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# Open Mathematics

### formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo

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Volume 16, Issue 1

# On the recursive properties of one kind hybrid power mean involving two-term exponential sums and Gauss sums

Shimeng Shen
Published Online: 2018-08-24 | DOI: https://doi.org/10.1515/math-2018-0081

## Abstract

The main purpose of this paper is to study the computational problem of one kind hybrid power mean involving two-term exponential sums and quartic Gauss sums using the analytic method and the properties of the classical Gauss sums, and to prove some interesting fourth-order linear recurrence formulae for this problem. As an application of our result, we can also obtain an exact computational formula for one kind congruence equation mod p, an odd prime.

MSC 2010: 11L05; 11L07

## 1 Introduction

Let p ≥ 3 be an odd prime. For any integer m with (m,p) = 1, the quartic Gauss sums B(m) = B(m, p) is defined as

$B(m)=∑a=0p−1ema4p,$

where as usual, e(y) = e2πiy.

Recently, some scholars have studied the hybrid power mean problems of various trigonometric sums, and obtained many interesting results. For example, Chen Li and Hu Jiayuan [1] studied the computational problem of the hybrid power mean

$Sk(p)=∑m=1p−1∑a=0p−1ema3pk⋅∑c=1p−1emc+c¯p2,$

where c denotes the multiplicative inverse of c mod p. That is, cc ≡ 1 mod p.

For p ≡ 1 mod 3, they used the elementary method to obtain an interesting third-order linear recurrence formula for Sk(p).

Li Xiaoxue and Hu Jiayuan [2] studied the computational problem of the hybrid power mean

$∑b=1p−1∑a=0p−1eba4p2⋅∑c=1p−1ebc+c¯p2,$(1)

and proved an exact computational formula for (1).

Zhang Han and Zhang Wenpeng [3] proved the identity

$∑m=1p−1∑a=0p−1ema3+nap4=2p3−p2if3∤p−1,2p3−7p2if3|p−1.$

Other related results can also be found in references [4,5,6,7,8,9,10,11,12,13].

In this paper, we will consider the calculating problem of the following hybrid power mean:

$Vk(p)=∑m=1p−1∑a=0p−1ema4pk⋅∑b=0p−1emb4+bp3,$(2)

where k ≥ 0 is an integer.

If p = 4h + 3, then from the properties of the Legendre’s symbol mod p we have (see [14], formula (30) in Chapter 9)

$∑a=0p−1ema4p=1+∑a=1p−11+χ2(a)ema2p=∑a=0p−1ema2p=iχ2(m)p,$

where χ2 = $\begin{array}{}\left(\frac{\ast }{p}\right)\end{array}$ denotes the Legendre’s symbol mod p.

So in this case, the problem we considered in (2) is trivial. If p = 4h + 1, then the situation is more complicated. We will use the analytic method and the properties of classical Gauss sums to study this problem, and prove some new interesting fourth-order linear recurrence formulae for (2) with p = 4 h + 1. That is, we will give the following four results.

#### Theorem 1.1

Let p be a prime with p = 24 h + 1. Then for any integer k ≥ 4, we have the fourth-order linear recurrence formula

$Vk(p)=6pVk−2(p)+8pαVk−3(p)−pp−4α2Vk−4(p),$

where the first four values are V0(p) = p2 – 6, V1(p) = p(p2 – 16p – 4α2), V2(p) = p2(2 + 3p – 58α) and V3(p) = p2(7p2 + 4 – 92p – 72α2), α= α(p) = $\begin{array}{}\sum _{a=1}^{\frac{p-1}{2}}\left(\frac{a+\overline{a}}{p}\right)\end{array}$ is an integer, which satisfies the identity (see Theorem 4-11 in [15])

$p=α2+β2≡∑a=1p−12a+a¯p2+∑a=1p−12a+ra¯p2,$

which r is any quadratic non-residue mod p.

