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BY 4.0 license Open Access Published by De Gruyter Open Access December 17, 2022

On the reciprocal sum of the fourth power of Fibonacci numbers

  • WonTae Hwang , Jong-Do Park EMAIL logo and Kyunghwan Song
From the journal Open Mathematics

Abstract

Let f n be the n th Fibonacci number with f 1 = f 2 = 1 . Recently, the exact values of k = n 1 f k s 1 have been obtained only for s = 1 , 2 , where x is the floor function. It has been an open problem for s 3 . In this article, we consider the case of s = 4 and show that

k = n 1 f k 4 1 = f n 4 f n 1 4 + 2 ( 1 ) n 5 f 2 n 1 n + 2 5 ,

where { x } x x .

1 Introduction and statement of main results

Let f 0 = 0 , f 1 = 1 , and f n = f n 1 + f n 2 for all n 2 . The number f n is called the n th Fibonacci number. The Fibonacci sequence is important and widely used in many mathematical areas, especially related to number theory and combinatorics. For more detailed information, see [1]. Also, many generalizations of the Fibonacci numbers have been introduced in [26]. For the recent results in view of the classical identities such as Binet’s formula, the generating function, Catalan’s identity, Cassini’s identity, and some binomial sums, see [710].

By contrast, the study on the reciprocal of the sum of convergent series is a relatively new research field, see the references [6,1121]. For example, the reciprocal sum of Fibonacci numbers was studied in 2008 [17]. By Binet’s formula, f n can be written as the explicit form f n = α n β n α β for any nonnegative integer n , where α = 1 + 5 2 and β = 1 5 2 are two solutions of x 2 x 1 = 0 . Since f n is comparable to α n for sufficiently large n , we see that

F ( s ) k = 1 1 f k s

converges for all s 1 . It follows that

lim n k = n 1 f k s 1 = .

Motivated by the above observation, in [17], the floor functions of k = n 1 / f k 1 and k = n 1 / f k 2 1 have been obtained as follows: for any n N ,

(1.1) k = n 1 f k 1 = f n 2 , n 2 is even ; f n 2 1 , n 3 is odd ,

and

(1.2) k = n 1 f k 2 1 = f n 1 f n 1 , n 2 is even ; f n 1 f n , n 1 is odd ,

where is the floor function. In this context, it is natural to consider the following problem.

Problem 1.1

For an integer s 3 , find explicit formulas for k = n 1 f k s 1 .

The above problem has been an open problem.

In this article, we give an answer for Problem 1.1 for s = 4 . It turns out that the formula itself is more complicated in the sense that it includes some unexpected terms, and the method of the proof is essentially different from those of (1.1) and (1.2). For all n N , define

(1.3) g n f n 4 f n 1 4 + 2 ( 1 ) n 5 f 2 n 1 + 2 5 75 .

In Section 6, we explain how we can find g n .

The following is the main theorem of this article.

Theorem 1.2

Let c n = 1 f 2 n 1 . Then,

  1. g n < k = n 1 f k 4 1 < g n + c n when n is even;

  2. g n c n < k = n 1 f k 4 1 < g n when n is odd.

As a consequence of Theorem 1.2, we have the following.

Theorem 1.3

For any positive integer n, we have

k = n 1 f k 4 1 = f n 4 f n 1 4 + 2 ( 1 ) n 5 f 2 n 1 n + 2 5 ,

where { x } x [ x ] is the decimal part of x.

In Section 2, we introduce the basic properties of the Fibonacci numbers and Lucas numbers, which are defined as L 0 = 2 , L 1 = 1 , and L n = L n 1 + L n 2 for all n 2 . These Lucas numbers are used for the proof of our main result. In Section 3, we introduce the basic identities, which are used directly to prove our main theorem. The key idea is that the principal part

( g n + 2 g n ) f n 4 f n + 1 4 ( f n 4 + f n + 1 4 ) g n g n + 2

can be expressed as a polynomial with respect to one variable L 2 n + 1 (see Proposition 3.3). In Section 4, we prove Theorem 1.2 using the identities proved in Section 3. In Section 5, we use the periodicity of f n modulo 5 to prove Theorem 1.3. In Section 6, we explain how we obtain the formula g n .

Remark 1.4

Our main theorem is the first result regarding Problem 1.1. We expect that our result and method can be used to solve the problem for any s 3 and any linear recurrence sequences.

2 Fibonacci numbers and Lucas numbers

The Fibonacci numbers satisfy the following well-known useful identities.

Lemma 2.1

[1] For any positive integers n and k , we have

  1. f n 2 = f n + k f n k + ( 1 ) n + k f k 2 ,

  2. f 2 n + 1 = f n 2 + f n + 1 2 .

It is more efficient to use the Lucas numbers when we compute the Fibonacci numbers. The Lucas numbers are defined as follows:

L n = L n 1 + L n 2 for all n 3 with L 0 = 2 , L 1 = 1 , L 2 = 3 .

Then, the Lucas numbers are related to the Fibonacci numbers as follows:

(2.1) f 2 n = f n L n , L n = f n 1 + f n + 1

for any positive integer n .

Now, we explain the basic properties of Fibonacci numbers and Lucas numbers.

Lemma 2.2

[1,22] For any m , n N , we have

  1. f m f n = 1 5 ( L m + n ( 1 ) n L m n ) .

  2. L m L n = L m + n + ( 1 ) n L m n .

  3. f n + 4 m + f n = L 2 m f n + 2 m .

  4. f n + 4 m f n = f 2 m L n + 2 m .

  5. L n 2 = 5 f n 2 + 4 ( 1 ) n .

  6. L 2 n = L n 2 2 ( 1 ) n .

Now, we obtain the equivalent forms of g n defined as in (1.3). Note that

g n = ( f n 2 + f n 1 2 ) ( f n + f n 1 ) ( f n f n 1 ) + 2 ( 1 ) n 5 f 2 n 1 + γ = f 2 n 1 f n + 1 f n 2 + 2 ( 1 ) n 5 f 2 n 1 + γ ,

where γ = 2 5 75 . By Lemma 2.2 (i), we have

f n + 1 f n 2 = 1 5 ( L 2 n 1 ( 1 ) n 2 L 3 ) = 1 5 L 2 n 1 4 ( 1 ) n 5 .

By the first relation (2.1), we have

g n = 1 5 f 2 n 1 L 2 n 1 2 ( 1 ) n 5 f 2 n 1 + γ = 1 5 f 4 n 2 2 ( 1 ) n 5 f 2 n 1 + γ .

3 Basic identities

In Section 4, we compute

1 g n 1 f n 4 + 1 f n + 1 4 + 1 g n + 2

for the proof of Theorem 1.2. In this section, we compute

F 0 ( g n + 2 g n ) f n 4 f n + 1 4 ( f n 4 + f n + 1 4 ) g n g n + 2 .

In Section 2, we obtained the following two equivalent forms of g n :

(3.1) g n = 1 5 f 2 n 1 ( L 2 n 1 2 ( 1 ) n ) + γ

and

(3.2) g n = 1 5 f 4 n 2 2 ( 1 ) n 5 f 2 n 1 + γ .

In fact, three terms g n + 2 g n , f n 4 f n + 1 4 , and f n 4 + f n + 1 4 in the definition of F 0 can be expressed in terms of L 2 n + 1 .

Proposition 3.1

For any positive integer n, we have

  1. g n + 2 g n = 1 5 ( 3 L 2 n + 1 2 2 ( 1 ) n L 2 n + 1 + 6 ) .

  2. f n 4 f n + 1 4 = 1 625 ( L 2 n + 1 ( 1 ) n ) 4 .

  3. f n 4 + f n + 1 4 = 1 25 ( 3 L 2 n + 1 2 + 4 ( 1 ) n L 2 n + 1 + 18 ) .

Proof

Here, we use the form (3.2) of g n .

(i) By Lemma 2.2 (iv) and (vi), we have

g n + 2 g n = 1 5 ( f 4 n + 6 f 4 n 2 ) 2 5 ( 1 ) n ( f 2 n + 3 f 2 n 1 ) = 3 5 L 4 n + 2 2 5 ( 1 ) n L 2 n + 1 = 1 5 ( 3 L 2 n + 1 2 2 ( 1 ) n L 2 n + 1 + 6 ) .

(ii) By Lemma 2.2 (i), we have

f n 4 f n + 1 4 = ( f n + 1 f n ) 4 = 1 5 ( L 2 n + 1 ( 1 ) n ) 4 .

(iii) By Lemmas 2.1 (ii) and 2.2 (v), we have

f n 4 + f n + 1 4 = ( f n 2 + f n + 1 2 ) 2 2 f n 2 f n + 1 2 = f 2 n + 1 2 2 ( f n f n + 1 ) 2 = 1 25 ( 3 L 2 n + 1 2 + 4 ( 1 ) n L 2 n + 1 + 18 ) .

