The matrix sequences created from special integer sequences are very interesting topics in number and matrix theory. The Pell sequence is formed by adding twice the previous term to the term before that. When this sequence and its generalized sequences are expressed in matrices, some properties of the sequences can be obtained using matrix theory. Some papers focused on generalized Pell matrix sequences and their properties can be seen in [1], [2]. These papers give an idea for discovering potential applications of these sequences. For instance, Binet formula or generating functions and various relations are used to find the general behavior of the sequences. The Pell numbers pn are defined by the recurrence relation pn = 2pn—1 + pn—2 for n ≥ 2, beginning with the values p0 = 0, p1 = 1 in [1,2] [3], [4]. The generalized Pell sequence, called (s, t)-Pell sequence depending on two real parameters s, t and s2 + t > 0 is introduced as

pn(s, t) = 2spn−1(s,t) + tpn−2(s, t), p0(s, t) = 0, p1(s, t) = 1, x1−x2

( ) √ 2

for n ≥ 2. The Binet formula of the (s, t)-Pell sequence is Pn s, t = , where x1 = s + s + t, x2 = s −

x1−x2

√s + t, where x1and x2are the roots of the characteristic equation of the recurrence formula of the (s, t)-Pell sequenc. The main target of the paper is to apply various binomial transforms to the (s, t)-Pell matrix sequence and find some relations and properties of the new binomial transform sequences. Prodinger investigsted binomial transform in [5]. Chen [6] found some properties about the binomial transform in. Falcon and Plaza [7] investigated the binomial transforms of k-Fibonacci sequence. The authors studied the binomial transforms of the k-Lucas sequence in [8]. Yilmaz and Taskara [9] studied binomial transforms of the Padovan and Perrin matrix sequences. The authors investigated different binomial transforms of k-Jacobsthal sequences in [10]. In [11], Binomial transform of quadrapell sequences and quadrapell matrix sequences are dealt with. Uygun [12] computed the binomial transforms of the generalized (s, t)-Jacobsthal matrix sequencein. Kaplan and Özkoç Öztürk [13] studied on the binomial transforms of the Horadam quaternion sequences. Kwon [14] gave the binomial transforms of the modified k-Fibonacci-like sequence in. Soykan [15–21] sudied binomial transforms of the generalized Tribonacci sequence, the generalized third order Pell sequence, the generalized fourth order Pell sequence, the generalized fifth order Pell sequence, the generalized Narayana sequence, the generalized Pentanacci sequence, the binomial transform of the generalized Jacobsthal-Padovan numbers.

Definition 1. The (s, t)-Pell matrix sequence is defined as

Pn+1(s, t) = 2sPn(s, t) + tPn−1(s, t),

1 0 2s 1 (1)

P0(s, t) = ( ) , P1 (s, t) = ( ).

0 1 t 0

For n ≥ 1, any positive integer, s, t real numbers such that s + t > 0.

Definition 2. The binomial transform of (s, t)-Pell matrix sequence indicated as {Bn(s, t)}n∈ℕ is defined as

Bn(s, t) = ∑ ( ) Pi(s, t). (2)

For any positive integer s, t.

Lemma 1. The binomial transform of (s, t)-Pell matrix sequence satisfies the relation.

Bn+1(s, t) = ∑ ( ) [Pi(s, t) + Pi+1(s, t)].

n+1 n n n

Proof: By the property ( ) = ( ) + ( ) and ( ) = 0 and Eq. (2), we have

i i i−1 n+1

n+1 n+1

n + 1 n n Bn+1(s, t) = ∑ ( ) Pi(s, t) = P0(s, t) + ∑ [( ) + ( )] Pi(s, t) i i i − 1

n n+1

= P0(s, t) + ∑ ( ) Pi (s, t) + ∑ ( ) Pi(s, t)

= ∑ ( ) Pi(s, t) + ∑ ( ) Pi + 1 (s, t).

Theorem 1. The recurrence relation of the binomial transform of (s, t)-Pell matrix sequence is as follows: Bn+1(s, t) = (2 + 2s)Bn(s, t) + (t − 2s − 1)Bn−1(s, t). (3)

Proof: By Lemma 1 and Eq. (1), it is obtained that

Bn+1(s, t) = ∑ ( ) (Pi(s, t) + Pi+1(s, t))

= P0(s, t) + P1(s, t) + ∑ ( ) (Pi(s, t) + Pi+1(s, t))

= P0(s, t) + P1(s, t) + ∑ ( ) (Pi (s, t) + 2sPi(s, t) + tPi−1(s, t))

= P0(s, t) + P1(s, t) + (1 + 2s) ∑ ( ) Pi (s, t) + t ∑ Pi−1 (s, t).

Then, by the definition of the binomial transform of (s, t)-Pell matrix sequence, we get

Bn+1(s, t) = −2sP0(s, t) + P1(s, t) + (1 + 2s)Bn(s, t) + t ∑ ( ) Pi−1(s, t). (4)

Let’s substitute for n in place of n + 1 in the last Equality (4).

Bn(s, t) = −2sP0(s, t) + P1(s, t) + (1 + 2s)Bn−1(s, t)

n − 1 +t ∑ ( ) Pi−1(s, t)

= −2sP0(s, t) + P1(s, t) + 2sBn−1(s, t)

n n−1

n − 1 n − 1 + ∑ ( ) Pi−1(s, t) + t ∑ ( ) Pi−1(s, t).

i − 1 i

By ( ) = 0, it is obtained that

Bn(s, t) = −2sP0(s, t) + P1(s, t) + 2sBn−1(s, t)

n − 1 n − 1 + ∑ [t ( ) + ( )] Pi−1(s, t)

= −2sP0(s, t) + P1(s, t) + 2sBn−1(s, t)

n − 1 n − 1 n − 1 n − 1 + ∑ [t ( ) + ( ) + t ( ) − t ( )] Pi−1(s, t) i i − 1 i − 1 i − 1

= −2sP0(s, t) + P1(s, t) + 2sBn−1(s, t)

n − 1 n + ∑ [(1 − t) ( ) + t ( )] Pi−1(s, t) i − 1 i

= −2sP0(s, t) + P1(s, t) + 2sBn−1(s, t) n n−1

n n − 1 +t ∑ ( ) Pi−1(s, t) + (1 − t) ∑ ( ) Pi(s, t).

Bn(s, t) = −2sP0(s, t) + P1(s, t) + (2s + 1 − t)Bn−1(s, t) + t ∑ ( ) Pi(s, t). (5)

By substituting the Eq. (4) into Eq. (5), and by some algebraic operations, the proof is completed as

Bn+1(s, t) = (2 + 2s)Bn(s, t) + (t − 2s − 1)Bn−1(s, t).

Theorem 2. The Binet formula of the binomial transform of (s, t)-Pell matrix sequence is computed as

[(1 − b2)P0(s, t) + P1(s, t)]b1 − [(1 − b1)P0(s, t) + P1(s, t)]b2 Bn(s, t) = ,

b1 − b2

where b1 = s + 1 + √s + t, b2 = s + 1 − √s + t.

Proof: The characteristic polynomial equation of the recurrence Relation (3) is x − (2 + 2s)x − (t − 2s − 1) = 0, whose solutions are b1 and b2. Assume that Bn(s, t) = c1b1 + c2b2. By definition, we find that B0(s, t) = P0(s, t) and B1(s, t) = P1(s, t). Let us substitute for n = 0 and n = 1 in this equality, then we deduce that c1 =

P0(a,t)+P1(a,t)−P0(a,t)b2 P0(a,t)b1−P1(a,t)−P0(a,t)

, c2 = . After substituting the values of c1, c2, we get the result.

b1−b2 b1−b2

Theorem 3. The generating function of the binomial transform of (s, t)-Pell matrix sequence is obtained as

B0(s, t) + x(B1(s, t) − (2 + 2s)B0(s, t)) Bn(s, t, x) = Bn = ∑ Bi(s, t)x = .

