A Parameter for Ramanujan's Function χ(q): Its Explicit Values and Applications

Abstract

We define a new parameter 𝐼_𝑘,𝑛 involving quotient of Ramanujan's function 𝜒(𝑞) for positive real numbers 𝑘 and 𝑛 and study its several properties. We prove some general theorems for the explicit evaluations of the parameter 𝐼_𝑘,𝑛 and find many explicit values. Some values of 𝐼_𝑘,𝑛 are then used to find some new and known values of Ramanujan's class invariant 𝐺_𝑛.

1. Introduction

In Chapter 16 of his second notebook [1], Ramanujan develops the theory of theta-function. Ramanujan's general theta-function is defined by 𝑓(𝑎,𝑏)=^∞_𝑛_=−∞𝑎^{𝑛(𝑛+1)/2}𝑏^{𝑛(𝑛−1)/2},||𝑎𝑏||<1.(1) After Ramanujan, for |𝑞|<1, we define 𝑓(−𝑞)∶=𝑓−𝑞,−𝑞²=^∞_𝑛_=−∞(−1)^𝑛𝑞^{𝑛(3𝑛−1)/2}=(𝑞;𝑞)_∞,(2) where (𝑎;𝑞)_∞∶=∏^∞𝑛=0(1−𝑎𝑞^𝑛). If 𝑞=𝑒^{2𝜋𝑖𝑧} with Im(𝑧)>0, then 𝑓(−𝑞)=𝑞^−1/24𝜂(𝑧), where 𝜂(𝑧) denotes the classical Dedekind eta function.

Ramanujan's function 𝜒(𝑞) is defined by 𝜒(𝑞)∶=𝑓(𝑞)𝑓−𝑞²=−𝑞;𝑞²_∞.(3) The function 𝜒(𝑞) is intimately connected to Ramanujan's class invariants 𝐺_𝑛 and 𝑔_𝑛, which are defined by 𝐺_𝑛=2^−1/4𝑞^−1/24𝜒(𝑞),𝑔_𝑛=2^−1/4𝑞^−1/24𝜒(−𝑞),(4) where 𝑞∶=𝑒^{√−𝜋𝑛} and 𝑛 is a positive rational number. Since from [2, page 124, Entry 12(v) & (vi)] 𝜒(𝑞)=2^1/6𝛼(1𝑞−𝛼)^−1/24𝜒(−𝑞)=2_1/6,(1−𝛼)^1/12𝛼𝑞^−1/24,(5) it follows from (4) that 𝑔_𝑛𝐺_𝑛={4𝛼(1−𝛼)}^−1/24,=2^−1/12(1−𝛼)^1/12𝛼^−1/24.(6) Also, if 𝛽 has degree 𝑟 over 𝛼, then 𝐺_𝑟2𝑛={4𝛽(1−𝛽)}^−1/24,𝑔_𝑟2𝑛=2^−1/12(1−𝛽)^1/12𝛽^−1/24.(7) In his notebooks [1] and paper [3], Ramanujan recorded a total of 116 class invariants or monic polynomials satisfied by them. The table at the end of Weber's book [4, page 721–726] also contains the values of 107 class invariants. Weber primarily was motivated to calculate class invariants so that he could construct Hilbert class fields. On the other hand Ramanujan calculated class invariants to approximate 𝜋 and probably for finding explicit values of Rogers-Ramanujan continued fractions, theta-functions, and so on. An account of Ramanujan's class invariants and applications can be found in Berndt's book [5]. For further references, see [6–12].

Ramanujan and Weber independently and many others in the literature calculated class invariants 𝐺_𝑛 for odd values of 𝑛 and 𝑔_𝑛 for even values of 𝑛. For the first time, Yi [13] calculated some values of 𝑔_𝑛 for odd values of 𝑛 by finding explicit values of the parameter 𝑟_𝑘,𝑛 (see [13, page 11, (2.1.1)] or [14, page 4, (1.11)]) defined by 𝑟_𝑘,𝑛∶=𝑓(−𝑞)𝑘^1/4𝑞^{(𝑘−1)/24}𝑓−𝑞^𝑘,𝑞=𝑒^{√−2𝜋𝑛/𝑘}.(8) In particular, she established the result [13, page 18, Theorem 2.2.3] 𝑔_𝑛=𝑟_2,𝑛/2.(9) However, the values of 𝐺_𝑛 for even values of 𝑛 have not been calculated. The main objective of this paper is to evaluate some new values of 𝐺_𝑛 for even values of 𝑛. We also prove some known values of 𝐺_𝑛. For evaluation of class invariant 𝐺_𝑛 in this paper, we introduce the parameter 𝐼_𝑘,𝑛, which is defined as 𝐼_𝑘,𝑛∶=𝜒(𝑞)𝑞^{(−𝑘+1)/24}𝜒𝑞^𝑘,𝑞=𝑒^{√−𝜋𝑛/𝑘},(10) where 𝑘 and 𝑛 are positive real numbers.

