Relations Established Between Hypergeometric Functions and Some Special Number Sequences

Abstract

In this paper, we establish new hypergeometric representations for two classical integer sequences, namely the Pell and Jacobsthal sequences. Motivated by Dilcher’s hypergeometric formulations of the Fibonacci sequence, we extend this framework to other second-order linear recurrence sequences with distinct characteristic structures. By employing Binet-type formulas, recurrence relations, Chebyshev polynomial connections, and classical transformation properties of Gauss hypergeometric functions, we derive several explicit and alternative representations for the Pell and Jacobsthal numbers. These representations unify known identities, yield new closed-form expressions, and reveal deeper structural parallels between hypergeometric functions and linear recurrence sequences. The results demonstrate that hypergeometric functions provide a systematic and versatile analytical tool for studying special number sequences beyond the Fibonacci case, and they suggest potential extensions to broader families such as Horadam-type sequences and their generalizations.

1. Introduction

Hypergeometric functions, with their serial definitions, parametric structures, and transformation properties, hold a central position in many branches of mathematics [1]. This field, pioneered by Gauss and Kummer, encompasses general solution approaches beyond classical methods and is now considered a powerful analytical tool not only in theoretical mathematics but also in many scientific fields, such as physics, biology, economics, numerical analysis, and cryptography [1,2,3,4,5,6,7].

The analytical power of hypergeometric functions is not limited to their respective differential equations, but also manifests in the relationships that can be established with various special number sequences. First, attention was focused on the Fibonacci numbers, which enables the derivation of various hypergeometric representations [8]. This seemingly unnatural relationship has been made compatible with hypergeometric structures through parametric transformations and mappings. However, the applicability of a similar approach to structurally similar Pell and Jacobsthal sequences has not been encountered.

Studies on the Fibonacci sequence are particularly prominent. Dilcher’s work details how hypergeometric functions can represent the Fibonacci sequence under different transformations and offers various representations using transformation techniques of classical series. While hypergeometric representations of integer sequences have been studied for the Fibonacci sequence, focusing exclusively on this case obscures the broader analytical potential of hypergeometric functions. Many classical integer sequences, such as the Pell and Jacobsthal sequences, satisfy structurally similar second-order linear recurrence relations but possess distinct characteristic roots and algebraic behaviors. Investigating their hypergeometric representations allows for a unified framework that highlights common structural features while simultaneously revealing sequence-specific properties. Beyond their intrinsic theoretical interest, such representations have potential applications in combinatorics through the derivation of new summation formulas and identities, in cryptography via structured integer sequences with predictable growth and transformation properties, and in approximation theory by providing alternative analytic expressions suitable for asymptotic analysis and numerical approximation schemes.

This paper demonstrates that hypergeometric functions are also powerful tools for the analytical representation of sequences of integers. Pell and Jacobsthal sequences are reconstructed from a hypergeometric function using classical transformations, series expansions, Binet formulas, and recurrence relations.

2. Materials and Methods

The gamma function is given as

$$\Gamma \left(\mathrm{y}\right)={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{x}}{\mathrm{x}}^{\mathrm{y}-1}\mathrm{d}\mathrm{x},\mathrm{y}>0$$

with the following properties

$$\Gamma \left(\mathrm{y}\right)=\left(\mathrm{y}-1\right)!,\mathrm{y}\in \mathbb{N}$$

$$\Gamma \left(\mathrm{y}+1\right)=\mathrm{y}\Gamma \left(\mathrm{y}\right),\mathrm{y}>0$$

$$\Gamma (\mathrm{y})\Gamma (1-\mathrm{y})={\displaystyle \frac{\pi }{\mathrm{s}\mathrm{i}\mathrm{n}\pi \mathrm{y}}}$$

(1)

$$\Gamma (\mathrm{y})\Gamma (\mathrm{y}+{\displaystyle \frac{1}{2}})=2^{1-2\mathrm{y}}\sqrt{\pi }\Gamma \left(2\mathrm{y}\right)$$

(2)

Given that m is a constant complex number and $n\in \mathbb{N}$, the Pochhammer symbol${\left(m\right)}_n$ is defined as

$${\left(m\right)}_n=m\left(m+1\right)\left(m+2\right)\dots \left(m+n-1\right)$$

(3)

The Pochhammer symbol is named after the 19th-century German mathematician, Leo August Pochhammer. It is used as a fundamental tool in numerous theoretical and applied studies involving hypergeometric functions, orthogonal polynomials, special number sequences and series solutions. We have

$${\left(m\right)}_0=\mathrm{and} {\left(1\right)}_n=n!$$

$${\left(m\right)}_n={\displaystyle \frac{\Gamma \left(m+n\right)}{\Gamma \left(m\right)}}$$

(4)

$${\left(m\right)}_{n+1}=m{\left(m+1\right)}_n$$

(5)

$${\left(m\right)}_{n-k}={\displaystyle \frac{{\left(m\right)}_n}{{\left(m+n-k\right)}_k}}$$

The Jacobsthal numbers $J_n$ create a special sequence of integers demonstrated recursively $J_0=0, J_1=1, J_n=J_{n-1}+2J_{n-2}, (n\ge 2)$. The Binet formula of the Jacobsthal sequence is $J_n={\displaystyle \frac{2^n-{(-1)}^n}{3}}$ [9].

The Pell numbers $P_n$ create a special sequence of integers demonstrated by $P_0=0, P_1=1, P_n=2P_{n-1}+P_{n-2}, \left(n\ge 2\right).$ Pell sequence is given as $P_n={\displaystyle \frac{1}{2\sqrt{2}}}\left({\left(1+\sqrt{2}\right)}^n-{\left(1-\sqrt{2}\right)}^n\right)$ [10].

Hypergeometric differential equation is defined as

$$t(1-t)y^{″}+\left[c-(a+b+1)t\right]y^{\prime }-aby=0$$

(6)

The solutions of the hypergeometric differential equation at x = 0 regular singular point are called hypergeometric functions:

$$y=AF\left(a,b,c;t\right)+B{t}^{1-c}F(a-c+1,b-c+\mathrm{1,2}-c;t)$$

where A and B any real numbers. The general solution is valid for $\left|x\right|<1$.

The representation of the hypergeometric function is:

$$F\left(a,b,c;t\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(a\right)}_n{\left(b\right)}_n}{{\left(c\right)}_{n}n!}}{t}^n.$$

Some functions can be expressed by hypergeometric function in the following:

$$F\left(\mathrm{1,1},1;t\right)={\displaystyle \frac{1}{1-t}}$$

$$F\left(\mathrm{1,1},1;t\right)={\displaystyle \frac{1}{1-t}}$$

$$F\left(\mathrm{1,2},1;t\right)={\displaystyle \frac{1}{{\left(1-t\right)}^2}}$$

$$F\left(-n,1;1;1-t\right){=t}^n n=\mathrm{0,1},2\dots ,$$

$$F\left(-n,b,b;-t\right)={\left(1+t\right)}^n$$

$$F\left(\mathrm{1,1},2;-t\right)={\displaystyle \frac{\ln \left(1+t\right)}{t}}$$

$$e^t=\underset{b\to \infty }{\lim }F\left(1,b,1;{\displaystyle \frac{t}{b}}\right).$$

Lemma 1

 [1]. Hypergeometric function holds the following property.

$$\mathrm{F}\left(\mathrm{a},{\displaystyle \frac{1}{2}}+\mathrm{a},{\displaystyle \frac{3}{2}};{\mathrm{t}}^2\right)={\displaystyle \frac{1}{2\mathrm{t}\left(1-2\mathrm{a}\right)}}\left[{\left(1+\mathrm{t}\right)}^{1-2\mathrm{a}}-{\left(1-\mathrm{t}\right)}^{1-2\mathrm{a}}\right].$$

(7)

Lemma 2

 [1]. Hypergeometric function holds the following property.

$$\mathrm{F}\left(\mathrm{a},\mathrm{b},\mathrm{c};\mathrm{t}\right)={\left(1-\mathrm{t}\right)}^{-\mathrm{a}}\mathrm{F}\left(\mathrm{a},\mathrm{c}-\mathrm{b},\mathrm{c};{\displaystyle \frac{\mathrm{t}}{\mathrm{t}-1}}\right).$$

(8)

Lemma 3

 [1]. The following property is used for the hypergeometric function.

$$\mathrm{F}\left(\mathrm{a},\mathrm{b},\mathrm{c};\mathrm{t}\right)={\left(1-\mathrm{t}\right)}^{-\mathrm{b}}\mathrm{F}\left(\mathrm{b},\mathrm{c}-\mathrm{a},\mathrm{c};{\displaystyle \frac{\mathrm{t}}{\mathrm{t}-1}}\right)$$