However, g n g n + 2 can be written in terms of L 2 n + 1 and f 2 n + 1 .□

Proposition 3.2

For any positive integer n, we have

g n g n + 2 = 1 125 ( L 2 n + 1 2 + 9 ) ( L 2 n + 1 2 6 ( 1 ) n L 2 n + 1 1 ) + A ,

where

(3.3) A γ 5 f 2 n + 1 ( 7 L 2 n + 1 6 ( 1 ) n ) + γ 2 with γ = 2 5 75 .

Proof

By (3.1), it follows that

g n g n + 2 = 1 25 f 2 n 1 f 2 n + 3 ( L 2 n 1 2 ( 1 ) n ) ( L 2 n + 3 2 ( 1 ) n ) + γ 2 + γ 5 { f 2 n 1 ( L 2 n 1 2 ( 1 ) n ) + f 2 n + 3 ( L 2 n + 3 2 ( 1 ) n ) } .

By Lemmas 2.1 (i) and 2.2 (v), we have

f 2 n 1 f 2 n + 3 = f 2 n + 1 2 + 1 = 1 5 ( L 2 n + 1 2 + 9 ) .

By Lemmas 2.2 (ii) and (vi), we have

L 2 n 1 L 2 n + 3 = L 4 n + 2 7 = L 2 n + 1 2 5 .

Note that

L 2 n 1 + L 2 n + 3 = 3 L 2 n + 1 .

Combining the above two identities, we have

( L 2 n 1 2 ( 1 ) n ) ( L 2 n + 3 2 ( 1 ) n ) = L 2 n 1 L 2 n + 3 2 ( 1 ) n ( L 2 n 1 + L 2 n + 3 ) + 4 = L 2 n + 1 2 6 ( 1 ) n L 2 n + 1 1 .

By (2.1) and Lemma 2.2 (iii), we have

f 2 n 1 ( L 2 n 1 2 ( 1 ) n ) + f 2 n + 3 ( L 2 n + 3 2 ( 1 ) n ) = ( f 4 n 2 + f 4 n + 6 ) 2 ( 1 ) n ( f 2 n 1 + f 2 n + 3 ) = 7 f 2 n + 1 L 2 n + 1 6 ( 1 ) n f 2 n + 1 = f 2 n + 1 ( 7 L 2 n + 1 6 ( 1 ) n ) .

If we combine the above formulas, the proof is done.□

To simplify our formulas, we write

x L 2 n + 1 .

Then, Propositions 3.1 and 3.2 can be summarized as follows:

g n + 2 g n = 1 5 ( 3 x 2 2 ( 1 ) n x + 6 ) , f n 4 f n + 1 4 = 1 625 ( x ( 1 ) n ) 4 , f n 4 + f n + 1 4 = 1 25 ( 3 x 2 + 4 ( 1 ) n x + 18 ) , g n g n + 2 = 1 125 ( x 2 + 9 ) ( x 2 6 ( 1 ) n x 1 ) + A .

Proposition 3.3

Let F 0 = ( g n + 2 g n ) f n 4 f n + 1 4 ( f n 4 + f n + 1 4 ) g n g n + 2 . Then,

F 0 = 1 3,125 { 14 x 4 + 190 ( 1 ) n x 3 + 146 x 2 + 982 ( 1 ) n x + 168 } 1 25 { 3 x 2 + 4 ( 1 ) n x + 18 } A ,

where x = L 2 n + 1 and A is the form defined as in (3.3).

Here, the term A cannot be written in terms of x = L 2 n + 1 . Instead, we obtain the bound of A as follows.

Lemma 3.4

In fact, A satisfies the inequality

2 375 x ( 7 x 6 ( 1 ) n ) + 4 1,125 A 2 375 ( x + 1 ) ( 7 x 6 ( 1 ) n ) + 4 1,125 .

Proof

Note that L 2 n + 1 2 < 5 f 2 n + 1 2 = L 2 n + 1 2 + 4 < ( L 2 n + 1 + 1 ) 2 .□

4 Proof of Theorem 1.2

In this section, we prove the main theorem using the identities of the previous section. The key idea is that all important terms can be written as the function of the variable x = L 2 n + 1 .

Theorem 4.1

(Theorem 1.2 (i)) Let n N be even and let c n = 1 f 2 n 1 .

g n < k = n 1 f k 4 1 < g n + c n .

Proof

Note that

(4.1) 1 g n 1 f n 4 + 1 f n + 1 4 + 1 g n + 2 = F 0 g n g n + 2 f n 4 f n + 1 4 ,

where F 0 = ( g n + 2 g n ) f n 4 f n + 1 4 ( f n 4 + f n + 1 4 ) g n g n + 2 , which is the same as in Section 3.