#### Theorem 1.2

Let p be a prime with p = 24h + 17. Then for any integer k ≥ 4, we have the fourth-order linear recurrence formula

$Vk(p)=6pVk−2(p)+8pαVk−3(p)−pp−4α2Vk−4(p),$

where the first four values are V0(p) = – p2 – 6, V1(p) = p(p2 – 18p – 4α2), V2(p) = p2(2 – 3p – 62α) and V3(p) = p2(7p2 – 4 – 106p – 72α2).

#### Theorem 1.3

Let p be a prime with p = 24h + 5. Then for any integer k ≥ 4, we have the fourth-order linear recurrence formula

$Vk(p)=−2pVk−2(p)+8pαVk−3(p)−p9p−4α2Vk−4(p),$

where the first four terms are V0(p) = –(p2 + 6), V1(p) = – p(p2 – 8p + 4α2), V2(p) = – p2(2p – 22α) and V3(p) = p2(5p2 – 6 – 28p – 36α2).

#### Theorem 1.4

Let p be a prime with p = 24h + 13. Then for any integer k ≥ 4, we have the fourth-order linear recurrence formula

$Vk(p)=−2pVk−2(p)+8pαVk−3(p)−p9p−4α2Vk−4(p),$

where the first four terms are V0(p) = p2 – 6, V1(p) = – p(p2 – 6p + 4α2), V2(p) = – p2(2 + p – 18α) and V3(p) = p2(5p2 + 6 – 18p – 36α2).

From our theorems we may immediately deduce the following:

#### Corollary 1.5

Let p be a prime with p ≡ 1 mod 4, then we have the identity

$∑m=1p−1∑a=0p−1ema4p3⋅∑b=0p−1emb4+bp3=p27p2+4pα−92p−72α2ifp=24h+1,p27p2−4pα−106p−72α2ifp=24h+17,p25p2−6pα−28p−36α2ifp=24h+5,p25p2+6pα−18p−36α2ifp=24h+13.$

Note that the estimate |α| ≤ $\begin{array}{}\sqrt{p},\end{array}$ from Corollary 1.5 we also have the following:

#### Corollary 1.6

Let p be a prime with p ≡ 1 mod 8, then we have the asymptotic formula

$∑m=1p−1∑a=0p−1ema4p3⋅∑b=0p−1emb4+bp3=7p4+Op72.$

#### Corollary 1.7

Let p be a prime with p ≡ 5 mod 8, then we have the asymptotic formula

$∑m=1p−1∑a=0p−1ema4p3⋅∑b=0p−1emb4+bp3=5p4+Op72.$

For any prime p with p ≡ 1 mod 4 and any positive integer k, let Mk(p) denote the number of the solutions of the congruence equation

$x14+x24+⋯+xk4+y14+y24+y34≡0modp,y1+y2+y3≡0modp,$

where 0 ≤ xi, yjp – 1, i = 1, 2, …, k, j = 1, 2, 3.

Then from our theorems we can give an exact computational formula for Mk(p). For example, let Hs(p) denote the number of the congruence equation

$x14+x24+⋯+xs4≡0modp,0≤xi≤p−1,i=1,2,⋯s.$

Then we have the identity

$Vk(p)=p2p−1⋅Mk(p)−pp−1⋅Hk(p).$

Since Hk(p) has a fourth-order linear recurrence formula (see [8]), so from the above formula and our theorems we can deduce the exact value of Mk(p).

## 2 Several lemmas

To complete the proofs of our theorems, we need to prove four simple lemmas. Hereafter, we will use many properties of the classical Gauss sums and the fourth-order character mod p, all of which can be found in books concerning Elementary Number Theory or Analytic Number Theory, such as references [7], [14] or [15]. Some important results related to Gauss sums can also be found in [16] and [17]. These contents will not be repeated here. First we have the following:

#### Lemma 2.1

Let p be a prime with p ≡ 1 mod 4, λ be any fourth-order character mod p, then we have

$τ2(λ)+τ2λ¯=p⋅∑a=1p−1a+a¯p=2p⋅α,$

where$\begin{array}{}\tau \left(\lambda \right)=\sum _{a=1}^{p-1}\lambda \left(a\right)e\left(\frac{a}{p}\right)\end{array}$ denotes the classical Gauss sums, and $\begin{array}{}\left(\frac{\ast }{p}\right)\end{array}$ is the Legendre’s symbol mod p.