(9)

Lemma 4

  [1]. The following property is valid for the hypergeometric function.

$$\mathrm{F}\left(\mathrm{a},\mathrm{b},\mathrm{c};\mathrm{t}\right)={\left(1-\mathrm{t}\right)}^{\mathrm{c}-\mathrm{a}-\mathrm{b}}\mathrm{F}\left(\mathrm{c}-\mathrm{a},\mathrm{c}-\mathrm{b},\mathrm{c};\mathrm{t}\right).$$

(10)

Lemma 5

 [1]. The following property is true for the hypergeometric function.

$$\begin{gathered} \mathrm{F}\left(\mathrm{a},\mathrm{b},\mathrm{c};\mathrm{t}\right)={\left(1-t\right)}^{-a}{\displaystyle \frac{\Gamma \left(\mathrm{c}\right)\Gamma \left(\mathrm{b}-\mathrm{a}\right)}{\Gamma \left(\mathrm{c}-\mathrm{a}\right)\Gamma \left(\mathrm{b}\right)}}\mathrm{F}\left(\mathrm{a},\mathrm{c}-\mathrm{b},\mathrm{a}-\mathrm{b}+1;{\displaystyle \frac{1}{1-t}}\right) \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}+{\left(1-\mathrm{t}\right)}^{-\mathrm{b}}{\displaystyle \frac{\Gamma \left(\mathrm{c}\right)\Gamma \left(\mathrm{a}-\mathrm{b}\right)}{\Gamma \left(\mathrm{a}\right)\Gamma \left(\mathrm{c}-\mathrm{b}\right)}}\mathrm{F}(\mathrm{b},\mathrm{c}-\mathrm{a},\mathrm{b}-\mathrm{a}+1;{\displaystyle \frac{1}{1-t}}) \end{gathered}$$

(11)

Definition 1.

Let $n\ge 2$ be any integer: the first kind ${\left\{T_n\right\}}_{n>0}$ and second kind   ${\left\{U_n\right\}}_{n>0}$ Chebyshev polynomial sequences are defined by the following recurrence relations

$$T_n=2x{T}_{n-1}-T_{n-2}, T_0=1, T_1=x,$$

$$U_n=2x{U}_{n-1}-U_{n-2}, U_0=1, U_1=2x,$$

respectively.

Chebyshev polynomials are also defined by $T_n\left(cos\phi \right)=cosn\phi , U_n\left(cos\phi \right)={\displaystyle \frac{sinn\phi }{sin\phi }},$ $n\in Z^{+}, sin\phi \ne 0$[11].

Lemma 6

 [8]. Chebyshev polynomials are denoted using hypergeometric function as follows.

$$U_n(x)=(n+1)F(-n, n+2, {\displaystyle \frac{3}{2}}; {\displaystyle \frac{1-x}{2}})$$

(12)

Lemma 7

 [1]. The following property is used for the hypergeometric function.

$$\mathrm{F}(\mathrm{a},\mathrm{b},\mathrm{c};\mathrm{z})=\mathrm{F}(\mathrm{a},\mathrm{b},\mathrm{a}+\mathrm{b}+\mathrm{m};\mathrm{z})$$

(13)

Lemma 8

 [1]. Hypergeometric function holds the following identity.

$$\begin{gathered} \mathrm{F}(\mathrm{a},\mathrm{b},\mathrm{a}+\mathrm{b}+\mathrm{m};\mathrm{z})={\displaystyle \frac{\Gamma (\mathrm{a}+\mathrm{b}+\mathrm{m})\Gamma (\mathrm{m})}{\Gamma (\mathrm{b}+\mathrm{m})\Gamma (\mathrm{a}+\mathrm{m})}}\mathrm{F}(\mathrm{a},\mathrm{b},-\mathrm{m}+1;1-\mathrm{z}) \\ +{(1-\mathrm{z})}^{\mathrm{m}}{\displaystyle \frac{\Gamma (\mathrm{a}+\mathrm{b}+\mathrm{m})\Gamma (-\mathrm{m})}{\Gamma (\mathrm{a})\Gamma (\mathrm{b})}}\mathrm{F}(\mathrm{b}+\mathrm{m},\mathrm{a}+\mathrm{m},\mathrm{m}+1;1-\mathrm{z}) \end{gathered}$$

(14)

3. Main Results with the Jacobsthal Sequence

This section comprehensively covers the analytic relationships between hypergeometric functions and the Jacobsthal sequence similar to the connection between the Fibonacci sequence and hypergeometric series is based on Dilcher’s work.

Theorem 1.

The Jacobsthal sequence is denoted by hypergeometric function as

$$J_m=2^{{\displaystyle \frac{m-1}{2}}}{i}^{m-1}mF\left(1-m, 1+m, {\displaystyle \frac{3}{2}}; {\displaystyle \frac{2\sqrt{2}-i}{4\sqrt{2}}}\right)$$

where $i=\sqrt{-1}.$

Proof. 

The Jacobstral sequence is expressed by the second type of the Chebyshev polynomials.

$$J_m={\left(2i^2\right)}^{{\displaystyle \frac{m-1}{2}}}{U}_{m-1}\left({\displaystyle \frac{1}{2\sqrt{2}i}}\right).$$

(15)

Ref. [11]. If we substitute $\mathrm{n} = \mathrm{m} - 1 \mathrm{and} \mathrm{t} = {\displaystyle \frac{1}{2\sqrt{2}i}}$ in the representation of the second type of the Chebyshev polynomial with hypergeometric function in Equation (12), we get

$$\begin{gathered} U_{m-1}({\displaystyle \frac{1}{2\sqrt{2}i}})=(m-1+1)F\left(-(m-1), m-1+2, {\displaystyle \frac{3}{2}};{\displaystyle \frac{1-\frac{1}{2\sqrt{2}i}}{2}}\right) \\ =mF\left(1-m, 1+m, {\displaystyle \frac{3}{2}}; {\displaystyle \frac{2\sqrt{2}i-1}{4\sqrt{2}i}}\right) \end{gathered}$$

(16)

By Equation (15), the result is obtained. □

Theorem 2.

The mth element of the Jacobsthal sequence is expressed by hypergeometric function as

$$J_m=2^{{\displaystyle \frac{m-1}{2}}}{i}^{m-1}\sum _{k=0}^{m-1}{(-{\displaystyle \frac{i}{\sqrt{2}}}-2)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m+k}{2k+1}}\right)$$

Proof. 

By Theorem 1 and the expansion of the series of the hypergeometric function, the following is obtained

$$\begin{gathered} J_m=2^{{\displaystyle \frac{m-1}{2}}}{i}^{m-1}mF\left(1-m, 1+m, {\displaystyle \frac{3}{2}}; {\displaystyle \frac{2\sqrt{2}-i}{4\sqrt{2}}}\right) \\ =2^{{\displaystyle \frac{m-1}{2}}}{i}^{m-1}m\sum _{\mathrm{k}=0}^{\mathrm{\infty }}{\displaystyle \frac{{(-\mathrm{m}+1)}_{\mathrm{k}}{(\mathrm{m}+1)}_{\mathrm{k}}}{{\left(\frac{3}{2}\right)}_{\mathrm{k}}}}{\displaystyle \frac{{(\frac{2\sqrt{2}i-1}{4\sqrt{2}i})}^{\mathrm{k}}}{\mathrm{k}!}}. \end{gathered}$$

□

Let us compute the following equalities

$${(1-m)}_k={(-1)}^k{\displaystyle \frac{(m-1)!}{(m-1-k)!}},$$

(17)

$${(1+m)}_k={\displaystyle \frac{(m+k)!}{m!}}.$$

(18)

If the Equalities (17) and (18) are substituted in the series above, the following is obtained

$$J_m=2^{{\displaystyle \frac{m-1}{2}}}{i}^{m-1}\sum _{k=0}^{m-1}{\displaystyle \frac{{\left(-1\right)}^k\left(m+k\right)!}{\left(m-1-k\right)!}}{\displaystyle \frac{1}{\left(2k+1\right)!}}{\displaystyle \frac{{\left(2\sqrt{2}i-1\right)}^k}{{\left(\sqrt{2}i\right)}^k}} .$$

For $\mathrm{m}-1-\mathrm{k}\ge 0,$ we have

$$\begin{gathered} J_m=2^{{\displaystyle \frac{m-1}{2}}}{i}^{m-1}\sum _{k=0}^{m-1}{\left(-1\right)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m+k}{2k+1}}\right){\displaystyle \frac{{\left(2\sqrt{2}i-1\right)}^k}{{\left(\sqrt{2}i\right)}^k}} \\ =2^{{\displaystyle \frac{m-1}{2}}}{i}^{m-1}\sum _{k=0}^{m-1}{(-{\displaystyle \frac{i}{\sqrt{2}}}-2)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m+k}{2k+1}}\right) \end{gathered}$$

Theorem 3.