(i) By Lemma 3.4, we have

A 2 375 ( x + 1 ) ( 7 x 6 ) + 4 1,125 .

Now, we will show that F 0 is positive for any even n N . More precisely, Proposition 3.3 gives the inequality

F 0 = 1 3,125 ( 14 x 4 + 190 x 3 + 146 x 2 + 982 x + 168 ) 1 25 ( 3 x 2 + 4 x + 18 ) A 2 28,125 ( 762 x 3 + 315 x 2 + 4,429 x + 1,044 ) > 0 .

Since F 0 > 0 for all n N , identity (4.1) gives the inequality

1 g n > 1 f n 4 + 1 f n + 1 4 + 1 g n + 2

for all n N . If we apply the above inequality repeatedly, it follows that

1 g n > 1 f n 4 + 1 f n + 1 4 + 1 g n + 2 > 1 f n 4 + 1 f n + 1 4 + 1 f n + 2 4 + 1 f n + 3 4 + 1 g n + 4 > > k = n 1 f k 4 ,

which proves the left inequality of our theorem.

(ii) Note that

(4.2) 1 g n + c n 1 f n 4 + 1 f n + 1 4 + 1 g n + 2 + c n + 2 = F ( g n + c n ) ( g n + 2 + c n + 2 ) f n 4 f n + 1 4 ,

where

F ( ( g n + 2 + c n + 2 ) ( g n + c n ) ) f n 4 f n + 1 4 ( f n 4 + f n + 1 4 ) ( g n + c n ) ( g n + 2 + c n + 2 ) .

Now, we will show that F is negative for all n N . We write F = F 0 F c , where

F 0 = ( g n + 2 g n ) f n 4 f n + 1 4 ( f n 4 + f n + 1 4 ) g n g n + 2

and

F c ( c n c n + 2 ) f n 4 f n + 1 4 + ( f n 4 + f n + 1 4 ) ( c n g n + 2 + c n + 2 g n + c n c n + 2 ) .

By Lemma 3.4, we have

A 2 375 x ( 7 x 6 ) + 4 1,125 .

It follows that

F 0 = 1 3,125 ( 14 x 4 + 190 x 3 + 146 x 2 + 982 x + 168 ) 1 25 ( 3 x 2 + 4 x + 18 ) A 2 5,625 ( 165 x 3 + 69 x 2 + 947 x + 144 ) .

Since c n > c n + 2 and g n + 2 > g n , we have

F c > ( f n 4 + f n + 1 4 ) ( c n g n + 2 + c n + 2 g n ) > ( f n 4 + f n + 1 4 ) ( c n g n + c n + 2 g n + 2 ) .

Note that

c n g n = 1 5 ( L 2 n 1 2 ) + α f 2 n 1 > 1 5 ( L 2 n 1 2 ) .

Since we have

c n g n + c n + 2 g n + 2 > 1 5 ( L 2 n 1 + L 2 n + 3 4 ) = 1 5 ( 3 L 2 n + 1 4 ) = 1 5 ( 3 x 4 ) ,

we obtain

F c > 1 125 ( 3 x 2 + 4 x + 18 ) ( 3 x 4 ) .

It follows that

F = F 0 F c < 1 5,625 ( 75 x 3 + 138 x 2 + 184 x + 3,528 ) < 0 .

Since F < 0 for all n N , identity (4.2) gives the inequality

1 g n + c n < 1 f n 4 + 1 f n + 1 4 + 1 g n + 2 + c n + 2

for all n N . If we apply the above inequality repeatedly, it follows that

1 g n + c n < 1 f n 4 + 1 f n + 1 4 + 1 g n + 2 + c n + 2 < 1 f n 4 + 1 f n + 1 4 + 1 f n + 2 4 + 1 f n + 3 4 + 1 g n + 4 + c n + 4 < < k = n 1 f k 4 ,

which proves the right inequality of Theorem 1.2(i).□

Theorem 4.2

(Theorem 1.2 (ii)) Let n N be odd and let c n = 1 f 2 n 1 .

g n c n < k = n 1 f k 4 1 < g n .

Proof

(i) By Lemma 3.4, we have

A 2 375 x ( 7 x + 6 ) + 4 1,125

for any odd n . It follows that

F 0 = 1 3,125 ( 14 x 4 190 x 3 + 146 x 2 982 x + 168 ) 1 25 ( 3 x 2 4 x + 18 ) A 2 5,625 ( 165 x 3 + 69 x 2 947 x + 144 ) < 0 .

