
first
transformation.
Theorem
1.
For
every
n
≥
k
≥
0
,
2
n
n
−
k
+
(
n
−
k
)
/
2
m
=1
(
−
1)
m
k
+
2
m
k
+
m
k
+
m
m

2
n
n
−
k
−
2
m
=
n
k
2
n
−
k
.
The
direct
transformation
of
Theorem
1
admits
the
following
inverse
form.
Corollary
2.
For
every
n
≥
k
≥
0
,
2
n
n
−
k
=
(
n
−
k
)
/
2
m
=0
k
+
2
m
m

n
k
+
2
m
2
n
−
k
−
2
m
.
For
d
≥
1,
the
coefficients
appearing
in
Theorem
1
are
precisely
those
occurring
in
the
classical
explicit
expansion
[6,
Chap.
2]
2
T
d
x
2
=
d/
2
m
=0
(
−
1)
m
d
d
−
m
d
−
m
m
x
d
−
2
m
(1)
M.
Merca
153
of
the
Chebyshev
polynomials
of
the
first
kind.
In
fact,
Theorem
1
is
only
the
first
member
of
a
broader
family
of
Chebyshev–binomial
identities.
The
even-indexed
and
odd-indexed
Cheby-
shev
families
give
rise
to
analogous
summation
formulas
whose
evaluations
are
remarkably
simple
and
involve
the
factor
n
k
4
n
−
k
.
We
state
two
repre-
sentative
examples
below.
Theorem
3.
For
every
n
≥
k
≥
0
,
2
n
n
−
k
+
n
−
k
m
=1
2
k
+
2
m
2
k
+
m
2
k
+
m
m

2
n
n
−
k
−
m
=
n
k
4
n
−
k
.
As
in
the
case
of
Theorem
1,
the
transformation
of
Theorem
3
also
admits
an
explicit
inverse.
Corollary
4.
For
every
n
≥
k
≥
0
,
2
n
n
−
k
=
n
−
k
m
=0
(
−
1)
m
2
k
+
2
m
m

n
k
+
m
4
n
−
k
−
m
.
Theorem
3
is
induced
by
the
even-indexed
Chebyshev
polynomials
2
T
2
r
iv
2
,
whereas
Theorem
5
arises
from
the
odd-indexed
polynomials
2
T
2
r
+1
iv
2
,
v
=
z
−
z
−
1
.
The
imaginary
argument
converts
the
alternating
coefficients
in
the
standard
Chebyshev
expansions
into
the
positive
coefficients
occurring
in
Lemmas
7
and
8.
Theorem
5.
For
every
n
≥
k
≥
0
,
n
−
k
m
=0
(2
k
+
2
m
+
1)
2
(2
k
+
m
+
1)(
n
+
k
+
m
+
1)
2
k
+
m
+
1
m

2
n
n
+
k
+
m
=
n
k
4
n
−
k
.
Theorem
5
is
accompanied
by
the
following
inverse
transformation.
Corollary
6.
For
every
n
≥
k
≥
0
,
2
k
+
1
n
+
k
+
1
2
n
n
+
k
=
n
−
k
m
=0
(
−
1)
m
2
k
+
2
m
+
1
m

n
k
+
m
4
n
−
k
−
m
.
Invertible
Chebyshev–binomial
transformations
154
The
common
value
n
k
4
n
−
k
in
Theorems
3
and
5
has
an
independent
combinatorial
interpretation.
If
a
4
n
-element
set
is
partitioned
into
n
blocks
of
size
4,
then
this
quantity
counts
the
(
n
+
3
k
)-subsets
that
intersect
every
block
and
contain
exactly
k
complete
blocks.
One
chooses
the
complete
blocks
in
n
k
ways
and
one
element
from
each
remaining
block
in
4
n
−
k
ways.
The
remainder
of
the
paper
is
organized
as
follows.
Section
2
proves
The-
orem
1
by
combining
the
classical
expansion
of
the
Chebyshev
polynomials
with
a
coefficient-extraction
argument.
Sections
3
and
4
establish
Theo-
rems
3
and
5,
respectively,
by
deriving
analogous
expansions
for
the
even
and
odd
Chebyshev
families
and
applying
the
same
coefficient-extraction
principle.
In
Section
5
we
introduce
a
signed
refinement
of
the
classical
enu-
meration
of
pairs
of
disjoint
subsets.
This
refinement
yields
a
combinatorial
interpretation
of
n
−
2
k
n
n
k
2
n
−
k
,
and,
through
the
same
Laurent
coefficient-extraction
mechanism,
yields
an
additional
pair
of
mutually
inverse
Chebyshev–binomial
transformations.
2
First
Chebyshev–binomial
identity
In
this
section,
we
prove
Theorem
1
by
expressing
the
elements
of
the
stan-
dard
monomial
symmetric
Laurent
basis
{
1
}
∪
{
z
d
+
z
−
d
}
d
≥
1
as
scaled
Chebyshev
polynomials
of
the
first
kind
in
u
=
z
+
z
−
1
.
The
explicit
expansion
of
these
polynomials
provides
the
coefficients
occurring
in
the
left-hand
side
of
Theorem
1,
while
a
coefficient-extraction
argument
yields
the
desired
closed
form.
We
conclude
the
section
by
proving
the
inverse
transformation
stated
in
Corollary
2.
Proof
of
Theorem
1
.
Let
T
d
(
x
)
denote
the
Chebyshev
polynomial
of
the
first
kind,
defined
by
T
d
(cos
θ
)
=
cos(
dθ
)
.
Equivalently,
if
u
=
z
+
z
−
1
,
M.
Merca
155
then
z
d
+
z
−
d
=
2
T
d
u
2
.
We
shall
use
the
classical
explicit
expansion
of
the
Chebyshev
polyno-
mials
of
the
first
kind
(1).
Consequently,
for
k
≥
1,
m
≥
0,
and
d
=
k
+
2
m
,
the
coefficient
of
u
k
in
z
d
+
z
−
d
is
[
u
k
]
z
k
+2
m
+
z
−
(
k
+2
m
)
=
(
−
1)
m
k
+
2
m
k
+
m
k
+
m
m
.
Now
observe
that
z
−
n
(1
+
z
)
2
n
=
(
z
+
2
+
z
−
1
)
n
=
(
u
+
2)
n
.
On
the
other
hand,
z
−
n
(1
+
z
)
2
n
=
n
d
=
−
n
2
n
n
+
d
z
d
.
Grouping
the
terms
corresponding
to
d
and
−
d
,
we
obtain
(
u
+
2)
n
=
2
n
n
+
n
d
=1
2
n
n
−
d
z
d
+
z
−
d
.
Taking
the
coefficient
of
u
k
,
with
k
≥
1,
gives
[
u
k
](
u
+
2)
n
=
k
≤
d
≤
n
d
≡
k
(mod
2)
2
n
n
−
d
[
u
k
](
z
d
+
z
−
d
)
.
Writing
d
=
k
+
2
m
,
we
obtain
[
u
k
](
u
+
2)
n
=
(
n
−
k
)
/
2
m
=0
(
−
1)
m
k
+
2
m
k
+
m
k
+
m
m

2
n
n
−
k
−
2
m
.
Since
[
u
k
](
u
+
2)
n
=
n
k
2
n
−
k
,
the
desired
identity
follows
for
k
≥
1.
It
remains
to
consider
k
=
0.
From
(
u
+
2)
n
=
2
n
n
+
n
d
=1
2
n
n
−
d
(
z
d
+
z
−
d
)
,
Invertible
Chebyshev–binomial
transformations
156
only
the
terms
with
even
d
contribute
to
the
constant
term
in
u
.
For
d
=
2
m
,
m
≥
1,
the
Chebyshev
expansion
gives
[
u
0
](
z
2
m
+
z
−
2
m
)
=
2(
−
1)
m
.
Therefore,
[
u
0
](
u
+
2)
n
=
2
n
n
+
2
n/
2
m
=1
(
−
1)
m
2
n
n
−
2
m
.
On
the
other
hand,
[
u
0
](
u
+
2)
n
=
2
n
.
Hence
2
n
n
+
2
n/
2
m
=1
(
−
1)
m
2
n
n
−
2
m
=
2
n
,
which
is
precisely
the
case
k
=
0
of
Theorem
1.
This
completes
the
proof.
Proof
of
Corollary
2
.
Using
u
=
z
+
z
−
1
,
we
have
(
u
+
2)
n
=
(
z
+
z
−
1
+
2)
n
=
z
−
n
(1
+
z
)
2
n
.
Therefore,
[
z
k
](
u
+
2)
n
=
[
z
n
+
k
](1
+
z
)
2
n
=
2
n
n
+
k
=
2
n
n
−
k
.
On
the
other
hand,
(
u
+
2)
n
=
n
r
=0
n
r
2
n
−
r
u
r
,
and
u
r
=
(
z
+
z
−
1
)
r
=
r
j
=0
r
j
z
r
−
2
j
.
Hence
the
coefficient
of
z
k
in
u
r
is
nonzero
only
when
r
≡
k
(mod
2)
and
r
≥
k.
Writing
r
=
k
+
2
m,
M.
Merca
157
we
obtain
[
z
k
]
u
k
+2
m
=
k
+
2
m
m
.
Consequently,
(
u
+
2)
n
=
(
n
−
k
)
/
2
m
=0
n
k
+
2
m
2
n
−
k
−
2
m
[
z
k
]
u
k
+2
m
=
(
n
−
k
)
/
2
m
=0
k
+
2
m
m

n
k
+
2
m
2
n
−
k
−
2
m
.
Comparing
this
expression
with
[
z
k
](
u
+
2)
n
=
2
n
n
−
k
gives
the
desired
identity.
For
fixed
n
,
define
a
k
:=
2
n
n
−
k
,
b
k
:=
n
k
2
n
−
k
,
0
≤
k
≤
n.
Theorem
1
has
the
form
b
k
=
n
d
=
k
d
≡
k
(mod
2)
C
k,d
a
d
,
with
C
k,k
=
1
.
Consequently,
after
arranging
the
even
and
odd
indices
separately,
the
co-
efficient
matrix
becomes
block
diagonal,
and
each
diagonal
block
is
upper
unitriangular.
It
is
therefore
invertible.
Equivalently,
in
the
natural
order-
ing
0
,
1
,
.
.
.
,
n
,
the
full
matrix
is
upper
triangular
with
unit
diagonal
and
with
zero
entries
whenever
d
≡
k
(mod
2).
Corollary
2
provides
the
inverse
transformation
explicitly.
Thus,
the
di-
rect
and
inverse
identities
form
a
pair
of
mutually
inverse
triangular
trans-
formations
between

2
n
n
−
k

n
k
=0
and

n
k
2
n
−
k
n
k
=0
.
Invertible
Chebyshev–binomial
transformations
158
3
Even
Chebyshev
family
As
in
Section
2,
the
expansion
used
below
is
induced
by
the
Chebyshev
polynomials
of
the
first
kind,
now
evaluated
at
an
imaginary
argument.
Indeed,
from
2
T
d
w
+
w
−
1
2
=
w
d
+
w
−
d
,
upon
setting
w
=
iz
and
v
=
z
−
z
−
1
,
we
obtain
2
T
d
iv
2
=
(
iz
)
d
+
(
iz
)
−
d
.
In
particular,
z
2
r
+
z
−
2
r
=
(
−
1)
r
2
T
2
r
iv
2
.
Thus,
the
coefficients
occurring
in
the
expansion
of
z
2
r
+
z
−
2
r
in
powers
of
v
2
are
obtained
from
the
even-indexed
Chebyshev
polynomials
of
the
first
kind.
We
first
derive
the
corresponding
explicit
expansion.
It
is
then
combined
with
the
identity
(
z
+
z
−
1
)
2
n
=
(
v
2
+
4)
n
to
obtain
Theorem
3
by
coefficient
extraction.
Lemma
7.
Let
v
=
z
−
z
−
1
.
For
every
r
≥
0
,
we
have
z
2
r
+
z
−
2
r
=
2
,
r
=
0
,
r
j
=0
2
r
r
+
j
r
+
j
r
−
j
v
2
j
,
r
>
0
.
Proof.
For
r
=
0,
the
identity
is
immediate.
Let
r
≥
1.
Since
z
2
r
+
z
−
2
r
=
(
−
1)
r
2
T
2
r
iv
2
,
formula
(1)
gives
z
2
r
+
z
−
2
r
=
(
−
1)
r
r
m
=0
(
−
1)
m
2
r
2
r
−
m
2
r
−
m
m
(
iv
)
2
r
−
2
m
.
Setting
j
=
r
−
m
,
and
observing
that
(
−
1)
r
(
−
1)
r
−
j
i
2
j
=
1
,
M.
Merca
159
we
obtain
z
2
r
+
z
−
2
r
=
r
j
=0
2
r
r
+
j
r
+
j
r
−
j
v
2
j
,
as
required.
Proof
of
Theorem
3
.
Since
v
2
+
4
=
(
z
−
z
−
1
)
2
+
4
=
(
z
+
z
−
1
)
2
,
we
obtain
(
v
2
+
4)
n
=
(
z
+
z
−
1
)
2
n
.
Expanding
the
right-hand
side
gives
(
z
+
z
−
1
)
2
n
=
z
−
2
n
(1
+
z
2
)
2
n
=
z
−
2
n
2
n
m
=0
2
n
m
z
2
m
=
n
r
=
−
n
2
n
n
+
r
z
2
r
=
2
n
n
+
n
r
=1
2
n
n
−
r
z
2
r
+
z
−
2
r
.
Applying
Lemma
7,
we
obtain
(
v
2
+
4)
n
=
2
n
n
+
n
r
=1
2
n
n
−
r
r
j
=0
2
r
r
+
j
r
+
j
r
−
j
v
2
j
.
First
suppose
that
k
≥
1.
Extracting
the
coefficient
of
v
2
k
gives
[
v
2
k
](
v
2
+
4)
n
=
n
r
=
k
2
n
n
−
r
2
r
r
+
k
r
+
k
r
−
k
.
Replacing
r
by
r
+
k
,
we
obtain
[
v
2
k
](
v
2
+
4)
n
=
n
−
k
r
=0
2
n
n
−
k
−
r
2
k
+
2
r
2
k
+
r
2
k
+
r
r
.
On
the
other
hand,
the
binomial
expansion
(
v
2
+
4)
n
=
n
r
=0
n
r
4
n
−
r
v
2
r
Invertible
Chebyshev–binomial
transformations
160
gives
[
v
2
k
](
v
2
+
4)
n
=
n
k
4
n
−
k
.
Comparing
the
two
expressions
proves
the
identity
for
k
≥
1.
It
remains
to
consider
k
=
0.
Since
the
term
corresponding
to
r
=
0
in
the
Laurent
expansion
was
separated,
we
have
[
v
0
](
v
2
+
4)
n
=
2
n
n
+
n
r
=1
2
n
n
−
r
[
v
0
](
z
2
r
+
z
−
2
r
)
.
By
Lemma
7,
[
v
0
](
z
2
r
+
z
−
2
r
)
=
2
r
r
r
r
=
2
,
r
≥
1
.
Hence
[
v
0
](
v
2
+
4)
n
=
2
n
n
+
2
n
r
=1
2
n
n
−
r
.
On
the
other
hand,
[
v
0
](
v
2
+
4)
n
=
4
n
.
Therefore,
2
n
n
+
2
n
r
=1
2
n
n
−
r
=
4
n
,
which
is
precisely
the
case
k
=
0
of
Theorem
3.
This
completes
the
proof.
Proof
of
Corollary
4
.
Let
v
=
z
−
z
−
1
.
Since
v
2
+
4
=
(
z
+
z
−
1
)
2
,
we
have
(
v
2
+
4)
n
=
(
z
+
z
−
1
)
2
n
.
Therefore,
[
z
2
k
](
v
2
+
4)
n
=
[
z
2
k
](
z
+
z
−
1
)
2
n
=
2
n
n
+
k
=
2
n
n
−
k
.
M.
Merca
161
On
the
other
hand,
(
v
2
+
4)
n
=
n
j
=0
n
j
4
n
−
j
v
2
j
,
while
v
2
j
=
(
z
−
z
−
1
)
2
j
=
2
j
=0
(
−
1)
2
j
z
2
j
−
2
.
For
j
≥
k
,
the
coefficient
of
z
2
k
in
v
2
j
is
[
z
2
k
]
v
2
j
=
(
−
1)
j
−
k
2
j
j
−
k
.
Consequently,
[
z
2
k
](
v
2
+
4)
n
=
n
j
=
k
(
−
1)
j
−
k
2
j
j
−
k

n
j
4
n
−
j
.
Replacing
j
by
k
+
m
gives
[
z
2
k
](
v
2
+
4)
n
=
n
−
k
m
=0
(
−
1)
m
2
k
+
2
m
m

n
k
+
m
4
n
−
k
−
m
.
Comparing
the
two
expressions
for
[
z
2
k
](
v
2
+
4)
n
proves
the
result.
As
in
Section
2,
the
coefficient
matrix
in
Theorem
3
is
upper
triangu-
lar
with
unit
diagonal,
and
is
therefore
invertible.
Corollary
4
gives
its
inverse
explicitly.
Hence
these
two
identities
define
mutually
inverse
trian-
gular
transformations
between

2
n
n
−
k

n
k
=0
and

n
k
4
n
−
k
n
k
=0
.
4
Odd
Chebyshev
family
The
odd
family
is
obtained
from
the
same
Chebyshev
substitution.
Indeed,
setting
w
=
iz
in
2
T
d
w
+
w
−
1
2
=
w
d
+
w
−
d
Invertible
Chebyshev–binomial
transformations
162
and
writing
v
=
z
−
z
−
1
,
we
obtain
2
T
d
iv
2
=
(
iz
)
d
+
(
iz
)
−
d
.
For
d
=
2
r
+
1,
this
gives
2
T
2
r
+1
iv
2
=
i
(
−
1)
r
z
2
r
+1
−
z
−
2
r
−
1
,
or
equivalently,
z
2
r
+1
−
z
−
2
r
−
1
=
(
−
1)
r
i
2
T
2
r
+1
iv
2
.
Hence
the
expansion
used
in
this
section
is
induced
by
the
odd-indexed
Chebyshev
polynomials
of
the
first
kind.
After
establishing
the
corresponding
explicit
expansion,
we
combine
it
with
(
z
−
z
−
1
)(
z
+
z
−
1
)
2
n
=
v
(
v
2
+
4)
n
and
compare
the
coefficients
of
v
2
k
+1
.
Lemma
8.
Let
v
=
z
−
z
−
1
.
For
every
r
≥
0
,
we
have
z
2
r
+1
−
z
−
2
r
−
1
=
r
j
=0
2
r
+
1
r
+
j
+
1
r
+
j
+
1
r
−
j
v
2
j
+1
.
Proof.
From
z
2
r
+1
−
z
−
2
r
−
1
=
(
−
1)
r
i
2
T
2
r
+1
iv
2
and
(1),
we
obtain
z
2
r
+1
−
z
−
2
r
−
1
=
(
−
1)
r
i
r
m
=0
(
−
1)
m
2
r
+
1
2
r
+
1
−
m
2
r
+
1
−
m
m
(
iv
)
2
r
+1
−
2
m
.
Setting
j
=
r
−
m
,
the
signs
cancel,
and
therefore
z
2
r
+1
−
z
−
2
r
−
1
=
r
j
=0
2
r
+
1
r
+
j
+
1
r
+
j
+
1
r
−
j
v
2
j
+1
.
M.
Merca
163
Proof
of
Theorem
5
.
Since
v
=
z
−
z
−
1
and
v
2
+
4
=
(
z
+
z
−
1
)
2
,
we
have
v
(
v
2
+
4)
n
=
(
z
−
z
−
1
)(
z
+
z
−
1
)
2
n
.
Expanding
the
right-hand
side,
we
obtain
(
z
−
z
−
1
)(
z
+
z
−
1
)
2
n
=
(
z
−
z
−
1
)
n
r
=
−
n
2
n
n
+
r
z
2
r
=
n
r
=0

2
n
n
+
r
−
2
n
n
+
r
+
1

z
2
r
+1
−
z
−
2
r
−
1
.
Since
2
n
n
+
r
−
2
n
n
+
r
+
1
=
2
r
+
1
n
+
r
+
1
2
n
n
+
r
,
it
follows
that
v
(
v
2
+
4)
n
=
n
r
=0
2
r
+
1
n
+
r
+
1
2
n
n
+
r
z
2
r
+1
−
z
−
2
r
−
1
.
Applying
Lemma
8,
we
get
v
(
v
2
+
4)
n
=
n
r
=0
2
r
+
1
n
+
r
+
1
2
n
n
+
r
r
j
=0
2
r
+
1
r
+
j
+
1
r
+
j
+
1
r
−
j
v
2
j
+1
.
Extracting
the
coefficient
of
v
2
k
+1
gives
[
v
2
k
+1
]
v
(
v
2
+
4)
n
=
n
r
=
k
(2
r
+
1)
2
(
r
+
k
+
1)(
n
+
r
+
1)
r
+
k
+
1
r
−
k

2
n
n
+
r
.
Replacing
r
by
r
+
k
,
we
obtain
[
v
2
k
+1
]
v
(
v
2
+4)
n
=
n
−
k
r
=0
(2
k
+
2
r
+
1)
2
(2
k
+
r
+
1)(
n
+
k
+
r
+
1)
2
k
+
r
+
1
r

2
n
n
+
k
+
r
.
On
the
other
hand,
v
(
v
2
+
4)
n
=
v
n
r
=0
n
r
4
n
−
r
v
2
r
,
Invertible
Chebyshev–binomial
transformations
164
and
therefore
[
v
2
k
+1
]
v
(
v
2
+
4)
n
=
n
k
4
n
−
k
.
Comparing
the
two
expressions
for
the
coefficient
of
v
2
k
+1
completes
the
proof.
Proof
of
Corollary
6
.
Let
v
=
z
−
z
−
1
.
Since
v
2
+
4
=
(
z
+
z
−
1
)
2
,
we
have
v
(
v
2
+
4)
n
=
(
z
−
z
−
1
)(
z
+
z
−
1
)
2
n
.
Expanding
the
right-hand
side,
we
obtain
(
z
−
z
−
1
)(
z
+
z
−
1
)
2
n
=
(
z
−
z
−
1
)
n
r
=
−
n
2
n
n
+
r
z
2
r
.
Therefore,
v
(
v
2
+
4)
n
=
2
n
n
+
k
−
2
n
n
+
k
+
1
=
2
k
+
1
n
+
k
+
1
2
n
n
+
k
.
On
the
other
hand,
v
(
v
2
+
4)
n
=
n
j
=0
n
j
4
n
−
j
v
2
j
+1
,
while
v
2
j
+1
=
(
z
−
z
−
1
)
2
j
+1
=
2
j
+1
=0
(
−
1)
2
j
+
1
z
2
j
+1
−
2
.
For
j
≥
k
,
the
coefficient
of
z
2
k
+1
in
v
2
j
+1
is
[
z
2
k
+1
]
v
2
j
+1
=
(
−
1)
j
−
k
2
j
+
1
j
−
k
.
Consequently,
[
z
2
k
+1
]
v
(
v
2
+
4)
n
=
n
j
=
k
(
−
1)
j
−
k
2
j
+
1
j
−
k

n
j
4
n
−
j
.
M.
Merca
165
Replacing
j
by
k
+
m
,
we
obtain
[
z
2
k
+1
]
v
(
v
2
+
4)
n
=
n
−
k
m
=0
(
−
1)
m
2
k
+
2
m
+
1
m

n
k
+
m
4
n
−
k
−
m
.
Comparing
the
two
expressions
for
[
z
2
k
+1
]
v
(
v
2
+
4)
n
proves
the
result.
For
fixed
n
,
set
a
odd
k
:=
2
k
+
1
n
+
k
+
1
2
n
n
+
k
=
2
n
n
+
k
−
2
n
n
+
k
+
1
,
and
b
k
:=
n
k
4
n
−
k
.
Then
Theorem
5
and
Corollary
6
define
mutually
inverse
triangular
trans-
formations
between
the
arrays
a
odd
k
n
k
=0
and
{
b
k
}
n
k
=0
.
The
corresponding
coefficient
matrices
are
triangular
with
unit
diagonal.
5
Signed
refinement
of
a
classical
enumeration
Recall
that
n
k
2
n
−
k
counts
the
ordered
pairs
(
A,
B
)
of
disjoint
subsets
of
an
n
-element
set
X
satisfying
|
A
|
=
k
.
We
now
introduce
a
signed
refinement
of
this
enumeration.
Let
x
be
a
distinguished
element
of
X
.
Instead
of
counting
every
pair
(
A,
B
)
with
weight
1,
we
assign
a
sign
according
to
whether
the
distinguished
element
belongs
to
A
.
More
precisely,
define
w
x
(
A,
B
)
=
1
,
x
/
∈
A,
−
1
,
x
∈
A.
The
following
result
shows
that
the
resulting
signed
enumeration
is
given
by
a
simple
modification
of
the
classical
counting
formula.
Invertible
Chebyshev–binomial
transformations
166
Theorem
9.
Let
n
≥
1
and
0
≤
k
≤
n
be
integers.
Let
X
be
an
n
-element
set,
and
let
x
∈
X
be
a
distinguished
element.
Then
A,B
⊆
X
A
∩
B
=
∅
,
|
A
|
=
k
w
x
(
A,
B
)
=
n
−
2
k
n
n
k
2
n
−
k
.
Proof.
The
pairs
with
positive
weight
are
precisely
those
for
which
x
/
∈
A
.
There
are
n
−
1
k
ways
to
choose
A
,
and
after
A
has
been
chosen,
each
of
the
remaining
n
−
k
elements
may
either
belong
to
B
or
remain
outside
A
∪
B
.
Hence
the
total
positive
contribution
equals
n
−
1
k
2
n
−
k
.
Similarly,
the
pairs
with
negative
weight
are
those
for
which
x
∈
A
.
In
this
case
the
remaining
k
−
1
elements
of
A
are
chosen
from
the
other
n
−
1
elements,
giving
n
−
1
k
−
1
2
n
−
k
such
pairs.
Therefore,
A,B
⊆
X
A
∩
B
=
∅
,
|
A
|
=
k
w
x
(
A,
B
)
=

n
−
1
k
−
n
−
1
k
−
1

2
n
−
k
.
Finally,
using
n
−
1
k
−
n
−
1
k
−
1
=
n
−
2
k
n
n
k
,
we
obtain
A,B
⊆
X
A
∩
B
=
∅
,
|
A
|
=
k
w
x
(
A,
B
)
=
n
k
n
−
2
k
n
2
n
−
k
,
as
claimed.
Combining
the
signed
enumeration
of
Theorem
9
with
the
coefficient-
extraction
argument
underlying
Theorem
1
leads
to
the
following
identity.
M.
Merca
167
Theorem
10.
For
every
n
≥
k
≥
1
,
(
n
−
k
)
/
2
m
=0
(
−
1)
m
k
+
2
m
k
+
m
k
+
m
m
n
−
2(
k
+
2
m
)
2
n
(2
n
−
1)
2
n
n
−
k
−
2
m
=
n
−
2
k
n
n
k
2
n
−
k
.
Proof.
Let
u
=
z
+
z
−
1
.
Then
u
+
2
=
z
+
z
−
1
+
2
=
(1
+
z
)
2
z
and
2
−
u
=
2
−
z
−
z
−
1
=
−
(
z
−
1)
2
z
.
Hence
(2
−
u
)(
u
+
2)
n
−
1
=
−
(
z
−
1)
2
(1
+
z
)
2
n
−
2
z
n
.
We
expand
the
right-hand
side
as
follows:
(2
−
u
)(
u
+
2)
n
−
1
=
−
z
−
n
(
z
2
−
2
z
+
1)(1
+
z
)
2
n
−
2
=
−
2
n
j
=0

2
n
−
2
j
−
2
−
2
2
n
−
2
j
−
1
+
2
n
−
2
j

z
j
−
n
,
where,
as
usual,
2
n
−
2
r
=
0
if
r
<
0
or
r
>
2
n
−
2.
Putting
s
=
j
−
n
,
we
get
(2
−
u
)(
u
+
2)
n
−
1
=
n
s
=
−
n
c
s
z
s
,
where
c
s
=
−

2
n
−
2
n
+
s
−
2
−
2
2
n
−
2
n
+
s
−
1
+
2
n
−
2
n
+
s

.
By
the
symmetry
of
binomial
coefficients,
c
s
=
c
−
s
.
Therefore
(2
−
u
)(
u
+
2)
n
−
1
=
c
0
+
n
s
=1
c
s
z
s
+
z
−
s
.
Invertible
Chebyshev–binomial
transformations
168
For
0
≤
s
≤
n
,
we
express
the
three
binomial
coefficients
in
terms
of
2
n
n
+
s
.
We
have
2
n
−
2
n
+
s
−
2
=
(
n
+
s
)(
n
+
s
−
1)
2
n
(2
n
−
1)
2
n
n
+
s
,
2
n
−
2
n
+
s
−
1
=
(
n
+
s
)(
n
−
s
)
2
n
(2
n
−
1)
2
n
n
+
s
,
and
2
n
−
2
n
+
s
=
(
n
−
s
)(
n
−
s
−
1)
2
n
(2
n
−
1)
2
n
n
+
s
.
Therefore,
c
s
=
2
n
n
+
s
2
n
(2
n
−
1)
−
(
n
+
s
)(
n
+
s
−
1)
+
2(
n
+
s
)(
n
−
s
)
−
(
n
−
s
)(
n
−
s
−
1)
.
The
expression
in
parentheses
simplifies
to
2
n
−
4
s
2
=
2(
n
−
2
s
2
)
.
Hence
c
s
=
n
−
2
s
2
n
(2
n
−
1)
2
n
n
+
s
.
Finally,
by
the
symmetry
of
the
binomial
coefficients,
2
n
n
+
s
=
2
n
n
−
s
,
and
thus
c
s
=
n
−
2
s
2
n
(2
n
−
1)
2
n
n
−
s
.
Thus
(2
−
u
)(
u
+
2)
n
−
1
=
c
0
+
n
s
=1
n
−
2
s
2
n
(2
n
−
1)
2
n
n
−
s
z
s
+
z
−
s
.
Now
let
1
≤
k
≤
n
.
Since
the
constant
Laurent
term
has
no
contribution
to
the
coefficient
of
u
k
,
we
obtain
[
u
k
](2
−
u
)(
u
+
2)
n
−
1
=
k
≤
s
≤
n
s
≡
k
(mod
2)
c
s
[
u
k
](
z
s
+
z
−
s
)
.
M.
Merca
169
Writing
s
=
k
+
2
m
,
we
have
[
u
k
](
z
k
+2
m
+
z
−
k
−
2
m
)
=
(
−
1)
m
k
+
2
m
k
+
m
k
+
m
m
.
Therefore,
(2
−
u
)(
u
+
2)
n
−
1
=
(
n
−
k
)
/
2
m
=0
(
−
1)
m
k
+
2
m
k
+
m
k
+
m
m
×
n
−
2(
k
+
2
m
)
2
n
(2
n
−
1)
2
n
n
−
k
−
2
m
.
On
the
other
hand,
(2
−
u
)(
u
+
2)
n
−
1
=
2(
u
+
2)
n
−
1
−
u
(
u
+
2)
n
−
1
,
and
hence
(2
−
u
)(
u
+
2)
n
−
1
=
2
n
−
1
k
2
n
−
1
−
k
−
n
−
1
k
−
1
2
n
−
k
=

n
−
1
k
−
n
−
1
k
−
1

2
n
−
k
.
Using
n
−
1
k
−
n
−
1
k
−
1
=
n
−
2
k
n
n
k
,
we
obtain
the
desired
identity.
Remark.
The
case
k
=
0
requires
separate
treatment
because
the
factor
k
+
2
m
k
+
m
is
undefined
at
k
=
m
=
0.
Extracting
the
constant
term
in
u
from
(2
−
u
)(
u
+
2)
n
−
1
=
c
0
+
n
s
=1
c
s
(
z
s
+
z
−
s
)
,
we
obtain
[
u
0
](2
−
u
)(
u
+
2)
n
−
1
=
c
0
+
n/
2
m
=1
c
2
m
[
u
0
](
z
2
m
+
z
−
2
m
)
.
Invertible
Chebyshev–binomial
transformations
170
Since
c
0
=
1
2
n
−
1
2
n
n
and
[
u
0
](
z
2
m
+
z
−
2
m
)
=
2(
−
1)
m
,
m
≥
1
,
it
follows
that
1
2
n
−
1
2
n
n
+
2
n/
2
m
=1
(
−
1)
m
n
−
8
m
2
n
(2
n
−
1)
2
n
n
−
2
m
=
2
n
.
Indeed,
[
u
0
](2
−
u
)(
u
+
2)
n
−
1
=
2
n
.
Thus,
this
identity
is
the
constant-term
counterpart
of
Theorem
10.
If
the
distinguished
element
is
allowed
to
vary
over
all
elements
of
X
,
one
obtains
a
related
signed
enumeration.
Consider
triples
(
A,
B,
x
),
where
A,
B
⊆
X
,
A
∩
B
=
∅
,
|
A
|
=
k
,
and
x
∈
X
.
Assign
the
weight
ω
(
A,
B,
x
)
=
1
,
x
/
∈
A,
−
1
,
x
∈
A.
For
each
fixed
pair
(
A,
B
),
there
are
n
−
k
choices
of
x
with
positive
weight
and
k
choices
with
negative
weight.
Hence
the
total
contribution
of
each
pair
is
n
−
2
k
.
Since
the
number
of
such
pairs
(
A,
B
)
is
n
k
2
n
−
k
,
we
obtain
A,B
⊆
X,
x
∈
X
A
∩
B
=
∅
,
|
A
|
=
k
ω
(
A,
B,
x
)
=
(
n
−
2
k
)
n
k
2
n
−
k
.
This
observation
allows
us
to
rewrite
Theorem
10
in
the
equivalent
form
(
n
−
k
)
/
2
m
=0
(
−
1)
m
k
+
2
m
k
+
m
k
+
m
m
n
−
2(
k
+
2
m
)
2
2
n
−
1
2
n
n
−
k
−
2
m
=
(
n
−
2
k
)
n
k
2
n
−
k
,
whose
right-hand
side
admits
the
signed
combinatorial
interpretation
de-
scribed
above.
Theorem
10,
together
with
its
constant-term
counterpart
given
above,
defines
a
triangular
transformation
for
the
full
range
0
≤
k
≤
n
.
Reversing
the
coefficient-extraction
argument
yields
the
corresponding
inverse
trans-
formation.
M.
Merca
171
Corollary
11.
For
every
n
≥
1
and
0
≤
k
≤
n
,
n
−
2
k
2
2
n
−
1
2
n
n
−
k
=
(
n
−
k
)
/
2
m
=0
(
n
−
2
k
−
4
m
)
k
+
2
m
m

n
k
+
2
m
2
n
−
k
−
2
m
.
Proof.
Let
u
=
z
+
z
−
1
and
set
F
n
(
u
)
:=
(2
−
u
)(
u
+
2)
n
−
1
.
Expanding
F
n
(
u
)
in
powers
of
u
,
we
obtain
F
n
(
u
)
=
2(
u
+
2)
n
−
1
−
u
(
u
+
2)
n
−
1
=
n
r
=0