The element $J_{2m+1}$ of the Jacobsthal sequence is found by hypergeometric function as

$$J_{2m+1}={\left({\displaystyle \frac{2\sqrt{2}i+1}{4}}\right)}^{2m}\left(2m+1\right)F\left(-2m,-2m-{\displaystyle \frac{1}{2}}, {\displaystyle \frac{3}{2}};{\displaystyle \frac{1-2\sqrt{2}i}{2\sqrt{2}i+1}}\right).$$

Proof. 

In Theorem 1, we substitute 2m + 1 for m, we obtain

$$J_{2m+1}=2^m{i}^{2m}\left(2m+1\right)F\left(-2m, 2m+2, {\displaystyle \frac{3}{2}};{\displaystyle \frac{2\sqrt{2}i-1}{4\sqrt{2}i}}\right).$$

(19)

□

If we use Lemma 2, we have

$$\begin{gathered} \mathrm{F}\left(-2m, 2m+2, {\displaystyle \frac{3}{2}}; {\displaystyle \frac{2\sqrt{2}i-1}{4\sqrt{2}i}}\right)={\left({\displaystyle \frac{2\sqrt{2}i+1}{4\sqrt{2}i}}\right)}^{2\mathrm{m}}\mathrm{F}\left(-2m,-2m-{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}};{\displaystyle \frac{\frac{2\sqrt{2}i-1}{4\sqrt{2}i}}{\frac{2\sqrt{2}i-1}{4\sqrt{2}i}-1}}\right) \\ ={\left({\displaystyle \frac{2\sqrt{2}i+1}{4\sqrt{2}i}}\right)}^{2\mathrm{m}}\mathrm{F}\left(-2m,-2m-{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}};{\displaystyle \frac{1-2\sqrt{2}i}{2\sqrt{2}i+1}}\right). \end{gathered}$$

Then, we get the result by the above equality and Equation (19)

$$\begin{gathered} J_{2m+1}=2^m{i}^{2m}(2m+1){\left({\displaystyle \frac{2\sqrt{2}i+1}{4\sqrt{2}i}}\right)}^{2\mathrm{m}}\mathrm{F}\left(-2m,-2m-{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}};{\displaystyle \frac{1-2\sqrt{2}i}{2\sqrt{2}i+1}}\right) \\ ={\left(\sqrt{2}i\right)}^{2m}\left(2m+1\right){\left({\displaystyle \frac{2\sqrt{2}i+1}{4\sqrt{2}i}}\right)}^{2\mathrm{m}}\mathrm{F}\left(-2m,-2m-{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}};{\displaystyle \frac{1-2\sqrt{2}i}{2\sqrt{2}i+1}}\right). \end{gathered}$$

Theorem 4.

The 2m^th element of the Jacobstral sequence satisfies the following equality

$$J_{2m}={\left({\displaystyle \frac{8}{2\sqrt{2}i+1}}\right)}^{2\mathrm{m}+1}\mathrm{m}\mathrm{F}\left(1+2m,2m+{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}};{\displaystyle \frac{1-2\sqrt{2}i}{2\sqrt{2}i+1}}\right).$$

Proof. 

In Theorem 1, we substitute 2m for m:

$$J_{2m}=2^{{\displaystyle \frac{2m-1}{2}}}{i}^{2m-1}2mF\left(1-2m, 1+2m,{\displaystyle \frac{3}{2}};{\displaystyle \frac{2\sqrt{2}i-1}{4\sqrt{2}i}}\right).$$

(20)

□

If we use Lemma 3, we get

$$\begin{gathered} \mathrm{F}\left(1-2m,1+2m,{\displaystyle \frac{3}{2}};{\displaystyle \frac{2\sqrt{2}i-1}{4\sqrt{2}i}}\right) \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}={\left({\displaystyle \frac{2\sqrt{2}i+1}{4\sqrt{2}i}}\right)}^{-2\mathrm{m}-1}\mathrm{F}\left(1+2m,2m+{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}};{\displaystyle \frac{\frac{2\sqrt{2}i-1}{4\sqrt{2}i}}{\frac{2\sqrt{2}i-1}{4\sqrt{2}i}-1}}\right) \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}={\left({\displaystyle \frac{2\sqrt{2}i+1}{4\sqrt{2}i}}\right)}^{-2\mathrm{m}-1}\mathrm{F}\left(1+2m,2m+{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}};{\displaystyle \frac{2\sqrt{2}i-1}{-2\sqrt{2}i-1}}\right). \end{gathered}$$

By Equation (20) and after some basic operations, we obtained the proof:

$$\begin{array}{l}{J}_{2m}=2^{{\displaystyle \frac{2m-1}{2}}}{i}^{2m-1}2m{\left({\displaystyle \frac{2\sqrt{2}i+1}{4\sqrt{2}i}}\right)}^{-2\mathrm{m}-1}\mathrm{F}\left(1+2m,2m+{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}};{\displaystyle \frac{-2\sqrt{2}i+1}{2\sqrt{2}i+1}}\right)\hfill \\ =2^m{i}^{2m}{\displaystyle \frac{1}{\sqrt{2}i}}2m{\left({\displaystyle \frac{2\sqrt{2}i+1}{4\sqrt{2}i}}\right)}^{-2\mathrm{m}-1}\mathrm{F}\left(1+2m,2m+{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}};{\displaystyle \frac{-2\sqrt{2}i+1}{2\sqrt{2}i+1}}\right)\hfill \\ ={\sqrt{2}}^{2m}{i}^{2m}{\displaystyle \frac{1}{\sqrt{2}i}}2m{\left({\displaystyle \frac{4\sqrt{2}i}{2\sqrt{2}i+1}}\right)\left({\displaystyle \frac{4\sqrt{2}i}{2\sqrt{2}i+1}}\right)}^{2\mathrm{m}}\mathrm{F}\left(1+2m,2m+{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}};{\displaystyle \frac{-2\sqrt{2}i+1}{2\sqrt{2}i+1}}\right)\hfill \\ ={\displaystyle \frac{\sqrt{2}m}{i}}{\left({\displaystyle \frac{4\sqrt{2}i}{2\sqrt{2}i+1}}\right)\left({\displaystyle \frac{8}{2\sqrt{2}i+1}}\right)}^{2\mathrm{m}}\mathrm{F}\left(1+2m,2m+{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}};{\displaystyle \frac{-2\sqrt{2}i+1}{2\sqrt{2}i+1}}\right)\hfill \\ =m{\left({\displaystyle \frac{8}{2\sqrt{2}i+1}}\right)}^{2\mathrm{m}+1}\mathrm{F}\left(1+2m,2m+{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}};{\displaystyle \frac{-2\sqrt{2}i+1}{2\sqrt{2}i+1}}\right).\hfill \end{array}$$

Theorem 5.

The (2m+1)^th element of the Jacobsthal sequence can also be demonstrated as follows:

$$J_{2m+1}=\left(2\mathrm{m}+1\right){\left(\sqrt{2}i\right)}^{2\mathrm{m}}\sqrt{{\displaystyle \frac{4\sqrt{2}i}{2\sqrt{2}i+1}}}\mathrm{F}\left({\displaystyle \frac{3}{2}}+2m,-2m-{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}};{\displaystyle \frac{2\sqrt{2}i-1}{4\sqrt{2}i}}\right).$$

Proof. 

If the linear transformation Equation (10) is applied to the hypergeometric expression in Equation (19), we find

$$\begin{array}{c}F\left(-2m, 2m+2, {\displaystyle \frac{3}{2}};{\displaystyle \frac{2\sqrt{2}i-1}{4\sqrt{2}i}}\right)\\ ={\left(1-{\displaystyle \frac{2\sqrt{2}i-1}{4\sqrt{2}i}}\right)}^{{\displaystyle \frac{3}{2}}-(-2m)-(2m+2)}\mathrm{F}\left({\displaystyle \frac{3}{2}}-(-2m),{\displaystyle \frac{3}{2}}-(2m+2),{\displaystyle \frac{3}{2}};{\displaystyle \frac{2\sqrt{2}i-1}{4\sqrt{2}i}}\right)\\ ={\left({\displaystyle \frac{2\sqrt{2}i+1}{4\sqrt{2}i}}\right)}^{{\displaystyle \frac{-1}{2}}}\mathrm{F}\left(\frac{3}{2}+2m,-2m-{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}};{\displaystyle \frac{2\sqrt{2}i-1}{4\sqrt{2}i}}\right)\end{array}$$

□

If we use this Equality (19), we compute

$$\begin{array}{c}{J}_{2m+1}=2^m{i}^{2m}(2m+1){\left({\displaystyle \frac{2\sqrt{2}i+1}{4\sqrt{2}i}}\right)}^{{\displaystyle \frac{-1}{2}}}\mathrm{F}\left({\displaystyle \frac{3}{2}}+2m,-2m-{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}};{\displaystyle \frac{2\sqrt{2}i-1}{4\sqrt{2}i}}\right)\\ =\sqrt{{\displaystyle \frac{4\sqrt{2}i}{2\sqrt{2}i+1}}}(2\mathrm{m}+1){\left(\sqrt{2}i\right)}^{2\mathrm{m}}\mathrm{F}\left({\displaystyle \frac{3}{2}}+2m,-2m-{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}},{\displaystyle \frac{2\sqrt{2}i-1}{4\sqrt{2}i}}\right)\end{array}$$

Theorem 6.

The (2m)^th element of the Jacobsthal sequence is denoted using the hypergeometric function

$$J_{2m}=2\mathrm{m}{\left(\sqrt{2}i\right)}^{2\mathrm{m}-1}\sqrt{{\displaystyle \frac{4\sqrt{2}i}{2\sqrt{2}i+1}}}\mathrm{F}\left({\displaystyle \frac{1}{2}}+2m,-2m+{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}};{\displaystyle \frac{2\sqrt{2}i-1}{4\sqrt{2}i}}\right).$$

Proof. 

If the linear transformation Equation (10) is applied to the hypergeometric expression Equation (20), we find

$$\begin{array}{c}F\left(1-2m, 2m+1, {\displaystyle \frac{3}{2}};{\displaystyle \frac{2\sqrt{2}i-1}{4\sqrt{2}i}}\right)\\ ={\left(1-{\displaystyle \frac{2\sqrt{2}i-1}{4\sqrt{2}i}}\right)}^{{\displaystyle \frac{3}{2}}-\left(1-2m\right)-\left(2m+1\right)}\mathrm{F}\left({\displaystyle \frac{3}{2}}-\left(1-2m\right),{\displaystyle \frac{3}{2}}-\left(2m+1\right),{\displaystyle \frac{3}{2}};{\displaystyle \frac{2\sqrt{2}i-1}{4\sqrt{2}i}}\right)\\ ={\left({\displaystyle \frac{2\sqrt{2}i+1}{4\sqrt{2}i}}\right)}^{{\displaystyle \frac{-1}{2}}}\mathrm{F}\left(2m+{\displaystyle \frac{1}{2}},-2m+{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}};{\displaystyle \frac{2\sqrt{2}i-1}{4\sqrt{2}i}}\right).\end{array}$$

□

If we use this Equality (20), we evaluate

$$\begin{array}{c}{J}_{2m}=2^{m-{\displaystyle \frac{1}{2}}}{i}^{2m-1}2m{\left({\displaystyle \frac{2\sqrt{2}i+1}{4\sqrt{2}i}}\right)}^{{\displaystyle \frac{-1}{2}}}\mathrm{F}\left(2m+{\displaystyle \frac{1}{2}},-2m+{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}};{\displaystyle \frac{2\sqrt{2}i-1}{4\sqrt{2}i}}\right)\\ =\sqrt{{\displaystyle \frac{4\sqrt{2}i}{2\sqrt{2}i+1}}}\left(2\mathrm{m}\right){\left(\sqrt{2}i\right)}^{2\mathrm{m}-1}\mathrm{F}\left(2m+{\displaystyle \frac{1}{2}},-2m+{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}};{\displaystyle \frac{2\sqrt{2}i-1}{4\sqrt{2}i}}\right).\end{array}$$

Theorem 7.

The element $J_{2m+1}$ of the Jacobsthal sequence is computed as

$$J_{2m+1}={\left(4\sqrt{2}i-7\right)}^m\mathrm{F}\left(-2m,-{\displaystyle \frac{1}{2}}-2m,-4m-1;{\displaystyle \frac{4\sqrt{2}i}{1+2\sqrt{2}i}}\right).$$

Proof. 

If we apply the equality in Lemma 5 to the hypergeometric expression in Equation (19),

$$\begin{array}{c}F\left(-2m, 2m+2, {\displaystyle \frac{3}{2}}; {\displaystyle \frac{2\sqrt{2}i-1}{4\sqrt{2}i}}\right)\\ ={\left({\displaystyle \frac{1+2\sqrt{2}i}{4\sqrt{2}i}}\right)}^{2m}{\displaystyle \frac{\Gamma \left(\frac{3}{2}\right)\Gamma \left(2m+2+2m\right)}{\Gamma \left(\frac{3}{2}+2m\right)\Gamma \left(2m+2\right)}}\mathrm{F}\left(-2m,{\displaystyle \frac{3}{2}}-2m-2,-2m-2m-2+1;{\displaystyle \frac{4\sqrt{2}i}{1+2\sqrt{2}i}}\right)\\ +{\left({\displaystyle \frac{1+2\sqrt{2}i}{4\sqrt{2}i}}\right)}^{-2m-2}{\displaystyle \frac{\Gamma \left(\frac{3}{2}\right)\Gamma (-2m-2m-2)}{\Gamma (-2m)\Gamma \left(\frac{3}{2}-2m-2\right)}}\mathrm{F}\left(2m+2,{\displaystyle \frac{3}{2}}+2m,2m+2+2m+1;{\displaystyle \frac{4\sqrt{2}i}{1+2\sqrt{2}i}}\right)\\ ={\left({\displaystyle \frac{1+2\sqrt{2}i}{4\sqrt{2}i}}\right)}^{2m}{\displaystyle \frac{\Gamma \left(\frac{3}{2}\right)\Gamma (4m+2)}{\Gamma \left(\frac{3}{2}+2m\right)\Gamma (2m+2)}}\mathrm{F}\left(-2m,{\displaystyle \frac{-1}{2}}-2m,-4m-1;{\displaystyle \frac{4\sqrt{2}i}{1+2\sqrt{2}i}}\right)\\ +{\left({\displaystyle \frac{1+2\sqrt{2}i}{4\sqrt{2}i}}\right)}^{-2m-2}{\displaystyle \frac{\Gamma \left(\frac{3}{2}\right)\Gamma \left(-4m-2\right)}{\Gamma \left(-2m\right)\Gamma \left(\frac{-1}{2}-2m\right)}}\mathrm{F}\left(2m+2,{\displaystyle \frac{3}{2}}+2m,4\mathrm{m}+3;{\displaystyle \frac{4\sqrt{2}i}{1+2\sqrt{2}i}}\right).\end{array}$$

□

The second sum of the equality is undefined because of by$\Gamma \left(-\mathrm{m}\right)$. Therefore, we only use the first part. By Equation (2), we get the following equalities:

$$\Gamma \left(4m+2\right)={\displaystyle \frac{2^{1+4\mathrm{m}}}{\sqrt{\pi }}}\Gamma \left(2\mathrm{m}+1\right)\Gamma \left({\displaystyle \frac{3}{2}}+2\mathrm{m}\right),$$

$$\Gamma (2m+2)={\displaystyle \frac{2^{1+2\mathrm{m}}}{\sqrt{\pi }}}\Gamma (\mathrm{m}+1)\Gamma \left({\displaystyle \frac{3}{2}}+\mathrm{m}\right),$$

$$\Gamma (1+2\mathrm{m})={\displaystyle \frac{2^{2\mathrm{m}}}{\sqrt{\pi }}}\Gamma \left(m+1\right)\Gamma (m+{\displaystyle \frac{1}{2}})$$