Similarly, as the proof of the previous theorem, we obtain

1 g n < 1 f n 4 + 1 f n + 1 4 + 1 g n + 2

for any odd n . It completes the proof of the right inequality of our theorem.

(ii) Note that

(4.3) 1 g n c n 1 f n 4 + 1 f n + 1 4 + 1 g n + 2 c n + 2 = F ( g n c n ) ( g n + 2 c n + 2 ) f n 4 f n + 1 4 ,

where

F ( ( g n + 2 c n + 2 ) ( g n c n ) ) f n 4 f n + 1 4 ( f n 4 + f n + 1 4 ) ( g n c n ) ( g n + 2 c n + 2 ) .

We write F = F 0 + F c , where

F 0 = ( g n + 2 g n ) f n 4 f n + 1 4 ( f n 4 + f n + 1 4 ) g n g n + 2

and

F c ( c n c n + 2 ) f n 4 f n + 1 4 + ( f n 4 + f n + 1 4 ) ( c n g n + 2 + c n + 2 g n c n c n + 2 ) .

By Lemma 3.4, we have

A 2 375 ( x + 1 ) ( 7 x + 6 ) + 4 1,125 .

It follows that

F 0 = 1 3,125 ( 14 x 4 190 x 3 + 146 x 2 982 x + 168 ) 1 25 ( 3 x 2 4 x + 18 ) A 2 28,125 ( 888 x 3 + 375 x 2 5,041 x + 396 ) .

By Lemma 2.2 (iv), we have

c n c n + 2 = f 2 n + 3 f 2 n 1 f 2 n 1 f 2 n + 3 = 5 L 2 n + 1 L 2 n + 1 2 + 9 = 5 x x 2 + 9 5 x ( x + 1 ) 2 for x 4 .

Note that

c n g n + 2 + c n + 2 g n > c n g n + c n + 2 g n + 2 > 1 5 ( 3 L 2 n + 1 + 4 ) = 1 5 ( 3 x + 4 ) .

It follows that

F c > 5 x ( x + 1 ) 2 1 625 ( x + 1 ) 4 + 1 25 ( 3 x 2 4 x + 18 ) 1 5 ( 3 x + 4 ) 1 = 1 125 ( 10 x 3 13 x 2 + 59 x 18 ) .

Thus, we have

F = F 0 + F > 1 28,125 ( 474 x 3 2,175 x 2 + 3,193 x 3,528 ) > 0

for x 4 . Similarly as the proof of the previous theorem, we obtain

1 g n c n > 1 f n 4 + 1 f n + 1 4 + 1 g n + 2 c n + 2

for any odd n . It completes the proof of the left inequality of Theorem 1.2 (ii).□

Corollary 4.3

lim n k = n 1 f k 4 1 g n = 0 .

Proof

It follows from squeeze property, Theorems 4.1 and 4.2.□

5 Proof of Theorem 1.3

There are many previous results related to the periodicity modulo m N of various linear recurrence sequences such as the case for the Fibonacci sequence { f n } n = 0 [23], and other generalized cases for three-step Fibonacci sequence [24], ( a , b ) -Fibonacci sequence [25], and so on. We focus on the length of the period modulo m of the Fibonacci sequence, which is called the Pisano period π ( m ) after Fibonacci’s real name, Leonard Pisano. For example,

π ( 2 ) = 3 , π ( 3 ) = 8 , π ( 4 ) = 6 , π ( 5 ) = 20 , π ( 6 ) = 24 , .

Now, we compute the value of k = n 1 f k 4 1 using the periodicity of the Fibonacci sequence modulo 5.

Lemma 5.1

[23] The Pisano period π ( 5 ) is 20. Precisely, the Fibonacci sequence { f 0 , f 1 , f 2 , f 3 , , } modulo 5 has the repeating pattern

{ 0 , 1 , 1 , 2 , 3 , 0 , 3 , 3 , 1 , 4 , 0 , 4 , 4 , 3 , 2 , 0 , 2 , 2 , 4 , 1 }

of length 20. It is clear that π ( 5 ) = 20 , since f 20 0 ( mod 5 ) and f 21 1 ( mod 5 ) .

Proof of Theorem 1.3

By Theorem 1.2 and the inequality 2 5 75 ± 1 f 2 n 1 < 1 5 for n 4 , we have

(5.1) k = n 1 f k 4 1 = f n 4 f n 1 4 + 2 5 ( 1 ) n f 2 n 1 .