#### Proof

In fact this is Lemma 2 of [18], so its proof is omitted. □

#### Lemma 2.2

Let p be a prime with p ≡ 1 mod 4, then for any fourth-order character λ mod p, we have the identity

$∑m=1p−1λ(m)∑a=0p−1ema4+ap3=−5pτ(λ)−2pατλ¯ifp≡1mod8,−pτ(λ)−2pατλ¯ifp≡5mod8,$

where α is the same as in Lemma 2.1.

#### Proof

First applying trigonometric identity

$∑m=1qenmq=qifq∣n,0ifq∤n$(3)

and note that λ4 = χ0, the principal character mod p, we have

$∑m=1p−1λ(m)∑a=0p−1ema4+ap3=∑m=1p−1λ(m)∑a=0p−1ema4+ap2+∑m=1p−1λ(m)∑a=0p−1ema4+ap2∑a=1p−1ema4+ap=τ(λ)∑a=0p−1∑b=0p−1λ¯a4+b4+1∑c=1p−1ec(a+b+1)p+τ(λ)∑a=0p−1∑b=0p−1λ¯a4+b4ea+bp=τ(λ)p∑a=0p−1∑b=0p−1a+b+1≡0modp⁡λ¯a4+b4+1−τ(λ)∑a=0p−1∑b=0p−1λ¯a4+b4+1−τ(λ)+τ(λ)∑a=0p−1λ¯a4+1∑b=1p−1eb(a+1)p.$(4)

From (3) we have

$τ(λ)∑a=0p−1λ¯a4+1∑b=1p−1eb(a+1)p=λ¯(2)τ(λ)(p−1)−τ(λ)∑a=0p−2λ¯a4+1=λ¯(2)τ(λ)p−τ(λ)∑a=0p−1λ¯a4+1.$(5)

Note that the identity λχ2 = λ and

$B(m)=∑a=0p−1ema4p=χ2(m)p+λ¯(m)τ(λ)+λ(m)τλ¯.$(6)

From (6) we have

$τ(λ)∑a=0p−1λ¯a4+1=∑b=1p−1λ(b)∑a=0p−1eba4+1p=∑b=1p−1λ(b)χ2(b)p+λ¯(b)τ(λ)+λ(b)τλ¯ebp=pτλ¯−τ(λ)+pτλ¯=2pτλ¯−τ(λ).$(7)

If p ≡ 5 bmod 8, then note that λ(–1) = –1 and, τ(λ)τ(λ) = –p, applying (6) and Lemma 2.1 we also have

$∑a=0p−1∑b=0p−1λ¯a4+b4+1=1τ(λ)∑c=1p−1λ(c)∑a=0p−1∑b=0p−1eca4+cb4+cp=1τ(λ)∑c=1p−1λ(c)ecp∑a=0p−1eca4p2=1τ(λ)∑c=1p−1λ(c)ecpχ2(b)p+λ¯(b)τ(λ)+λ(b)τλ¯2=1τ(λ)∑c=1p−1λ(c)2χ2(c)pα−p+2λ(c)pτ(λ)+2λ¯(c)pτλ¯ecp=p+2pα−1τλ¯τ(λ).$(8)

Note that λ2 = χ2 = λ2 and the congruence, a + b + 1 ≡ 0 mod p implies the congruence a4 + b4 + 1 ≡ 2(a2 + a + 1)2 mod p. So we have

$∑a=0p−1∑b=0p−1a+b+1≡0modp⁡λ¯a4+b4+1=∑a=0p−1λ¯2a2+a+12=λ¯(2)∑a=0p−1χ2a2+a+1=λ¯(2)∑a=0p−1χ24a2+4a+4=λ¯(2)∑a=0p−1χ2(2a+1)2+3=λ¯(2)∑a=0p−1χ2a2+3=−λ¯(2).$(9)