By these equalities above, we have

$$\begin{array}{c}F\left(-2m, 2m+2, {\displaystyle \frac{3}{2}}; {\displaystyle \frac{2\sqrt{2}i-1}{4\sqrt{2}i}}\right)\\ ={\left({\displaystyle \frac{2\sqrt{2}i+1}{4\sqrt{2}i}}\right)}^{2m}{\displaystyle \frac{\frac{\sqrt{\pi }}{2}\frac{2^{4\mathrm{m}+1}}{\sqrt{\pi }}\frac{2^{2\mathrm{m}}}{\sqrt{\pi }}\Gamma \left(\frac{1}{2}+\mathrm{m}\right)\Gamma (m+1)\Gamma \left(2\mathrm{m}+\frac{3}{2}\right)}{\Gamma (\frac{3}{2}+2m)\frac{2^{2\mathrm{m}+1}}{\sqrt{\pi }}\Gamma (\mathrm{m}+1)\Gamma \left(\mathrm{m}+\frac{3}{2}\right)}}\mathrm{F}\left(-2m,-{\displaystyle \frac{1}{2}}-2m,-4m-1;{\displaystyle \frac{4\sqrt{2}i}{1+2\sqrt{2}i}}\right)\\ ={\left({\displaystyle \frac{1+2\sqrt{2}i}{4\sqrt{2}i}}\right)}^{2m}{\displaystyle \frac{\frac{\sqrt{\pi }}{2}\frac{2^{4\mathrm{m}+1}}{\sqrt{\pi }}\frac{2^{2\mathrm{m}}}{\sqrt{\pi }}}{\frac{2^{2\mathrm{m}+1}}{\sqrt{\pi }}\left(\mathrm{m}+\frac{1}{2}\right)}}\mathrm{F}\left(-2m,-{\displaystyle \frac{1}{2}}-2m,-4m-1;{\displaystyle \frac{4\sqrt{2}i}{1+2\sqrt{2}i}}\right)\end{array}$$

If we substitute the equality of $F(-2m, 2m+2, {\displaystyle \frac{3}{2}}; {\displaystyle \frac{2\sqrt{2}i-1}{4\sqrt{2}i}})$ in Equation (19), we have

$$\begin{array}{c}{J}_{2m+1}=2^m{i}^{2m}(2m+1){\left({\displaystyle \frac{1+2\sqrt{2}i}{4\sqrt{2}i}}\right)}^{2m}{\displaystyle \frac{\frac{\sqrt{\pi }}{2}\frac{2^{4\mathrm{m}+1}}{\sqrt{\pi }}\frac{2^{2\mathrm{m}}}{\sqrt{\pi }}}{\frac{2^{2\mathrm{m}+1}}{\sqrt{\pi }}\left(\mathrm{m}+\frac{1}{2}\right)}}\mathrm{F}\left(-2m,-{\displaystyle \frac{1}{2}}-2m,-4m-1;{\displaystyle \frac{4\sqrt{2}i}{1+2\sqrt{2}i}}\right)\\ =(2m+1){\left(4\sqrt{2}i-7\right)}^m{\displaystyle \frac{1}{2\mathrm{m}+1}}\mathrm{F}\left(-2m,-{\displaystyle \frac{1}{2}}-2m,-4m-1;{\displaystyle \frac{4\sqrt{2}i}{1+2\sqrt{2}i}}\right)\end{array}$$

Theorem 8.

The 2m^th element of the Jacobsthal sequence is computed using the hypergeometric function with the following equality.

$$\begin{array}{c}{J}_{2m}={\displaystyle \frac{1}{8^m}}{\displaystyle \frac{1}{1+2\sqrt{2}i}}{\left({\displaystyle \frac{1+2\sqrt{2}i}{4\sqrt{2}}}\right)}^{2m}\mathrm{F}\left(1-2m,{\displaystyle \frac{1}{2}}-2m,-4m+1;{\displaystyle \frac{4\sqrt{2}i}{1+2\sqrt{2}i}}\right)\\ -{\left({\displaystyle \frac{4\sqrt{2}}{1+2\sqrt{2}i}}\right)}^{2\mathrm{m}}\mathrm{F}\left(1+2m,{\displaystyle \frac{1}{2}}+2\mathrm{m},4m+1;{\displaystyle \frac{4\sqrt{2}i}{1+2\sqrt{2}i}}\right).\end{array}$$

Proof. 

If we apply Lemma 5 to the hypergeometric expression in Equation (20), we get

$$\begin{array}{c}F\left(1-2m, 1+2m,{\displaystyle \frac{3}{2}};{\displaystyle \frac{2\sqrt{2}i-1}{4\sqrt{2}i}}\right)\\ ={\left({\displaystyle \frac{2\sqrt{2}i+1}{4\sqrt{2}i}}\right)}^{2m-1}{\displaystyle \frac{\Gamma \left(\frac{3}{2}\right)\Gamma (2m+2\mathrm{m})}{\Gamma \left(\frac{3}{2}-1+2m\right)\Gamma (1+2m)}}\mathrm{F}\left(1-2m,{\displaystyle \frac{3}{2}}-1-2m,1-2m-1-2m+1;{\displaystyle \frac{4\sqrt{2}i}{1+2\sqrt{2}i}}\right)\\ +{\left({\displaystyle \frac{2\sqrt{2}i+1}{4\sqrt{2}i}}\right)}^{-1-2\mathrm{m}}{\displaystyle \frac{\Gamma \left(\frac{3}{2}\right)\Gamma (-2m-2m)}{\Gamma (1-2m)\Gamma \left(\frac{3}{2}-1-2m\right)}}\mathrm{F}\left(1+2m,{\displaystyle \frac{3}{2}}-1+2\mathrm{m},1+2m-1+2\mathrm{m}+1;{\displaystyle \frac{4\sqrt{2}i}{1+2\sqrt{2}i}}\right)\\ ={\left({\displaystyle \frac{2\sqrt{2}i+1}{4\sqrt{2}i}}\right)}^{2m-1}{\displaystyle \frac{\Gamma \left(\frac{3}{2}\right)\Gamma (4m)}{\Gamma \left(\frac{1}{2}+2m\right)\Gamma (1+2m)}}\mathrm{F}\left(1-2m,{\displaystyle \frac{1}{2}}-2m,-4m+1;{\displaystyle \frac{4\sqrt{2}i}{1+2\sqrt{2}i}}\right)\\ +{\left({\displaystyle \frac{2\sqrt{2}i+1}{4\sqrt{2}i}}\right)}^{-1-2\mathrm{m}}{\displaystyle \frac{\Gamma \left(\frac{3}{2}\right)\Gamma (-4m)}{\Gamma (1-2m)\left(\frac{1}{2}-2m\right)}}\mathrm{F}\left(1+2m,{\displaystyle \frac{1}{2}}+2\mathrm{m},4m+1;{\displaystyle \frac{4\sqrt{2}i}{1+2\sqrt{2}i}}\right)\end{array}$$

(21)

□

By Equation (2), we get $\Gamma \left(4m\right)={\displaystyle \frac{2^{4\mathrm{m}-1}}{\sqrt{\pi }}}\Gamma \left(2\mathrm{m}\right)\Gamma \left({\displaystyle \frac{1}{2}}+2\mathrm{m}\right)$ and $\Gamma \left(-4m\right)={\displaystyle \frac{2^{-4\mathrm{m}-1}}{\sqrt{\pi }}}\Gamma \left(-2\mathrm{m}\right)\Gamma \left({\displaystyle \frac{1}{2}}-2\mathrm{m}\right).$

By that property $\Gamma \left(\mathrm{y}+1\right)=\mathrm{y}\Gamma \left(\mathrm{y}\right),$ we have $\Gamma \left({\displaystyle \frac{3}{2}}\right)={\displaystyle \frac{\sqrt{\pi }}{2}},\Gamma \left(1-2m\right)=-2\mathrm{m}\Gamma \left(-2\mathrm{m}\right),$ and $\Gamma \left(2m+1\right)=2\mathrm{m}\Gamma \left(2\mathrm{m}\right).$