Now, we divide into the integer part and the decimal part of

2 5 ( 1 ) n f 2 n 1 = 2 5 ( 1 ) n f 2 n 1 + 2 5 ( 1 ) n f 2 n 1 .

By Lemma 5.1, the sequence { f 2 n 1 } n = 1 modulo 5 has the repeating pattern

{ 1 , 2 , 0 , 3 , 4 , 4 , 3 , 0 , 2 , 1 }

of length 10.

(i) If n = 4 , 6 , 8 , 10 , 12 then the sequence { 2 f 7 , 2 f 11 , 2 f 15 , 2 f 19 , 2 f 23 } modulo 5 has the repeating pattern

{ 1 , 3 , 0 , 2 , 4 }

of length 5. It means that

(5.2) 2 5 f 2 n 1 = n + 2 5

when n is even.

(ii) If n = 5 , 7 , 9 , 11 , 13 , then the sequence { 2 f 9 , 2 f 13 , 2 f 17 , 2 f 21 , 2 f 25 , } modulo 5 has the repeating pattern

{ 3 , 1 , 4 , 2 , 0 }

of length 5. It means that

(5.3) 2 5 f 2 n 1 = n + 2 5

when n is odd.

By (5.2) and (5.3), we have

(5.4) 2 5 ( 1 ) n f 2 n 1 = 2 5 ( 1 ) n f 2 n 1 + n + 2 5

for all n 4 . Combining (5.1) and (5.4), we conclude that

(5.5) k = n 1 f k 4 1 = f n 4 f n 1 4 + 2 ( 1 ) n 5 f 2 n 1 n + 2 5

for n 4 . We can easily show that identity (5.5) holds for n 3 by direct computation. It completes the proof of Theorem 1.3.□

6 How to obtain the formula g n

In this final section, we need to explain how we guess the formula

g n = f n 4 f n 1 4 + 2 5 ( 1 ) n f 2 n 1 + 2 5 75 .

By using a computer software program, we have constructed the following table for the values of

h n k = n 1 f k 4 1 ( f n 4 f n 1 4 ) .

One can see that h n is positive for even n and negative for odd n in Table 1. Since h n goes to infinity as n , we must calculate h n . Some h n ’s can be written as follows:

h 3 = 1.9674 = 2 + 0.0325 h 4 = 5.2698 = 5 + 0.2698 h 5 = 13.5442 = 14 + 0.4557 h 6 = 35.6611 = 35 + 0.6611 h 7 = 93.1409 = 94 + 0.8590 h 8 = 244.0598 = 243 + 1.0598 .

Table 1

The values of hn for 3 ≤ n ≤ 6

n k = n 1 f k 4 1 f n 4 f n 1 4 k = n 1 f k 4 1 ( f n 4 f n 1 4 )
3 13.032567 15 1.967432
4 70.269868 65 5.269869
5 530.455702 544 13.544298
6 3506.661126 3,471 35.661126

The above observation gives us the following expectation:

h n = d n + 1 5 ( n 3 ) + γ n ,

where d n Z and γ n approaches 2 75 5 = 0.0596284793999943 . See the following values of d n in Table 2.

Table 2

The values of dn for 1 ≤ n ≤ 18

n d n n d n n d n
1 1 7 94 13 30,012
2 0 8 243 14 78,565
3 2 9 640 15 205,694
4 5 10 1,671 16 538,505
5 14 11 4,380 17 1,409,834
6 35 12 11,461 18 3,690,983

Finally, we obtain the explicit form of d n , which is the main part of this section.

Theorem 6.1

For any positive integer n, we have

d n = 2 ( 1 ) n 5 f 2 n 1 1 5 ( n 3 ) .

By Theorem 6.1, it follows that

g n = f n 4 f n 1 4 + 2 ( 1 ) n 5 + γ ,

where γ = 2 5 75 . From the above table, the sequence { d n } satisfies the relation

(6.1) d n = 2 d n 1 + 2 d n 2 + d n 3 1 , n 6

with

(6.2) d 3 = 2 , d 4 = 5 , d 5 = 14 .

If we define e n d n d n 1 , then the relation (6.1) is reduced to

e n = 3 e n 1 e n 2 1

with

e 4 = 7 , e 5 = 19 .

Note that

e 6 = ( 3 e 5 + e 4 + 1 ) = ( f 4 e 5 + f 2 e 4 + f 1 f 2 ) e 7 = 8 e 5 + 3 e 4 + 2 = f 6 e 5 + f 4 e 4 + f 2 f 3 e 8 = ( 21 e 5 + 8 e 4 + 6 ) = ( f 8 e 5 + f 6 e 4 + f 3 f 4 ) .