Combining (4), (5), (7), (8) and (9) we have the identity

$∑m=1p−1λ(m)∑a=0p−1ema4+ap3=−pτ(λ)−2pατλ¯.$(10)

If p ≡ 1 mod 8, then λ(–1) = 1 and τ(λ)τ(λ) = p, from the method of proving (8) we have

$∑a=0p−1∑b=0p−1λ¯a4+b4+1=1τ(λ)∑c=1p−1λ(c)∑a=0p−1∑b=0p−1eca4+cb4+cp=1τ(λ)∑c=1p−1λ(c)ecpχ2(b)p+λ¯(b)τ(λ)+λ(b)τλ¯2=1τ(λ)∑c=1p−1λ(c)3p+2χ2(c)pα+2λ(c)pτ(λ)+2λ¯(c)pτλ¯ecp=5p+2pα−1τλ¯τ(λ).$(11)

Combining (4), (5), (7), (8) and (11) we have the identity

$∑m=1p−1λ(m)∑a=0p−1ema4+ap3=−5pτ(λ)−2pατλ¯.$(12)

Now Lemma 2.2 follows from (10) and (12). □

#### Lemma 2.3

Let p be a prime with p ≡ 1 mod 4, then we have the identity

$∑m=1p−1∑a=0p−1ema4+ap3=p2−6pαifp=24h+1orp=24h+13,−p2−6pαifp=24h+5orp=24h+17.$

#### Proof

From (3) we have

$∑m=1p−1∑a=0p−1ema4+ap3=p∑a=0p−1∑b=0p−1∑c=0p−1a4+b4+c4≡0modp⁡ea+b+cp=p∑a=0p−1∑b=0p−1a4+b4≡0modp⁡ea+bp+p∑a=0p−1∑b=0p−1a4+b4+1≡0modp⁡∑c=1p−1ec(a+b++1)p=p∑a=0p−1a4≡0modp⁡eap+p∑a=0p−1a4+1≡0modp⁡∑b=1p−1eb(a+1)p+p2∑a=0p−1∑b=0p−1a4+b4+1≡0modpa+b+1≡0modp⁡1−p∑a=0p−1∑b=0p−1a4+b4+1≡0modp⁡1=p−p∑a=0p−1a4+1≡0modp⁡1+p2∑a=0p−1∑b=0p−1a4+b4+1≡0modpa+b+1≡0modp⁡1−p∑a=0p−1∑b=0p−1a4+b4+1≡0modp⁡1.$(13)

Now we calculate each term in (13). If p ≡ 5 mod 8, then note that λ(–1) = –1 we have

$p∑a=0p−1a4+1≡0modp⁡1=0.$(14)

Applying (6) and Lemma 2.1 we have

$p∑a=0p−1∑b=0p−1a4+b4+1≡0modp⁡1=∑a=0p−1∑b=0p−1∑m=0p−1ema4+b4+1p=p2+∑m=1p−1∑a=0p−1ema4p2emp=p2+∑m=1p−12χ2(c)pα−p+2λ(c)pτ(λ)+2λ¯(c)pτλ¯emp=p2+2pα+p+2pτ2(λ)+2pτ2λ¯=p2+p+6pα.$(15)

It is clear that the congruences a4 + b4 + 1 ≡ 0 mod p and a + b + 1 ≡ 0 mod p implies that ab ≡ 1 mod p and a3 ≡ b3 ≡ 1 mod p with ab. So we have

$p2∑a=0p−1∑b=0p−1a4+b4+1≡0modpa+b+1≡0modp⁡1=p2∑a=2p−1∑b=2p−1a3≡b3≡1modpab≡1modp⁡1=2p2ifp≡1mod3,0ifp≡2mod3.$(16)

Applying (13), (14), (15) and (16) we have the identity

$∑m=1p−1∑a=0p−1ema4+ap3=p2−6pαifp=24h+13,−p2−6pαifp=24h+5.$(17)

If p ≡ 1 mod 8, then we also have

$p∑a=0p−1a4+1≡0modp⁡1=4p.$(18)