If we substitute these values in Equation (21), we obtain

$$\begin{array}{c}F(1-2m, 1+2m,{\displaystyle \frac{3}{2}};{\displaystyle \frac{2\sqrt{2}i-1}{4\sqrt{2}i}})\\ ={\left({\displaystyle \frac{2\sqrt{2}i+1}{4\sqrt{2}i}}\right)}^{2m-1}{\displaystyle \frac{\frac{\sqrt{\pi }}{2}\frac{2^{4\mathrm{m}-1}}{\sqrt{\pi }}\Gamma \left(2\mathrm{m}\right)\Gamma \left(2m+\frac{1}{2}\right)}{\Gamma \left(\frac{1}{2}+2m\right)2\mathrm{m}\Gamma \left(2\mathrm{m}\right)}}\mathrm{F}\left(1-2m,{\displaystyle \frac{1}{2}}-2m,-4m+1;{\displaystyle \frac{4\sqrt{2}i}{2\sqrt{2}i+1}}\right)\\ +{\left({\displaystyle \frac{1+2\sqrt{2}i}{4\sqrt{2}i}}\right)}^{-1-2\mathrm{m}}{\displaystyle \frac{\frac{\sqrt{\pi }}{2}\frac{2^{-4\mathrm{m}-1}}{\sqrt{\pi }}\Gamma \left(\frac{1}{2}-2m\right)\Gamma (-2\mathrm{m})}{-2\mathrm{m}\Gamma (-2\mathrm{m})\Gamma \left(\frac{1}{2}-2m\right)}}\mathrm{F}\left(1+2m,{\displaystyle \frac{1}{2}}+2\mathrm{m},4m+1;{\displaystyle \frac{4\sqrt{2}i}{1+2\sqrt{2}i}}\right)\\ ={\left({\displaystyle \frac{1+2\sqrt{2}i}{4\sqrt{2}i}}\right)}^{2m-1}{\displaystyle \frac{\frac{\sqrt{\pi }}{2}\frac{2^{4\mathrm{m}-1}}{\sqrt{\pi }}\Gamma (2\mathrm{m})\Gamma \left(\frac{1}{2}+2m\right)}{\Gamma \left(\frac{1}{2}-2m\right)2\mathrm{n}\Gamma (2\mathrm{n})}}\mathrm{F}\left(1-2m,{\displaystyle \frac{1}{2}}-2m,-4m+1;{\displaystyle \frac{4\sqrt{2}i}{1+2\sqrt{2}i}}\right)\\ +{\left({\displaystyle \frac{1+2\sqrt{2}i}{4\sqrt{2}i}}\right)}^{-1-2\mathrm{m}}{\displaystyle \frac{(\frac{\sqrt{\pi }}{2}\frac{2^{-4\mathrm{m}-1}}{\sqrt{\pi }})\Gamma (\frac{1}{2}-2\mathrm{m})\Gamma (-2\mathrm{m})}{-2\mathrm{m}\Gamma (-2\mathrm{m})\Gamma (\frac{1}{2}-2m)}}\mathrm{F}\left(1+2m,{\displaystyle \frac{1}{2}}+2\mathrm{m},4m+1;{\displaystyle \frac{4\sqrt{2}i}{1+2\sqrt{2}i}}\right).\end{array}$$

If we substitute these equalities in Equation (20), we get the following equality for $J_{2m}$

$$\begin{array}{c}{J}_{2m}=2^{{\displaystyle \frac{2m-1}{2}}}{i}^{2m-1}2m{\left({\displaystyle \frac{1+2\sqrt{2}i}{4\sqrt{2}i}}\right)}^{2m-1}{\displaystyle \frac{\frac{\sqrt{\pi }}{2}\frac{2^{4\mathrm{m}-1}}{\sqrt{\pi }}\Gamma \left(\frac{1}{2}+2m\right)\Gamma \left(2\mathrm{m}\right)}{\Gamma \left(\frac{1}{2}+2m\right)2\mathrm{m}\Gamma \left(2\mathrm{m}\right)}}\mathrm{F}\left(1-2m,{\displaystyle \frac{1}{2}}-2m,-4m+1;{\displaystyle \frac{4\sqrt{2}i}{2\sqrt{2}i+1}}\right)\\ +{\left({\displaystyle \frac{1+2\sqrt{2}i}{4\sqrt{2}i}}\right)}^{-1-2\mathrm{m}}{\displaystyle \frac{\frac{\sqrt{\pi }}{2}\frac{2^{-4\mathrm{m}-1}}{\sqrt{\pi }}\Gamma \left(\frac{1}{2}-2m\right)\Gamma (-2\mathrm{m})}{-2\mathrm{m}\Gamma (-2\mathrm{m})\Gamma \left(\frac{1}{2}-2m\right)}}\mathrm{F}\left(1+2m,{\displaystyle \frac{1}{2}}+2\mathrm{m},4m+1;{\displaystyle \frac{4\sqrt{2}i}{1+2\sqrt{2}i}}\right)\\ ={\displaystyle \frac{1}{8^{\mathrm{m}}}}{\displaystyle \frac{1}{2\sqrt{2}i+1}}{\left({\displaystyle \frac{1+2\sqrt{2}i}{4\sqrt{2}}}\right)}^{2m}\mathrm{F}\left(1-2m,{\displaystyle \frac{1}{2}}-2m,-4m+1;{\displaystyle \frac{4\sqrt{2}i}{1+2\sqrt{2}i}}\right)\\ -{\left({\displaystyle \frac{4\sqrt{2}}{1+2\sqrt{2}i}}\right)}^{2\mathrm{m}}\mathrm{F}\left(1+2m,{\displaystyle \frac{1}{2}}+2\mathrm{m},4m+1;{\displaystyle \frac{4\sqrt{2}i}{1+2\sqrt{2}i}}\right)\end{array}$$

4. Main Results with the Pell Sequence

This section comprehensively covers the analytic relationships between hypergeometric functions and the Pell sequence similar to the connection between the Fibonacci sequence and hypergeometric series is based on Dilcher’s work.

Theorem 9.

The Pell numbers verifies the following result.

$$P_m=mF\left({\displaystyle \frac{1-m}{2}}, 1-{\displaystyle \frac{m}{2}}, {\displaystyle \frac{3}{2}}; 2\right).$$

Proof. 

The Binet formula of the Pell sequence is $P_n={\displaystyle \frac{{(1+\sqrt{2})}^m-{(1-\sqrt{2})}^m}{2\sqrt{2}}}.$ If we substitute $1-2\mathrm{a}=\mathrm{m} \mathrm{a}\mathrm{n}\mathrm{d} \mathrm{t}=\sqrt{2}$ in Equation (7), we get

$$\mathrm{F}({\displaystyle \frac{1-\mathrm{m}}{2}},1-{\displaystyle \frac{\mathrm{m}}{2}},{\displaystyle \frac{3}{2}};2)={\displaystyle \frac{1}{2\sqrt{2}\mathrm{m}}}\left[{(1+\sqrt{2})}^{\mathrm{m}}-{(1-2\sqrt{2})}^{\mathrm{m}}\right]$$

□

Theorem 10.

The (2m+1)^th element of the Pell sequence is denoted using the hypergeometric function as follows:

$$P_{2m+1}=\left(2m+1\right){\left(-1\right)}^{m}F\left(-m,1+m,{\displaystyle \frac{3}{2}};2\right).$$

Proof. 

We know that by Lemma 2

$$\mathrm{F}\left(\mathrm{a},\mathrm{b},\mathrm{c};\mathrm{z}\right)={\left(1-\mathrm{z}\right)}^{-\mathrm{a}}\mathrm{F}\left(\mathrm{a},\mathrm{c}-\mathrm{b},\mathrm{c};{\displaystyle \frac{\mathrm{z}}{\mathrm{z}-1}}\right).$$

□

By Theorem 9, we get

$$P_{2m+1}=\left(2m+1\right)F\left({\displaystyle \frac{1-2m-1}{2}},1-{\displaystyle \frac{2m+1}{2}},{\displaystyle \frac{3}{2}};2\right).$$

By Equation (8), we have

$$P_{2m+1}=\left(2m+1\right)F\left(-m,{\displaystyle \frac{1}{2}}-m,{\displaystyle \frac{3}{2}};2\right)=\left(2m+1\right){\left(-1\right)}^{m}F\left(-m,1+m,{\displaystyle \frac{3}{2}};2\right).$$

Theorem 11.

The 2m^th element of the Pell sequence is demonstrated with the hypergeometric function.

$$P_{2m}={\left(-1\right)}^{m}2mF\left(1-m,1+m,{\displaystyle \frac{3}{2}};2\right).$$

Proof. 

By Theorem 9 and Equation (9), the following is satisfied.

$$\begin{gathered} P_{2m}=2mF\left({\displaystyle \frac{1}{2}}-m,1-m,{\displaystyle \frac{3}{2}};2\right) \\ ={\left(-1\right)}^{m}2mF\left(1-m,1+m,{\displaystyle \frac{3}{2}};2\right). \end{gathered}$$

□

Theorem 12.

The Pell sequence also holds the following representation.

$$P_m=2^{m-1}F\left({\displaystyle \frac{1-m}{2}}, {\displaystyle \frac{2-m}{2}}, 1-m;-1\right).$$

Proof. 