( ) ( ) 2

1 − 2 + 2s x − t − 2s − 1 x

Proof: The generating function is a power series whose coefficients are the binomial transform of the (s, t)- Pell matrix sequence centered at the origin. By multiplying Bn(s, t, x) by −(2 + 2s)x and −(t − 2s − 1)x , it is obtained that

−(2 + 2s)xBn = −(2 + 2s)xB0(s, t) − (2 + 2s)x B1(s, t) + ⋯

−(t − 2s − 1)x Bn = −(t − 2s − 1)x B0(s, t) − (t − 2s − 1)x B1(s, t) + ⋯ .

By these equalities and the recurrence Relation (3), it is computed that

[1 − (2 + 2s)x − (t − 2s − 1)x ]Bn =B0(s, t) + x(B1(s, t) − (2 + 2s)B0(s, t)) +x (B2(s, t) − (2 + 2s)B1(s, t) − (t − 2s − 1)B0(s, t)) + ⋯ =B0(s, t) + x[B1(s, t) − (2 + 2s)B0(s, t)].

The proof is completed.

Theorem 4. Let n be a positive integer. Then the sum of the binomial transform of (s, t)-Pell sequence is given as

Bn(s, t) − Bmp+n(s, t) − (2s − t + 1) Bm−n(s, t) +(2s − t + 1) Bm(p−1)+n(s, t)

∑ Bmi+n(s, t) = .

m m2 m

1 − (b1 + b2 ) + (2s − t + 1)

Proof: By the Binet formula of the binomial transform of (s, t)-Pell matrix sequence and the geometric sum property, it is computed as p−1 p−1 p−1 p−1 mi+n mi+n n mi n mi

∑ Bmi+n(s, t) = ∑ c1b1 + c2b2 = c1b1 ∑ b1 + c2b2 ∑ b2

i=0 i=0 i=0 i=0 mp mp

1 − b1 1 − b2

= c1b1 ( mp) + c2b2 ( m )

1 − b 1 − b2

n n m n m n

(c1b1 + c2b2) − c1b2 b1 − c2b1 b2

mp+n mp+n mp+n mn mp+n m

−c1b1 − c2b2 + c2b2 b1 + c1b1 b2 = m m

( ) ( )m

1 − b1 + b2 + b1b2

n n mp+n mp+n

(c1b1 + c2b2) − (c1b + c2b )

n m−n m−n m m(p−1)+n m(p−1)+n

−(b1b2) [c2b2 + c1b1 ] − (b1b2) [c2b + c1b ] = m m .

( ) ( )m

1 − b1 + b2 + b1b2 We know that b1b2 = 2s − t + 1. And by the Binet formula of the sequence, the result is obtained.

Definition 3. The s-binomial transform of the (s, t)-Pell matrix sequence {On(s, t)}n∈N is demonstrated by

On(s, t) = ∑ ( ) s Pi(s, t). (6)

It is easily seen that On(s, t) = s Bn(s, t).

Lemma 2. The following relation for the s-binomial transform of the (s, t)-Pell matrix sequence verifies.

On+1(s, t) = ∑ ( ) s [Pi(s, t) + Pi+1(s, t)].

n+1 n n n

Proof: By the properties ( ) = ( ) + ( ), ( ) = 0, we get

i i i−1 n+1

n + 1 On+1(s, t) = ∑ ( ) s Pi(s, t)

n+1 n+1

= s P0(s, t) + ∑ [( ) + ( )] s Pi(s, t)

n+1 n+1

= P0(s, t) + ∑ ( ) Pi(s, t) + ∑ ( ) Pi(s, t)

n+1 n+1

= ∑ ( ) s Pi(s, t) + ∑ ( ) s Pi+1(s, t).

Theorem 5. The recurrence formula of s-the binomial transform of the (s, t) Pell matrix sequence is demonstrated as

O (s, t) = s(2 + 2s)O (s, t) + s (t − 2s − 1)O (s, t). (7)

n+1 n n−1

Proof: The initial conditions are found by Eq. (6) as O0(s, t) = P0(s, t) and O1(s, t) = sP1(s, t). By Definition 1, we obtain On+1(s, t) = ∑ ( ) s (Pi(s, t) + Pi+1(s, t))

=s (P0(s, t) + P1(s, t))

+ ∑ ( ) s (Pi(s, t) + Pi+1(s, t))

=s (P0(s, t) + P1(s, t))

+ ∑ ( ) s (Pi(s, t) + 2sPi(s, t) + tPi−1(s, t))

=s (P0(s, t) + P1(s, t))

n+1 n+1

+(1 + 2s) ∑ ( ) s Pi(s, t) + t ∑ ( ) s Pi−1(s, t).

On+1(s, t) =s (P1(s, t) − 2sP0(s, t)) + s(1 + 2s)On(s, t)

n (8)

+t ∑ ( ) s Pi−1(s, t).

If we replace n by n + 1 in Eq. (8), it is obtained that

On(s, t) =s (P1(s, t) − 2sP0(s, t)) + s(1 + 2s)On−1(s, t) n − 1 +t ∑ ( ) s Pi−1(s, t)

n − 1

n 2 n

=s (P1(s, t) − 2sP0(s, t)) + 2s On−1(s, t) + ∑ ( ) s Pi−1(s, t)

n − 1 +t ∑ ( ) s Pi−1(s, t)

n − 1

n 2 n

=s (P1(s, t) − 2sP0(s, t)) + 2s On−1(s, t) + ∑ ( ) s Pi(s, t) i − 1

n − 1 +t ∑ ( ) s Pi−1(s, t).

Then, we get

On(s, t) = s (P1(s, t) − 2sP0(s, t)) + 2s On−1(s, t)

n − 1 n − 1 + ∑ [t ( ) + ( )] s Pi−1(s, t)

= s (P1(s, t) − 2sP0(s, t)) + 2s On−1(s, t)

n − 1 n − 1 n − 1 n − 1 + ∑ [t ( ) + ( ) + t ( ) − t ( )] s Pi−1(s, t) i i − 1 i − 1 i − 1 = s (P1(s, t) − 2sP0(s, t)) + 2s On−1(s, t)

n − 1 n + ∑ [(1 − t) ( ) + t ( )] s Pi−1(s, t) i − 1 i

= s (P1(s, t) − 2sP0(s, t)) + 2s On−1(s, t)

n n−1

n n − 1 +t ∑ ( ) s Pi−1(s, t) + (1 − t) ∑ ( ) s Pi(s, t).

On(s, t) = s (P1(s, t) − 2sP0(s, t)) + 2s On−1(s, t)

n (9)

+t ∑ ( ) s Pi−1(s, t) + (1 − t)sOn−1(s, t).

By substituting the Equality (8) into Equality (9), we get

n+1 3

sOn(s, t) = s (P1(s, t) − 2sP0(s, t)) + 2s On−1(s, t) + On+1(s, t)

n+1 2

− s (P1(s, t) − 2sP0(s, t)) − (1 + 2s)sOn(s, t) + (1 − t)s On−1(s, t) = (2s − ts + s )On−1(s, t) − (1 + 2s)On(s, t) + On+1(s, t).

The proof is found after some simple calculations as

On+1(s, t) = s(2 + 2s)On(s, t) + s (t − 2s − 1)On−1(s, t).

Theorem 6. The Binet formula of the s-binomial transform of (s, t)-Pell matrix sequence is demonstrated by

[(s − o2)P0(s, t) + sP1(s, t)]o1 − [(s − o1)P0(s, t) + sP1(s, t)]o2 On(s, t) = , o1 − o2

where o1 = s(s + 1) + s√s + t, o2 = s(s + 1) − s√s + t.

Proof: The characteristic equation of the recurrence Eq. (7) is obtained as x − s(2 + 2s)x + s (2s + t − 1) = 0, whose roots are o1 and o2 where o1 = s(s + 1) + s√s + t, o2 = s(s + 1) − s√s + t. Assume that On(s, t) = c1o1 + c2o2. By definition O0(s, t) = P0(s, t) and O1(s, t) = sP1(s, t). Let us substitute for n = 0 and n = 1 in

(s−o2)P0(s,t)+sP1(s,t) (s−o1)P0(s,t)+sP1(s,t)

the equality On(s, t), then we deduce that c1 = , c2 = . We can easily see

o1−o2 O1−o2

the result after substituting these values into On(s, t).

Theorem 7. The generating function for the s-binomial transform of (s, t) Pell matrix sequence is obtained as O0(s, t) + x[O1(s, t) − s(2 + 2s)O0(s, t)] On(s, t, x) = On = ∑ Oi(s, t)x = .

1 − s(2 + 2s)x + s (2s − t + 1)x

Proof: By multiplication On(s, t, x) by −s(2 + 2s)x and s (2s − t + 1)x , the following equlities are obtained

−s(2 + 2s)xOn = −s(2 + 2s)xO0(s, t) − s(2 + 2s)x O1(s, t) + ⋯

2 2 2 2 2 3

s (2s − t + 1)x On = s (2s − t + 1)x O0(s, t) + s (2s − t + 1)x O1(s, t) + ⋯.

From these equalities and the recurrence Relation (7), the generating function is obtained.

Theorem 8. Let m, n be any positive integers. Then the sum of the s-binomial transform of (s, t)-Pell sequence is given as On(s, t) − Omp+n(s, t) − [s (2s − t + 1)] Om−n(s, t) +[s (2s − t + 1)] Om(p−1)+n(s, t) ∑ Omi+n(s, t) = .

m m 2 m

1 − (o + o ) + [s (2s − t + 1)]

Proof: By the Binet formula of the s-binomial transform of (s, t)-Pell matrix sequence and the geometric sum property, the result is computed by using the same method in Theorem 4.

Definition 4. The rising binomial transform of the (s, t)-Pell matrix sequence {In(s, t)}n∈N is defined by the following formula

i (10) In(s, t) = ∑ ( ) s Pi(s, t).

Lemma 3 ([2]). The Binet formula of the (s, t)-Pell matrix sequence is

Pn(s, t) = Mx1 − Nx2,

P1−x2P0 P1−x1P0

where M = , N = , x1 = s + √s + t, x2 = s − √s + t.

x1−x2 x1−x2

Theorem 9. The Binet formula for the rising binomial transform of the (s, t) Pell matrix sequence is

In(s, t) = M(sx1 + 1) − N(sx2 + 1) ,

where x1 = s − √s + t, x2 = s + √s + t.

Proof: In Eq. (14), we have

i i i i

In(s, t) = ∑ ( ) s Pi(s, t) = ∑ ( ) s (Mx1 − Nx2)

= M ∑ ( ) (sx1) − N ∑ ( ) (sx2)

= M(sx1 + 1) − N(sx2 + 1) ).

Theorem 10. For n ⩾ 1, the rising binomial transform of the (s, t)-Pell matrix sequence is a recurrence sequence such that

In+1(s, t) = (2s + 2)In(s, t) − (1 − s t + 2s )In−1(s, t).

with initial conditions In(s, t) = P0(s, t) and I1(s, t) = sP1(s, t).

Proof: By Binet formula for the rising binomial transform of the (s, t)-Pell matrix sequence, we get (2s + 2)In(s, t) − (1 − s t + 2s )In−1(s, t)

2 n n

=(2s + 2)[M(sx1 + 1) − N(sx2 + 1) ]

2 2 n−1 n−1

−(1 − s t + 2s )[M(sx1 + 1) − N(sx2 + 1) ]

n−1 2 2 2

M(sx1 + 1) [(2s + 2)(sx1 + 1) − (1 − s t + 2s )] ={ }

n−1 2 2 2

−N(sx2 + 1) [(2s + 2)(sx2 + 1) − (1 − s t + 2s )]

n−1 2

M(sx1 + 1) [s (2sx1 + t) + 2sx1 + 1] ={ }

n−1 2

−N(sx2 + 1) [s (2sx2 + t) + 2sx2 + 1]

n−1 2

M(sx1 + 1) [(sx1 + 1) ] ={ }

n−1 2

−N(sx2 + 1) [(sx2 + 1) ] =In+1(s, t).