$p∑a=0p−1∑b=0p−1a4+b4+1≡0modp⁡1=∑a=0p−1∑b=0p−1∑m=0p−1ema4+b4+1p=p2+∑m=1p−13p+2χ2(c)pα+2λ(c)pτ(λ)+2λ¯(c)pτλ¯emp=p2+2pα−3p+2pτ2(λ)+2pτ2λ¯=p2−3p+6pα.$(19)

$p2∑a=0p−1∑b=0p−1a4+b4+1≡0modpa+b+1≡0modp⁡1=p2∑a=2p−1∑b=2p−1a3≡b3≡1modpab≡1modp⁡1=2p2ifp=24h+1,0ifp=24h+17.$(20)

Applying (13), (18), (19) and (20) we have

$∑m=1p−1∑a=0p−1ema4+ap3=p2−6pα ifp=24h+1,−p2−6pα ifp=24h+17.$(21)

It is clear that Lemma 2.3 follows from (17) and (21). □

#### Lemma 2.4

Let p be a prime with p ≡ 1 mod 4, then we have the identity

$∑m=1p−1χ2(m)∑a=0p−1ema4+ap3=p32(p−6)if p=24h+1,p32(p−8)if p=24h+17,−p32(p−4)if p=24h+13,−p32(p−6)if p=24h+5.$

#### Proof

From the properties of the Legendre’s symbol mod p we have

$∑m=1p−1χ2(m)∑a=0p−1ema4+ap3=∑m=1p−1χ2(m)∑a=0p−1ema4+ap2+∑m=1p−1χ2(m)∑a=0p−1ema4+ap2∑c=1p−1emc4+cp=p∑a=1p−1eap+p∑a=0p−1χ2a4+1∑b=1p−1eb(a+1)p+p∑a=0p−1∑b=0p−1χ2a4+b4+1∑c=1p−1ec(a+b+1)p=−p+χ2(2)p32−p∑a=0p−1χ2a4+1−p∑a=0p−1∑b=0p−1χ2a4+b4+1+p32∑a=0p−1∑b=0p−1a+b+1≡0modp⁡χ2a4+b4+1.$(22)

From the properties of fourth-order mod p and Lemma 2.1 we have

$∑a=0p−1χ2a4+1=1+∑a=1p−1χ2(a+1)1+λ(a)+χ2(a)+λ¯(a)=1p⋅τ2(λ+τ2λ¯−1=2α−1.$(23)

$∑a=0p−1∑b=0p−1a+b+1≡0modp⁡χ2a4+b4+1=∑a=0p−1χ2a4+(a+1)4+1=χ2(2)∑a=0p−1χ2a4+2a3+3a2+2a+1=χ2(2)∑a=0p−1χ2a2+a+12=χ2(2)p ifp=12h+1,χ2(2)(p−2) ifp=12h+5.$(24)

Note that τ(λ)τ(λ) = –p, if p = 8h+5. τ(λ)τ(λ) = p, if p = 8h+1. From the method of proving (15) and (19) we have

$p∑a=0p−1∑b=0p−1χ2a4+b4+1=∑a=0p−1∑b=0p−1∑m=1p−1χ2(m)ema4+mb4+mp=∑m=1p−1χ2(m)χ2p+λ(m)τλ¯+λ¯(m)τ(λ)2emp=7p32−2pα if p=8h+1,−5p32−2pα if p=8h+5.$(25)

Combining (22), (23), (24) and (25) we have

$∑m=1p−1χ2(m)∑a=0p−1ema4+ap3=p32(p−6) if p=24h+1,p32(p−8) if p=24h+17,−p32(p−4) if p=24h+13,−p32(p−6) if p=24h+5.$

This proves Lemma 2.4. □

## 3 Proofs of the theorems

Now we prove our main results. First we prove Theorem 1.1. If p = 24h + 1, then from Lemmas 2.1, 2.2 and 2.4 we have