By Equation (13), we know that $\mathrm{F}\left(\mathrm{a},\mathrm{b},\mathrm{a}+\mathrm{b}+\mathrm{m};\mathrm{z}\right)=\mathrm{F}(\mathrm{a},\mathrm{b},\mathrm{c};\mathrm{z})$. if we choose $\mathrm{a}={\displaystyle \frac{1-\mathrm{m}}{2}},\mathrm{b}={\displaystyle \frac{2-\mathrm{m}}{2}},\mathrm{a}+\mathrm{b}+\mathrm{m}={\displaystyle \frac{3}{2}}, \mathrm{a}\mathrm{n}\mathrm{d} \mathrm{t}=2$ in Theorem 9, we get

$$P_m=mF\left({\displaystyle \frac{1-m}{2}},{\displaystyle \frac{2-m}{2}}, {\displaystyle \frac{3}{2}}; 2\right).$$

□

By Equation (14), we have

$$\begin{gathered} P_m=mF\left({\displaystyle \frac{1-m}{2}}, {\displaystyle \frac{2-m}{2}}, {\displaystyle \frac{3}{2}}; 2\right) \\ =m{\displaystyle \frac{\Gamma (\frac{3}{2})\Gamma (\mathrm{m})}{\Gamma (\frac{2+\mathrm{m}}{2})\Gamma (\frac{1+\mathrm{m}}{2})}}\left({\displaystyle \frac{1-m}{2}}, {\displaystyle \frac{2-m}{2}}, 1-m;-1\right). \end{gathered}$$

By the property of the gamma function Equation (2), the result is obtained as

$$P_m=m{\displaystyle \frac{\frac{1}{2}\sqrt{\pi }\frac{2^{\mathrm{m}-1}}{\sqrt{\pi }}\Gamma \left(\frac{\mathrm{m}}{2}\right)\Gamma \left(\frac{\mathrm{m}+1}{2}\right)}{\frac{\mathrm{m}}{2}\Gamma \left(\frac{\mathrm{m}}{2}\right)\Gamma \left(\frac{1+\mathrm{m}}{2}\right)}}F\left({\displaystyle \frac{1-m}{2}}, {\displaystyle \frac{2-m}{2}}, 1-m;-1\right).$$

Theorem 13.

The (2m+1)^th element of the Pell sequence is expressed by the following equality:

$$P_{2m+1}=(2m+1)F\left(-m, 1+m, {\displaystyle \frac{1}{2}};-1\right)$$

Proof. 

We use Lemmas 7 and 8 for the proof. In Theorem 10, if we use first sum of Lemma 7 and choose $\mathrm{a}=-\mathrm{m}$, $\mathrm{b}=1+\mathrm{m}$, $\mathrm{z}=2$ ve $\mathrm{a}+\mathrm{b}+\mathrm{m}={\displaystyle \frac{3}{2}}$, we get

$$\begin{gathered} P_{2m+1}=(2m+1){(-1)}^m{\displaystyle \frac{\Gamma \left(\frac{1}{2}\right)\Gamma \left(\frac{3}{2}\right)}{\Gamma \left(\frac{3}{2}+m\right)\Gamma \left(\frac{1}{2}-\mathrm{m}\right)}}F\left(-m, 1+m, {\displaystyle \frac{1}{2}};-1\right) \\ =(2m+1){\left(-1\right)}^m{\displaystyle \frac{\sqrt{\pi }\frac{1}{2}\sqrt{\pi }}{\frac{1}{2}\Gamma \left(\frac{1}{2}+\mathrm{m}\right)\Gamma \left(\frac{1}{2}-\mathrm{m}\right)}}F\left(-m, 1+m, {\displaystyle \frac{1}{2}};-1\right) \\ =\left(2m+1\right){\left(-1\right)}^m{\displaystyle \frac{\pi }{\frac{\pi }{cosm\pi }}}F\left(-m, 1+m, {\displaystyle \frac{1}{2}};-1\right). \end{gathered}$$

□

Theorem 14.

The 2m^th element of the Pell sequence is given as follows:

$$P_{2m}=-2mF\left(1-m, 1+m, {\displaystyle \frac{3}{2}};-1\right).$$

Proof. 

We use Lemmas 7 and 8 for the proof. In Theorem 11, if we use second sum of Lemma 7 and choose $\mathrm{a}=1-\mathrm{m}$, $\mathrm{b}=1+\mathrm{m}$, $\mathrm{z}=2$ ve $\mathrm{a}+\mathrm{b}+\mathrm{m}={\displaystyle \frac{3}{2}}$, we get

$$P_{2m}=(2m){(-1)}^n{\displaystyle \frac{\Gamma \left(-\frac{1}{2}\right)\Gamma \left(\frac{3}{2}\right)}{\frac{1}{2}\Gamma \left(\frac{1}{2}+\mathrm{m}\right)\Gamma \left(\frac{1}{2}-\mathrm{m}\right)}}F\left(1-m, 1+m, {\displaystyle \frac{3}{2}};-1\right)$$

□

By the property of the gamma function Equation (2), the result is obtained as

$$=\left(2m\right){\left(-1\right)}^m{\displaystyle \frac{-2\sqrt{\pi }\frac{1}{2}\sqrt{\pi }}{\frac{\pi }{cosm\pi }}}F\left(1-m, 1+m, {\displaystyle \frac{3}{2}};-1\right).$$

Theorem 15.

The 2m^th element of the Pell sequence is denoted as

$$P_{2m}=2mF\left(1+m, {\displaystyle \frac{1}{2}}+m, {\displaystyle \frac{3}{2}}; 2\right).$$

Proof. 

In Theorem 9, we substitute 2m for m, we obtain

$$P_{2m}=2mF\left({\displaystyle \frac{1}{2}}-m, 1-m, {\displaystyle \frac{3}{2}}; 2\right).$$

□

By Lemma 4, we get

$$\begin{gathered} \mathrm{F}({\displaystyle \frac{1}{2}}-m, 1-m, {\displaystyle \frac{3}{2}}; 2){=(-1)}^{{\displaystyle \frac{3}{2}}-{\displaystyle \frac{1-2m}{2}}-(1-m)}\mathrm{F}\left({\displaystyle \frac{3}{2}}-{\displaystyle \frac{1}{2}}-m,{\displaystyle \frac{3}{2}}-(1-m),{\displaystyle \frac{3}{2}};2\right) \\ {=(-1)}^{2\mathrm{m}}\mathrm{F}\left(1+m,{\displaystyle \frac{1}{2}}+\mathrm{m},{\displaystyle \frac{3}{2}};2\right) \end{gathered}$$

Theorem 16.

The (2m+1)^th element of the Pell sequence is demonstrated as

$$P_{2m+1}=\left(2m+1\right)\mathrm{F}\left({\displaystyle \frac{3}{2}}+m,2+m,{\displaystyle \frac{3}{2}};2\right).$$

Proof. 

In Theorem 9, we substitute 2m+1 for m, we get

$$P_{2m+1}=\left(2m+1\right)F\left(-m,-m-{\displaystyle \frac{1}{2}}, {\displaystyle \frac{3}{2}}; 2\right).$$

□

If we use Equation (10) for this equation, we establish

$$\begin{gathered} \mathrm{F}(-m,-m-{\displaystyle \frac{1}{2}}, {\displaystyle \frac{3}{2}}; 2){=(-1)}^{{\displaystyle \frac{3}{2}}+\mathrm{m}-(-m-{\displaystyle \frac{1}{2}})}\mathrm{F}\left({\displaystyle \frac{3}{2}}+m,{\displaystyle \frac{3}{2}}-(-m-{\displaystyle \frac{1}{2}}),{\displaystyle \frac{3}{2}};2\right) \\ {=(-1)}^{2\mathrm{m}+2}\mathrm{F}\left({\displaystyle \frac{3}{2}}+m,2+m,{\displaystyle \frac{3}{2}};2\right) \end{gathered}$$

Hence,

$$P_{2m+1}=(2m+1)F\left(-m,-m-{\displaystyle \frac{1}{2}}, {\displaystyle \frac{3}{2}}; 2\right)=\left(2m+1\right)\mathrm{F}\left({\displaystyle \frac{3}{2}}+m,2+m,{\displaystyle \frac{3}{2}};2\right).$$

Theorem 17.

The 2m^th element of the Pell sequence is found as

$$P_{2m}=2m\mathrm{F}\left(1-m,1+m,{\displaystyle \frac{3}{2}};-1\right).$$

Proof. 