One can easily prove the following by induction.

Proposition 6.2

For any n 6 , we have

e n = ( 1 ) n + 1 { f 2 n 8 e 5 + f 2 n 10 e 4 + f n 5 f n 4 } .

Lemma 6.3

Let m N .

  1. k = 1 m f 4 k 1 = f 2 m f 2 m + 1 , k = 1 m f 4 k + 1 = f 2 m f 2 m + 3 .

  2. f 1 2 + f 2 2 + + f m 2 = f m f m + 1 .

  3. f 1 2 + f 3 2 + + f 2 n 1 2 = 1 5 ( f 4 n + 2 n ) .

  4. f 2 2 + f 4 2 + + f 2 n 2 = 1 5 ( f 4 n + 2 2 n 1 ) .

  5. 5 f n 2 + f 2 n = 2 f 2 n + 1 2 ( 1 ) n .

Proof

The formulas (i), (ii), (iii), and (iv) are known. Here, we prove (v) only. By Lemma 2.2 (v), we have

5 f n 2 + f 2 n = L n 2 + f 2 n 4 ( 1 ) n .

Since L n = f n 1 + f n + 1 and f 2 n = f n L n , we have

5 f n 2 + f 2 n = 2 ( f n + 1 f n 1 + f n + 1 2 ) 4 ( 1 ) n .

By Lemma 2.1 (i) and (ii), we have

5 f n 2 + f 2 n = 2 ( f n 2 + f n + 1 2 ) 2 ( 1 ) n = 2 f 2 n + 1 2 ( 1 ) n .

Lemma 6.4

Let m be any positive integer. Then,

  1. k = 1 m ( 1 ) k f 2 k = ( 1 ) m f m f m + 1 .

  2. k = 1 m ( 1 ) k f 2 k + 2 = ( 1 ) m f m f m + 3 .

  3. k = 1 m ( 1 ) k f k f k + 1 = f 2 2 + f 4 2 + + f m 2 , m i s e v e n ( f 2 2 + f 4 2 + + f m 1 2 ) f m f m + 1 , m i s o d d .

Proof

Since the proofs are similar, we prove (i) only. If m is even, then by Lemma 6.3 (i), we have

k = 1 m ( 1 ) k f 2 k = ( f 2 + f 4 ) + ( f 6 + f 8 ) + + ( f 2 m 2 + f 2 m ) = f 3 + f 7 + + f 2 m 1 = f m f m + 1 .

If m is odd, then by Lemma 6.3 (i), we have

k = 1 m ( 1 ) k f 2 k = ( f 3 + f 7 + + f 2 m 3 ) f 2 m = f m 1 f m f m L m = f m f m + 1 .

It completes the proof of (i).□

Proof of Theorem 6.1

Note that for any positive integer n 6 , we have

d n = d 5 + k = 1 n 5 e k + 5 = 14 + 7 k = 1 n 5 ( 1 ) k f 2 k 19 k = 1 n 5 ( 1 ) k f 2 k + 2 + k = 1 n 5 ( 1 ) k f k f k + 1 .

(i) If n is odd, then by Lemma 6.4, we have

d n = 14 + 7 f n 5 f n 4 19 f n 5 f n 2 + ( f 2 2 + f 4 2 + + f n 5 2 ) = 14 f n 5 ( 19 f n 2 7 f n 4 ) + ( f 2 2 + f 4 2 + + f n 5 2 ) .

A direct computation shows that

f n 5 ( 19 f n 2 7 f n 4 ) = f n 5 ( f n + 4 f n 1 ) = f n 5 ( f n + 3 + f n + 1 + f n 3 + f n 4 ) = f n 1 2 + f n 2 2 + f n 4 2 14 + f n 5 f n 4 .

It follows that

d n = ( f n 1 2 + f n 2 2 + f n 4 2 ) f n 5 f n 4 + ( f 2 2 + f 4 2 + + f n 5 2 ) .

By Lemma 6.3 (ii), we have

f 1 2 + f 2 2 + + f n 5 2 = f n 5 f n 4 .

By Lemma 6.3 (iii) and (v), we have

d n = ( f 1 2 + f 3 2 + + f n 2 2 ) f n 1 2 = 1 5 ( f 2 n 2 + n 1 ) 1 5 ( 2 f 2 n 1 f 2 n 2 2 ) = 2 5 f 2 n 1 1 5 ( n 3 ) .