In Section 3, we study some properties of 𝐼_𝑘,𝑛 and also establish its relations with Ramanujan's class invariant 𝐺_𝑛. In Section 4, by employing Ramanujan's modular equations, we present some general theorems for the explicit evaluations of 𝐼_𝑘,𝑛 and find several explicit values of 𝐼_𝑘,𝑛. In Section 5, we establish some general theorems connecting the parameter 𝐼_𝑘,𝑛 and the class invariant 𝐺_𝑛. We also evaluate some explicit values of the product 𝐺_𝑛𝑘𝐺_𝑛/𝑘 by employing some values of 𝐼_𝑘,𝑛 evaluated in Section 4. Finally, in Section 6, we calculate new and known values of class invariant 𝐺_𝑛 by combining the explicit values of 𝐼_𝑘,𝑛 and the product 𝐺_𝑛𝑘𝐺_𝑛/𝑘 evaluated in Sections 4 and 5, respectively. Section 2 is devoted to record some preliminary results.

Since Ramanujan's modular equations are key in our evaluations of 𝐼_𝑘,𝑛 and 𝐺_𝑛, we complete this introduction by defining Ramanujan's modular equation from Berndt's book [2]. The complete elliptic integral of the first kind 𝐾(𝑘) is defined by 𝐾(𝑘)∶=^0𝜋/2𝑑𝜙√1−𝑘²sin²𝜙=𝜋2^∞_𝑛=0(1/2)^2𝑛(𝑛!)²𝑘^2𝑛=𝜋2²𝐹₁12,12;1;𝑘²,(11) where 0<𝑘<1, ₂𝐹₁ denotes the ordinary or Gaussian hypergeometric function, and (𝑎)_𝑛=𝑎(𝑎+1)(𝑎+2)⋯(𝑎+𝑛−1).(12) The number 𝑘 is called the modulus of 𝐾, and 𝑘^∶=√1−𝑘² is called the complementary modulus. Let 𝐾,𝐾^, 𝐿, and 𝐿^ denote the complete elliptic integrals of the first kind associated with the moduli 𝑘, 𝑘^, 𝑙, and 𝑙^, respectively. Suppose that the equality 𝑛𝐾′𝐾=𝐿𝐿′(13) holds for some positive integer 𝑛. Then, a modular equation of degree 𝑛 is a relation between the moduli 𝑘 and 𝑙, which is implied by (13).

If we set 𝑞=exp−𝜋𝐾^𝐾,𝑞^=exp−𝜋𝐿^𝐿,(14) we see that (13) is equivalent to the relation 𝑞^𝑛=𝑞^. Thus, a modular equation can be viewed as an identity involving theta-functions at the arguments 𝑞 and 𝑞^𝑛. Ramanujan recorded his modular equations in terms of 𝛼 and 𝛽, where 𝛼=𝑘² and 𝛽=𝑙². We say that 𝛽 has degree 𝑛 over 𝛼. The multiplier 𝑚 connecting 𝛼 and 𝛽 is defined by 𝑚=𝐾𝐿.(15) Ramanujan also established many “mixed” modular equations in which four distinct moduli appear, which we define from Berndt's book [2, page 325].

Let 𝐾, 𝐾^, 𝐿₁, 𝐿^1, 𝐿₂, 𝐿^2, 𝐿₃, and 𝐿^3 denote complete elliptic integrals of the first kind corresponding, in pairs, to the moduli √𝛼, √𝛽, √𝛾, and √𝛿 and their complementary moduli, respectively. Let 𝑛₁, 𝑛₂, and 𝑛₃ be positive integers such that 𝑛₃=𝑛₁𝑛₂. Suppose that the equalities 𝑛₁𝐾′𝐾=𝐿𝐿^1₁,𝑛₂𝐾′𝐾=𝐿𝐿^2₂,𝑛₃𝐾′𝐾=𝐿𝐿^3₃(16) hold. Then, a “mixed” modular equation is a relation between the moduli √𝛼, √𝛽, √𝛾, and √𝛿 that is induced by (16). In such an instance, we say that 𝛽, 𝛾, and 𝛿 are of degrees 𝑛₁, 𝑛₂, and 𝑛₃, respectively, over 𝛼 or 𝛼, 𝛽, 𝛾, and 𝛿 have degrees 1, 𝑛₁, 𝑛₂, and 𝑛₃, respectively. Denoting 𝑧_𝑟=𝜙²(𝑞^𝑟), where 𝑞=exp−𝜋𝐾^𝐾,𝜙(𝑞)=𝑓(𝑞,𝑞),||𝑞||<1(17) the multipliers 𝑚 and 𝑚^ associated with 𝛼, 𝛽, and 𝛾, 𝛿, respectively, are defined by 𝑚=𝑧₁/𝑧_𝑛1 and 𝑚′=𝑧_𝑛2/𝑧_𝑛3.

2. