Home Mathematics On the reciprocal sum of the fourth power of Fibonacci numbers
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On the reciprocal sum of the fourth power of Fibonacci numbers

  • WonTae Hwang , Jong-Do Park EMAIL logo and Kyunghwan Song
Published/Copyright: December 17, 2022
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Open Mathematics
From the journal Open Mathematics Volume 20 Issue 1

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 ⌋ .

Keywords: recurrence relation; Fibonacci number; Lucas number; Catalan’s identity; Pisano period
MSC 2010: Primary 11B37; 11B39; 11B50; Secondary 11Y55; 65Q30

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 [2–6]. 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 [7–10].

By contrast, the study on the reciprocal of the sum of convergent series is a relatively new research field, see the references [6,11–21]. 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

This work is licensed under the Creative Commons Attribution 4.0 International License.

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https://doi.org/10.1515/math-2022-0525
Keywords for this article
recurrence relation; Fibonacci number; Lucas number; Catalan’s identity; Pisano period
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  1. Regular Articles
  2. A random von Neumann theorem for uniformly distributed sequences of partitions
  3. Note on structural properties of graphs
  4. Mean-field formulation for mean-variance asset-liability management with cash flow under an uncertain exit time
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  54. Regularity of models associated with Markov jump processes
  55. Multiplicity solutions for a class of p-Laplacian fractional differential equations via variational methods
  56. Minimal period problem for second-order Hamiltonian systems with asymptotically linear nonlinearities
  57. Convergence rate of the modified Levenberg-Marquardt method under Hölderian local error bound
  58. Non-binary quantum codes from constacyclic codes over 𝔽q[u1, u2,…,uk]/⟨ui3 = ui, uiuj = ujui⟩
  59. On the general position number of two classes of graphs
  60. A posteriori regularization method for the two-dimensional inverse heat conduction problem
  61. Orbital stability and Zhukovskiǐ quasi-stability in impulsive dynamical systems
  62. Approximations related to the complete p-elliptic integrals
  63. A note on commutators of strongly singular Calderón-Zygmund operators
  64. Generalized Munn rings
  65. Double domination in maximal outerplanar graphs
  66. Existence and uniqueness of solutions to the norm minimum problem on digraphs
  67. On the p-integrable trajectories of the nonlinear control system described by the Urysohn-type integral equation
  68. Robust estimation for varying coefficient partially functional linear regression models based on exponential squared loss function
  69. Hessian equations of Krylov type on compact Hermitian manifolds
  70. Class fields generated by coordinates of elliptic curves
  71. The lattice of (2, 1)-congruences on a left restriction semigroup
  72. A numerical solution of problem for essentially loaded differential equations with an integro-multipoint condition
  73. On stochastic accelerated gradient with convergence rate
  74. Displacement structure of the DMP inverse
  75. Dependence of eigenvalues of Sturm-Liouville problems on time scales with eigenparameter-dependent boundary conditions
  76. Existence of positive solutions of discrete third-order three-point BVP with sign-changing Green's function
  77. Some new fixed point theorems for nonexpansive-type mappings in geodesic spaces
  78. Generalized 4-connectivity of hierarchical star networks
  79. Spectra and reticulation of semihoops
  80. Stein-Weiss inequality for local mixed radial-angular Morrey spaces
  81. Eigenvalues of transition weight matrix for a family of weighted networks
  82. A modified Tikhonov regularization for unknown source in space fractional diffusion equation
  83. Modular forms of half-integral weight on Γ0(4) with few nonvanishing coefficients modulo ℓ
  84. Some estimates for commutators of bilinear pseudo-differential operators
  85. Extension of isometries in real Hilbert spaces
  86. Existence of positive periodic solutions for first-order nonlinear differential equations with multiple time-varying delays
  87. B-Fredholm elements in primitive C*-algebras
  88. Unique solvability for an inverse problem of a nonlinear parabolic PDE with nonlocal integral overdetermination condition
  89. An algebraic semigroup method for discovering maximal frequent itemsets
  90. Class-preserving Coleman automorphisms of some classes of finite groups
  91. Exponential stability of traveling waves for a nonlocal dispersal SIR model with delay
  92. Existence and multiplicity of solutions for second-order Dirichlet problems with nonlinear impulses
  93. The transitivity of primary conjugacy in regular ω-semigroups