(ii) If n is even, then by Lemma 6.4, we have

d n = 14 + 7 k = 1 n 5 ( 1 ) k f 2 k 19 k = 1 n 5 ( 1 ) k f 2 k + 2 + k = 1 n 5 ( 1 ) k f k f k + 1 = 14 + f n 5 ( 19 f n 2 7 f n 4 ) + ( f 2 2 + f 4 2 + + f n 6 2 ) f n 5 f n 4 .

Similarly as in the case when n is odd, we obtain

f n 5 ( 19 f n 2 7 f n 4 ) = f n 1 2 + f n 2 2 + f n 4 2 + 14 + f n 5 f n 4 .

By Lemma 6.3 (iv) and (v), we conclude that

d n = f n 1 2 + f n 2 2 + f n 4 2 + ( f 2 2 + f 4 2 + + f n 6 2 ) = ( f 2 2 + f 4 2 + + f n 2 2 ) + f n 1 2 = 1 5 ( f 2 n 2 n + 1 ) + 1 5 ( 2 f 2 n 1 f 2 n 2 + 2 ) = 2 5 f 2 n 1 1 5 ( n 3 ) .

It completes the proof of Theorem 6.1.□

We will finish this section with proving Theorem 6.1 again using the theory of linear recurrence relations. Note that the characteristic equation of a nonhomogeneous linear recurrence relation (6.1) is ( x 1 ) ( x 2 + 3 x + 1 ) = 0 and has three distinct roots α 2 , β 2 , and 1, where α = 1 + 5 2 and β = 1 5 2 . It is easy to show that (6.1) has a particular solution d n = 1 5 n . Thus, there exist constants A , B , and C such that

d n = A ( α 2 ) n + B ( β 2 ) n + C 1 5 n .

From the conditions (6.2), the constants A , B , and C satisfy

α 6 A + β 6 B C = 7 5 , α 8 A + β 8 B + C = 29 5 , α 10 A + β 10 B C = 13 .

Solving the system of linear equations, we obtain

A = 2 ( 18 β 2 47 ) 5 α 6 ( α 2 + 1 ) ( α 2 β 2 ) , B = 2 ( 18 α 2 47 ) 5 β 6 ( β 2 + 1 ) ( β 2 α 2 ) , C = 7 α 2 β 2 + 29 α 2 + 29 β 2 65 5 ( α 2 + 1 ) ( β 2 + 1 ) .

The forms of A , B , and C are too complicated, but we can simplify them. Since α and β are roots of x 2 x 1 = 0 , we obtain

A = 2 5 α ( α β ) , B = 2 5 β ( α β ) , C = 3 5 .

It follows that

d n = 2 5 ( 1 ) n α 2 n 1 β 2 n 1 α β + 3 5 1 5 n .

This is exactly the same as the form in Theorem 6.1 by Binet’s formula.

7 Conclusion and future work

In this article, we find an explicit formula for k = n 1 f k 4 1 . It is given by the following expression:

f n 4 f n 1 4 + 2 ( 1 ) n 5 f 2 n 1 n + 2 5 .

To obtain the desired formula, we start with guessing the leading terms f n 4 f n 1 4 by observing

lim n k = n 1 f k 4 1 ( f n 4 f n 1 4 ) f n 4 = 0 .

After that, we note that the values of the sequence k = n 1 f k 4 1 ( f n 4 f n 1 4 ) are proportional to f n 2 as n increases, and we obtain the remaining term by direct computation. We give the detailed process in Section 6.

This work might help in finding the formula for k = n 1 f k s 1 for s = 3 or s 5 and might also be useful for the problem of obtaining the formula for the reciprocal of the sums of the sequences that have homogeneous and non-homogeneous recurrence relations with second-order constant coefficients and some other related problems.

  1. Funding information: This paper was supported by research funds for newly appointed professors of Jeonbuk National University in 2021. This work was supported by NRF-2018R1D1A1B07050044 from the National Research Foundation of Korea. Kyunghwan Song was supported by NRF-2020R1I1A1A01070546 from the National Research Foundation of Korea.

  2. Author contributions: All authors have accepted responsibility for the entire content of this manuscript and approved its submission.

  3. Conflict of interest: The authors state that there is no conflict of interest.

  4. Data availability statement: No data were used to support this study.

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Received: 2021-12-17
Revised: 2022-10-17
Accepted: 2022-10-24
Published Online: 2022-12-17

© 2022 the author(s), published by De Gruyter

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