n
−
1
r
−
n
−
1
r
−
1

2
n
−
r
u
r
=
n
r
=0
n
−
2
r
n
n
r
2
n
−
r
u
r
,
where
the
usual
convention
n
−
1
−
1
=
n
−
1
n
=
0
is
understood.
On
the
other
hand,
the
Laurent
expansion
used
in
the
proof
of
Theo-
rem
10
gives
F
n
(
u
)
=
c
0
+
n
s
=1
c
s
(
z
s
+
z
−
s
)
,
where
c
s
=
n
−
2
s
2
n
(2
n
−
1)
2
n
n
−
s
.
Hence
[
z
k
]
F
n
(
u
)
=
n
−
2
k
2
n
(2
n
−
1)
2
n
n
−
k
.
Since
u
r
=
(
z
+
z
−
1
)
r
=
r
j
=0
r
j
z
r
−
2
j
,
the
coefficient
of
z
k
in
u
r
is
nonzero
only
when
r
≥
k
and
r
≡
k
(mod
2).
Writing
r
=
k
+
2
m
,
we
have
[
z
k
]
u
k
+2
m
=
k
+
2
m
m
.
Invertible
Chebyshev–binomial
transformations
172
Consequently,
F
n
(
u
)
=
(
n
−
k
)
/
2
m
=0
k
+
2
m
m
n
−
2(
k
+
2
m
)
n
n
k
+
2
m
2
n
−
k
−
2
m
.
Comparing
the
two
expressions
for
[
z
k
]
F
n
(
u
)
and
multiplying
by
n
proves
the
result.
For
fixed
n
≥
1,
define
a
k
:=
n
−
2
k
2
n
(2
n
−
1)
2
n
n
−
k
,
b
k
:=
n
−
2
k
n
n
k
2
n
−
k
,
0
≤
k
≤
n.
Then
Theorem
10,
together
with
its
constant-term
counterpart,
gives
the
complete
direct
transformation
b
0
=
a
0
+
2
n/
2
m
=1
(
−
1)
m
a
2
m
,
and,
for
1
≤
k
≤
n
,
b
k
=
(
n
−
k
)
/
2
m
=0
(
−
1)
m
k
+
2
m
k
+
m
k
+
m
m
a
k
+2
m
.
On
the
other
hand,
Corollary
11
is
equivalently
written
as
a
k
=
(
n
−
k
)
/
2
m
=0
k
+
2
m
m
b
k
+2
m
,
0
≤
k
≤
n.
Thus,
after
arranging
the
even
and
odd
indices
separately,
the
coefficient
matrix
of
the
direct
transformation
becomes
block
diagonal,
and
each
di-
agonal
block
is
upper
unitriangular.
The
inverse
formula
above
gives
the
inverse
matrix
explicitly.
Hence
Theorem
10,
its
constant-term
counter-
part,
and
Corollary
11
together
define
a
pair
of
mutually
inverse
triangular
transformations
between
n
−
2
k
2
n
(2
n
−
1)
2
n
n
−
k

n
k
=0
and
n
−
2
k
n
n
k
2
n
−
k
n
k
=0
.
M.
Merca
173
6
Concluding
remarks
A
central
outcome
of
this
paper
is
that
the
explicit
expansions
of
the
Cheby-
shev
polynomials
of
the
first
kind
naturally
induce
several
invertible
tri-
angular
transformations
between
classical
families
of
binomial
coefficients.
Although
these
transformations
arise
from
different
Chebyshev
expansions,
they
are
all
obtained
by
the
same
coefficient-extraction
principle,
which
pro-
vides
a
unified
framework
for
deriving
both
direct
and
inverse
identities.
Beyond
their
algebraic
interest,
the
closed
forms
occurring
in
the
direct
transformations
admit
natural
combinatorial
interpretations.
They
connect
the
binomial
coefficients
2
n
n
−
k
with
the
classical
combinatorial
sequences
n
k
2
n
−
k
and
n
k
4
n
−
k
,
whose
enumerative
meanings
arise
from
disjoint
subsets
and
block
decompo-
sitions,
respectively.
Furthermore,
the
signed
refinement
introduced
in
Sec-
tion
5
shows
that
the
same
approach
naturally
extends
to
weighted
enumer-
ations,
yielding
an
additional
pair
of
direct
and
inverse
Chebyshev–binomial
transformations.
The
coefficient-extraction
approach
employed
here
is
not
restricted
to
the
particular
identities
considered
in
this
paper.
Whenever
suitable
explicit
connection
formulas
or
Laurent-type
representations
are
available,
the
same
method
may
produce
further
invertible
triangular
transformations
together
with
their
associated
combinatorial
interpretations.
It
would
therefore
be
interesting
to
investigate
analogous
constructions
for
other
classical
orthog-
onal
polynomials,
particularly
the
Jacobi
polynomials
and
their
Gegenbauer
specializations.
References
[1]
K.
Adegoke,
R.
Frontczak
and
T.
Goy,
Binomial
Fibonacci
sums
from
Chebyshev
polynomials,
J.
Integer
Sequences
26
(2023),
23.9.6.
[2]
R.
Bacher,
Chebyshev
polynomials,
quadratic
surds
and
a
variation
of
Pascal’s
triangle,
(2015),
Preprint,
arXiv:1509.09054.
[3]
J.
Cigler,
Some
remarks
about
q
-Chebyshev
polynomials
and
q
-Catalan
numbers
and
related
results,
(2013),
Preprint,
arXiv:1312.2767.
Invertible
Chebyshev–binomial
transformations
174
[4]
M.
Janji´
c
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B.
Petkovi´
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A
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some
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of
integers,
J.
Integer
Sequences
17
(2014),
14.3.5.
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T.H.
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J.C.
Mason
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D.C.
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Chapman
&
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T.J.
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John
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York,
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