$V1(p)=∑m=1p−1B(m)∑a=0p−1ema4+ap3=∑m=1p−1χ2(m)p+λ¯(m)τ(λ)+λ(m)τλ¯∑a=0p−1ema4+ap3=p2(p−6)−5p2−2pατ2λ−5p2−2pατ2λ¯=pp2−16p−4α2.$(26)

Applying Lemmas 2.12.4 we also have

$V2(p)=∑m=1p−1B2(m)∑a=0p−1ema4+ap3=∑m=1p−1χ2(m)p+λ¯(m)τ(λ)+λ(m)τλ¯2∑a=0p−1ema4+ap3=∑m=1p−13p+2χ2(m)pα+2λ(m)pτ(λ)+2λ¯(m)pτλ¯∑a=0p−1ema4+ap3=p22pα+3p−58α.$(27)

If p = 8h + 1, then from (6) we have

$B3(m)=χ2(m)p+λ¯(m)τ(λ)+λ(m)τλ¯3=7χ2(m)p32+4pα+5pλ¯(m)τ(λ)+λ(m)τλ¯ +2λ¯(m)τλ¯+λ(m)τλpα.$(28)

So if p = 24h + 1, then from (28), Lemmas 2.12.4 we have

$V3(p)=∑m=1p−1B3(m)∑a=0p−1ema4+ap3=7p3(p−6)+4pαp2−6pα−5pτλ¯5pτ(λ)+2pατλ¯−5pτλ5pτλ¯+2pατλ−2pατ(λ)5pτ(λ)+2pατλ¯−2pατλ¯5pτλ¯+2pατλ=p27p2+4pα−92p−72α2.$(29)

If p = 24h + 17, then from Lemmas 2.12.4 we have

$V1(p)=∑m=1p−1B(m)∑a=0p−1ema4+ap3=p2(p−8)−5p2−2pατ2λ−5p2−2pατ2λ¯=pp2−18p−4α2.$(30)

$V2(p)=∑m=1p−1B2(m)∑a=0p−1ema4+ap3=∑m=1p−13p+2χ2(c)pα+2λ(c)pτ(λ)+2λ¯(c)pτλ¯∑a=0p−1ema4+ap3=p22pα−3p−62α.$(31)

Applying (28) and the method of proving (29) we also have

$V3(p)=∑m=1p−1B3(m)∑a=0p−1ema4+ap3=7p3(p−8)−4pαp2+6pα−5pτλ¯5pτ(λ)+2pατλ¯−5pτλ5pτλ¯+2pατλ−2pατ(λ)5pτ(λ)+2pατλ¯−2pατλ¯5pτλ¯+2pατλ=p27p2−4pα−106p−72α2.$(32)

Similarly, if p = 24h + 5, then we have

$V1(p)=∑m=1p−1B(m)∑a=0p−1ema4+ap3=−p2(p−6)+p2−2pατ2λ+p2−2pατ2λ¯=−pp2−8p+4α2.$(33)

$V2(p)=∑m=1p−1B2(m)∑a=0p−1ema4+ap3=∑m=1p−1χ2(m)p+λ¯(m)τ(λ)+λ(m)τλ¯2∑a=0p−1ema4+ap3=∑m=1p−12χ2(c)pα−p+2λ(c)pτ(λ)+2λ¯(c)pτλ¯∑a=0p−1ema4+ap3=−p22pα−p−22α.$(34)

If p = 24h + 5, then from (6) we have

$B3(m)=χ2(m)p+λ¯(m)τ(λ)+λ(m)τλ¯3=−5χ2(m)p32+6pα+pλ¯(m)τ(λ)+λ(m)τλ¯+2λ¯(m)τλ¯+λ(m)τλpα.$(35)

So from (35) and the method of proving (29) we have

$V3(p)=∑m=1p−1B3(m)∑a=0p−1ema4+ap3=5p3(p−6)−6pαp2+6pα−pτλ¯pτ(λ)+2pατλ¯−pτλpτλ¯+2pατλ−2pατ(λ)pτ(λ)+2pατλ¯−2pατλ¯pτλ¯+2pατλ=p25p2−6pα−28p−36α2.$(36)