  94. Stability estimation of some Markov controlled processes
  95. On nonnil-coherent modules and nonnil-Noetherian modules
  96. N-Tuples of weighted noncommutative Orlicz space and some geometrical properties
  97. The dimension-free estimate for the truncated maximal operator
  98. A human error risk priority number calculation methodology using fuzzy and TOPSIS grey
  99. Compact mappings and s-mappings at subsets
  100. The structural properties of the Gompertz-two-parameter-Lindley distribution and associated inference
  101. A monotone iteration for a nonlinear Euler-Bernoulli beam equation with indefinite weight and Neumann boundary conditions
  102. Delta waves of the isentropic relativistic Euler system coupled with an advection equation for Chaplygin gas
  103. Multiplicity and minimality of periodic solutions to fourth-order super-quadratic difference systems
  104. On the reciprocal sum of the fourth power of Fibonacci numbers
  105. Averaging principle for two-time-scale stochastic differential equations with correlated noise
  106. Phragmén-Lindelöf alternative results and structural stability for Brinkman fluid in porous media in a semi-infinite cylinder
  107. Study on r-truncated degenerate Stirling numbers of the second kind
  108. On 7-valent symmetric graphs of order 2pq and 11-valent symmetric graphs of order 4pq
  109. Some new characterizations of finite p-nilpotent groups
  110. A Billingsley type theorem for Bowen topological entropy of nonautonomous dynamical systems
  111. F4 and PSp (8, ℂ)-Higgs pairs understood as fixed points of the moduli space of E6-Higgs bundles over a compact Riemann surface
  112. On modules related to McCoy modules
  113. On generalized extragradient implicit method for systems of variational inequalities with constraints of variational inclusion and fixed point problems
  114. Solvability for a nonlocal dispersal model governed by time and space integrals
  115. Finite groups whose maximal subgroups of even order are MSN-groups
  116. Symmetric results of a Hénon-type elliptic system with coupled linear part
  117. On the connection between Sp-almost periodic functions defined on time scales and ℝ
  118. On a class of Harada rings
  119. On regular subgroup functors of finite groups
  120. Fast iterative solutions of Riccati and Lyapunov equations
  121. Weak measure expansivity of C2 dynamics
  122. Admissible congruences on type B semigroups
  123. Generalized fractional Hermite-Hadamard type inclusions for co-ordinated convex interval-valued functions
  124. Inverse eigenvalue problems for rank one perturbations of the Sturm-Liouville operator
  125. Data transmission mechanism of vehicle networking based on fuzzy comprehensive evaluation
  126. Dual uniformities in function spaces over uniform continuity
  127. Review Article
  128. On Hahn-Banach theorem and some of its applications
  129. Rapid Communication
  130. Discussion of foundation of mathematics and quantum theory
  131. Special Issue on Boundary Value Problems and their Applications on Biosciences and Engineering (Part II)
  132. A study of minimax shrinkage estimators dominating the James-Stein estimator under the balanced loss function
  133. Representations by degenerate Daehee polynomials
  134. Multilevel MC method for weak approximation of stochastic differential equation with the exact coupling scheme
  135. Multiple periodic solutions for discrete boundary value problem involving the mean curvature operator
  136. Special Issue on Evolution Equations, Theory and Applications (Part II)
  137. Coupled measure of noncompactness and functional integral equations
  138. Existence results for neutral evolution equations with nonlocal conditions and delay via fractional operator
  139. Global weak solution of 3D-NSE with exponential damping
  140. Special Issue on Fractional Problems with Variable-Order or Variable Exponents (Part I)
  141. Ground state solutions of nonlinear Schrödinger equations involving the fractional p-Laplacian and potential wells
  142. A class of p1(x, ⋅) & p2(x, ⋅)-fractional Kirchhoff-type problem with variable s(x, ⋅)-order and without the Ambrosetti-Rabinowitz condition in ℝN
  143. Jensen-type inequalities for m-convex functions
  144. Special Issue on Problems, Methods and Applications of Nonlinear Analysis (Part III)
  145. The influence of the noise on the exact solutions of a Kuramoto-Sivashinsky equation
  146. Basic inequalities for statistical submanifolds in Golden-like statistical manifolds
  147. Global existence and blow up of the solution for nonlinear Klein-Gordon equation with variable coefficient nonlinear source term
  148. Hopf bifurcation and Turing instability in a diffusive predator-prey model with hunting cooperation
  149. Efficient fixed-point iteration for generalized nonexpansive mappings and its stability in Banach spaces
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