Theorem 11. The generating function of rising binomial transform of the (s, t) Pell matrix sequence is given as

I0(s, t) + x[I1(s, t) − (2s + 2)I0(s, t)] In(s, t, x) = In = .

( 2 ) ( 2 2) 2 1 − 2s + 2 x + 1 − s t + 2s x Proof: By following same procedure, we have

2 2 2 2

−(2s + 2)xIn = −(2s + 2)xI0(s, t) − (2s + 2)x I1(s, t) + ⋯

2 2 2 2 2 2 2 2 3

(1 − s t + 2s )x In = (1 − s t + 2s )x I0(s, t) + (1 − s t + 2s )x I1(s, t) + ⋯

2 2 2 2 2 2

[1 − (2s + 2)x + (1 − s t + 2s )x ]In = I0(s, t) − (2s + 2)xI0(s, t) + xI1(s, t) + x (0).

By the above computations, we obtain the generating function.

Definition 5. The falling binomial transform of the (s, t)-Pell matrix sequence {Dn(s, t)}n∈N is given by

Dn(s, t) = ∑ ( ) s Pi(s, t). (11)

Lemma 4. The falling binomial transform of the (s, t)-Pell matrix sequence verifies the relation.

Dn+1(s, t) = ∑ ( ) s [sPi(s, t) + Pi+1(s, t)],

with initial conditions D0(s, t) = P0(s, t) and D1(s, t) = sP0(s, t) + P1(s, t).

Proof. In Eq. (11), we take n + 1 in place of n. After some computations, we obtain the following:

n + 1

n+1−i

Dn+1(s, t) = ∑ ( ) s Pi(s, t)

n+1 n+1−i

= s P0(s, t) + ∑ [( ) + ( )] s Pi(s, t)

n n+1 n+1 n+1−i n−i

= s P0(s, t) + ∑ s ( ) Pi(s, t) + ∑ s ( ) Pi+1(s, t)

n+1−i n−i

= ∑ s ( ) Pi(s, t) + ∑ s ( ) Pi+1(s, t).

Theorem 12. The following recurrence relation is verified by the falling binomial transform of the (s, t)-Pell matrix sequence.

Dn+1(s, t) = 4sDn(s, t) + (3s − t)Dn−1(s, t).

Proof: The initial conditions are found by Definition 5 as D0(s, t) = P0(s, t) and D1(s, t) = sP0(s, t) + P1(s, t). By Lemma 4 and Eq. (1), it is obtained that

Dn+1(s, t) = ∑ ( ) s [sPi(s, t) + Pi+1(s, t)]

n n−i

= s [sP0(s, t) + P1(s, t)] + ∑ ( ) s [sPi(s, t) + Pi+1(s, t)]

(12)

n n−i n−i

= s [sP0(s, t) + P1(s, t)] + 3s ∑ ( ) s Pi(s, t) + t ∑ ( ) s Pi−1(s, t)

n n−i

= s [P1(s, t) − 2sP0(s, t)] + 3sDn(s, t) + t ∑ ( ) s Pi−1.(s, t).

Let's substitute in place of n + 1 in Equality (12).

Dn(s, t) =s [P1(s, t) − 2sP0(s, t)] + 3sDn−1(s, t) n − 1

n−1−i

+t ∑ ( ) s Pi−1(s, t)

n − 1

n−1 n−1−i

=s [P1(s, t) − 2sP0(s, t)] + 3s ∑ ( ) s Pi(s, t)

n − 1

n−1−i

+t ∑ ( ) s Pi−1(s, t)

n − 1

n−1 n−i

=s [P1(s, t) − 2sP0(s, t)] + 3s ∑ ( ) s Pi−1(s, t) i − 1

t n − 1 + ∑ ( ) s Pi−1(s, t).

We know that ( ) = 0, hence we get

Dn(s, t) = s [P1(s, t) − 2sP0(s, t)]

t n − 1 n − 1 + ∑ [ ( ) + 3s ( )] s Pi−1(s, t) s i i − 1 (13) = s [P1(s, t) − 2sP0(s, t)]

t n − 1 n − 1 t n − 1 t n − 1 + ∑ [ ( ) + 3s ( ) + ( ) − ( )] s Pi−1(s, t) s i i − 1 s i − 1 s i − 1 = s [P1(s, t) − 2sP0(s, t)]

t n − 1 t n + ∑ [(3s − ) ( ) + ( )] s Pi−1(s, t) s i − 1 s i

= s [P1(s, t) − 2sP0(s, t)]

n n−1 (13) t n t n − 1

n−i n−1−i

+ ∑ ( ) s Pi−1(s, t) + (3s − ) ∑ ( ) s Pi−1(s, t) s i s i

= s [P1(s, t) − 2sP0(s, t)]

t n t + ∑ ( ) s Pi−1(s, t) + (3s − ) Dn−1(s, t).

s i s

By substituting Eq. (12) into Eq. (13) and by some simple calculations, the proof is completed as

Dn+1(s, t) = 4sDn(s, t) + (3s − t)Dn−1(s, t). (14) Theorem 13. The Binet formula for the falling binomial transform of (s, t)-Pell matrix sequence is calculated as

[(s − d2)P0(s, t) + sP1(s, t)]d1 − [(s − d1)P0(s, t) + sP1(s, t)]d2 Dn(s, t) = , d1 − d2 where d1 = 2s + √s + t , d2 = 2s − √s + t.

Proof: The characteristic polynomial equation of recurrence Eq. (14) is x − 4sx + 3s − t = 0, whose

2 2 n n

solutions are d1 = 2s + √s + t, d2 = 2s − √s + t. Assume that Dn(s, t) = c1d1 + c2d2. By definition D0(s, t) = P0(s, t) and D1(s, t) = sP1(s, t). Let us substitute for n = 0 and n = 1 in this equality, then we deduce

(s−d2)P0(s,t)+sP1(s,t) (s−d1)P0(s,t)+sP1(s,t)

that c1 = , c2 = − . After substituting the values of c1 and c2, we get the

d1−d2 d1−d2

result.

Theorem 14. The generating function of the falling binomial transform of (s, t) Pell matrix sequence is as follows:

D0(s, t) + x[D1(s, t) − 4sD0(s, t)] Dn(s, t, x) = .

( 2 ) 2 1 − 4sx + 3s − t x Proof: If we product the equality Dn(s, t, x) by −4sx and (3s − t)x , it is obtained that

−4sxDn(s, t, x) = −4sxD0(s, t) − 4sx D1(s, t) + ⋯

2 2 2 2 2 3

(3s − t)x Dn(s, t, x) = (3s − t)x D0(s, t) + (3s − t)x D1(s, t) + ⋯ From the above equalities and Theorem 12, we get the desired result.

D0(s, t) + x[D1(s, t) − 4sD0(s, t)] Dn(s, t, x) = .

( 2 ) 2 1 − 4sx + 3s − t x Theorem 15. Assume that n is a positive integer. Then the sum of the binomial transform of (s, t)-Pell matrix sequence is given as Dn(s, t) − (3s − t) Dm−n − Dmp+n(s, t) +(3s − t) Dm(p−1)+n(s, t) ∑ Dmi+n(s, t) = .

m m 2 m

1 − (d1 + d2 ) + (3s − t)

In this study, various binomial-type transforms of the generalized (s, t)-Pell matrix sequence were systematically examined. Explicit recurrence relations, Binet formulas, generating functions, and summation identities were established for each transform, highlighting their algebraic consistency and structural richness. The results confirm that these transforms offer effective tools for generating new matrix sequences with predictable properties. Future research may focus on extending these transformations to higher-order or hybrid recursive sequences and exploring potential applications in combinatorics, coding theory, and related areas of applied mathematics.

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