If p = 24h + 13, then (35), Lemmas 2.12.4 we have

$V1(p)=∑m=1p−1B(m)∑a=0p−1ema4+ap3=−p2(p−4)+p2−2pατ2λ+p2−2pατ2λ¯=−pp2−6p+4α2.$(37)

$V2(p)=∑m=1p−1B2(m)∑a=0p−1ema4+ap3=∑m=1p−1χ2(m)p+λ¯(m)τ(λ)+λ(m)τλ¯2∑a=0p−1ema4+ap3=∑m=1p−12χ2(c)pα−p+2λ(c)pτ(λ)+2λ¯(c)pτλ¯∑a=0p−1ema4+ap3=−p22pα+p−18α.$(38)

$V3(p)=∑m=1p−1B3(m)∑a=0p−1ema4+ap3=5p3(p−4)+6pαp2−6pα−pτλ¯pτ(λ)+2pατλ¯−pτλpτλ¯+2pατλ−2pατ(λ)pτ(λ)+2pατλ¯−2pατλ¯pτλ¯+2pατλ=p25p2+6pα−18p−36α2.$(39)

Finally, note that if p = 8h + 1, then from (6) and direct calculation (or see Lemma 3 in [7]) we have the identity

$B4(m)=6pB2(m)+8pαB(m)−pp−4α2.$(40)

For any prime p = 24h + 1 and integer k ≥ 4, from (26), (27), (29) and (40) we may immediately deduce the fourth-order linear recurrence formula

$Vk(p)=∑m=1p−1Bk(m)∑a=0p−1ema4+ap3=∑m=1p−1Bk−4(m)6pB2(m)+8pαB(m)−pp−4α2∑a=0p−1ema4+ap3=6pVk−2(p)+8pαVk−3(p)−pp−4α2Vk−4(p),$

where the first four values V0(p) = p2 – 6p α, V1(p) = p(p2 – 16p – 4α2), V2(p) = p2(2+3p – 58α) and V3(p) = p2(7p2+4p α – 92p – 72α2).

This proves Theorem 1.1.

If p = 24h + 17, then from (30), (31), (32) and (40) we have

$Vk(p)=6pVk−2(p)+8pαVk−3(p)−pp−4α2Vk−4(p),$

where the first four values V0(p) = – p2 – 6p α, V1(p) = p(p2 – 18p – 4α2), V2(p) = p2(2p α – 3p – 62α) and V3(p) = p2(7p2 – 4p α – 106p – 72α2).

This proves Theorem 1.2.

If p = 8h + 5, then from (6) and direct calculation (or see Lemma 3 in [7]) we also have

$B4(m)=6pB2(m)+8pαB(m)−pp−4α2.$(41)

For any prime p = 24h + 5 and integer k ≥ 4, from (33), (34), (35) and (41) we can deduce the fourth-order linear recurrence formula

$Vk(p)=6pVk−2(p)+8pαVk−3(p)−pp−4α2Vk−4(p),$

where the first four terms are V0(p) = –(p2+6p α), V1(p) = – p(p2 – 8p + 4α2), V2(p) = –p2(2p αp – 22α) and V3(p) = p2(5p2 – 6p α – 28p – 36α2).

This proves Theorem 1.3.

If p = 24h + 13, then from (37), (38), (39) and (41) we also have

$Vk(p)=−2pVk−2(p)+8pαVk−3(p)−p9p−4α2Vk−4(p),$

where the first four terms are V0(p) = p2 – 6p α, V1(p) = – p(p2 – 6p + 4α2), V2(p) = –p2(2p α+p – 18α) and V3(p) = p2(5p2+6p α – 18p – 36α2).

This completes the proofs of our all results.

## Acknowledgement

The author would like to thank the referees for their very helpful and detailed comments, which have significantly contributed to improving the presentation of this paper. This work is supported by the N. S. F. (11771351) of P. R. China and Northwest University Graduate Innovation and Creativity Founds (YZZ17086).

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Accepted: 2018-06-29

Published Online: 2018-08-24

Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 955–966, ISSN (Online) 2391-5455,

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