If 2m is written instead of m in the equation of Theorem 9, we get

$$P_{2m}=2mF\left({\displaystyle \frac{1}{2}}-\mathrm{m}, 1-m, {\displaystyle \frac{3}{2}}; 2\right).$$

□

If we use Lemma 5 for this equality, we get

$$\begin{array}{c}F\left({\displaystyle \frac{1-2m}{2}}, 1-m, {\displaystyle \frac{3}{2}};2\right)\\ ={(-1)}^{{\displaystyle \frac{2m-1}{2}}}{\displaystyle \frac{\Gamma \left(1-m-\frac{1-2m}{2}\right)\Gamma \left(\frac{3}{2}\right)}{\Gamma \left(\frac{3}{2}-\frac{1-2m}{2}\right)\Gamma (1-m)}}\mathrm{F}\left({\displaystyle \frac{1-2m}{2}},{\displaystyle \frac{3}{2}}-(1-m),{\displaystyle \frac{1}{2}}-\mathrm{m}-(1-m)+1;-1\right)\\ +{(-1)}^{-(1-m)}{\displaystyle \frac{\Gamma \left(\frac{1}{2}-\mathrm{m}-(1-m)\right)\Gamma \left(\frac{3}{2}\right)}{\Gamma \left(\frac{1-2m}{2}\right)\Gamma \left(\frac{3}{2}-(1-m)\right)}}\mathrm{F}\left(1-m,{\displaystyle \frac{3}{2}}-{\displaystyle \frac{1-2m}{2}},(1-m)-{\displaystyle \frac{1-2m}{2}}+1;-1\right)\end{array}$$

If we substitute it in the above equality for $P_{2m}$, we get

$$\begin{gathered} P_{2m}=2m{(-1)}^{m-{\displaystyle \frac{1}{2}}}{\displaystyle \frac{\Gamma \left(\frac{1}{2}\right)\Gamma \left(\frac{3}{2}\right)}{\Gamma (1+m)\Gamma (1-m)}}\mathrm{F}\left({\displaystyle \frac{1}{2}}-\mathrm{m},\mathrm{m}+{\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{2}};-1\right) \\ +2m{\left(-1\right)}^{\mathrm{m}-1}{\displaystyle \frac{\Gamma \left(\frac{3}{2}\right)\Gamma \left(-\frac{1}{2}\right)}{\Gamma \left(\frac{1}{2}+m\right)\Gamma \left(\frac{1}{2}-\mathrm{m}\right)}}\mathrm{F}\left(1-m,1+m,{\displaystyle \frac{3}{2}};-1\right). \end{gathered}$$

By$\Gamma \left(1-\mathrm{m}\right),$ the first part of the equality is undefined. Therefore, we use only the second part.

We know that $\Gamma \left({\displaystyle \frac{-1}{2}}\right)=-2\sqrt{\pi }$, $\Gamma \left({\displaystyle \frac{3}{2}}\right)={\displaystyle \frac{\sqrt{\pi }}{2}}$ and by (2.1), we have $\Gamma ({\displaystyle \frac{1}{2}}-\mathrm{m})\Gamma ({\displaystyle \frac{1}{2}}+\mathrm{m})={\displaystyle \frac{\pi }{\mathrm{c}\mathrm{o}\mathrm{s}(\mathrm{m}\pi )}}$. Hence,

$$\begin{gathered} P_{2m}=2m{(-1)}^{\mathrm{m}-1}{\displaystyle \frac{-2\sqrt{\pi }\frac{\sqrt{\pi }}{2}}{\frac{\pi }{\mathrm{c}\mathrm{o}\mathrm{s}(\mathrm{m}\pi )}}}\mathrm{F}\left(1-m,1+m,{\displaystyle \frac{3}{2}};-1\right) \\ =2m\mathrm{F}\left(1-m,1+m,{\displaystyle \frac{3}{2}};-1\right). \end{gathered}$$

Theorem 18.

The (2m+1)^th element of the Pell sequence is expressed by hypergeometric function as

$$P_{2m+1}=\left(2m+1\right)\mathrm{F}\left(-\mathrm{m},2+\mathrm{m},{\displaystyle \frac{3}{2}};-1\right).$$

Proof. 

If we use Lemma 5 to the hypergeometric expression in Theorem 16, we obtain

$$\begin{array}{c}{P}_{2m+1}=\left(2m+1\right)F(-m,-m-{\displaystyle \frac{1}{2}}, {\displaystyle \frac{3}{2}}; 2)\\ =\left(2m+1\right){(-1)}^m{\displaystyle \frac{\Gamma \left(-m-\frac{1}{2}+\mathrm{m}\right)\Gamma \left(\frac{3}{2}\right)}{\Gamma \left(\frac{3}{2}+m\right)\Gamma \left(-m-\frac{1}{2}\right)}}\mathrm{F}\left(-\mathrm{m},{\displaystyle \frac{3}{2}}-(-m-{\displaystyle \frac{1}{2}}),-\mathrm{m}-(-m-{\displaystyle \frac{1}{2}})+1;-1\right)\\ +{\left(2m+1\right)(-1)}^{\mathrm{m}+{\displaystyle \frac{1}{2}}}{\displaystyle \frac{\Gamma \left(-\mathrm{m}-(-m-\frac{1}{2})\right)\Gamma \left(\frac{3}{2}\right)}{\Gamma (-\mathrm{m})\Gamma \left(\frac{3}{2}-(-m-\frac{1}{2})\right)}}\mathrm{F}\left(-m-{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}}+\mathrm{m},-m-{\displaystyle \frac{1}{2}}+\mathrm{m}+1;-1\right)\\ =\left(2m+1\right){(-1)}^m{\displaystyle \frac{\Gamma \left(\frac{-1}{2}\right)\Gamma \left(\frac{3}{2}\right)}{\Gamma \left(\frac{3}{2}+m\right)\Gamma \left(-m-\frac{1}{2}\right)}}\mathrm{F}\left(-\mathrm{m},2+\mathrm{m},{\displaystyle \frac{3}{2}};-1\right)\\ +\left(2m+1\right){\left(-1\right)}^{\mathrm{m}+{\displaystyle \frac{1}{2}}}{\displaystyle \frac{\Gamma \left(\frac{1}{2}\right)\Gamma \left(\frac{3}{2}\right)}{\Gamma \left(-\mathrm{m}\right)\Gamma \left(2+\mathrm{m}\right)}}\mathrm{F}\left(-m-{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}}+\mathrm{m},{\displaystyle \frac{1}{2}};-1\right).\end{array}$$

□

By$\Gamma \left(-\mathrm{m}\right),$ the second part of the equality is undefined. Therefore, we use only the first part. We know that $\Gamma ({\displaystyle \frac{-1}{2}})=-2\sqrt{\pi }$, $\Gamma ({\displaystyle \frac{3}{2}})={\displaystyle \frac{\sqrt{\pi }}{2}}$ and by Equation (1), $\Gamma (-{\displaystyle \frac{1}{2}}-\mathrm{m})\Gamma ({\displaystyle \frac{3}{2}}+\mathrm{m})=-{\displaystyle \frac{\pi }{\mathrm{c}\mathrm{o}\mathrm{s}(\mathrm{m}\pi )}}$. Therefore, the following is satisfied:

$$\begin{gathered} P_{2m+1}=(2m{+1\left)\right(-1)}^{\mathrm{m}}{\displaystyle \frac{-\pi }{-\frac{\pi }{\mathrm{c}\mathrm{o}\mathrm{s}(\mathrm{m}\pi )}}}\mathrm{F}\left(-\mathrm{m},2+\mathrm{m},{\displaystyle \frac{3}{2}};-1\right) \\ =\left(2m+1\right)\mathrm{F}\left(-\mathrm{m},2+\mathrm{m},{\displaystyle \frac{3}{2}};-1\right) \end{gathered}$$

5. Discussion and Conclusions

The results demonstrate that hypergeometric functions are not used in special function theory but also a powerful expression tool in areas such as integer sequences, combinatorics, and numerical analysis. The proposed representations provide analytical solutions that can be used in theoretical analysis and structures that can facilitate algorithmic computations.

The results obtained here demonstrate some equalities between hypergeometric functions and Pell, Jacobsthal number sequences, and suggest that further research is possible in this area. While Pell and Jacobsthal sequences are examined in detail, broader structures such as Horadam family are also open to investigation in terms of hypergeometric representations. Similar transformations can be developed for both classical and q-analog versions of these sequences.

Most of the proposed representations have been evaluated analytically. However, the numerical equivalents of these structures have not been investigated thoroughly. Future work could consider integrating the proposed representations into approximate computation methods, algorithmic typesetting, and computerized proof systems. Furthermore, the usability of the resulting transformations in applied fields such as cryptography, signal processing, and combinatorial structure analysis can be investigated.

The hypergeometric representations obtained in this study provide analytically explicit forms that are particularly well-suited for algorithmic and symbolic computation. Since hypergeometric functions are natively supported in computer algebra systems, such as Maple, Mathematica and the derived formulas can be implemented to automate identity verification, series transformations, and recurrence evaluations. In symbolic computation, these representations facilitate closed-form simplifications, asymptotic expansions, and exact summation procedures that are difficult to obtain using recurrence relations alone. From an applied perspective, the availability of hypergeometric expressions enables efficient approximation schemes, numerical evaluation with controlled error bounds, and potential applications in areas such as cryptography, signal processing, and discrete dynamical systems, where structured integer sequences and their analytic transformations play an essential role.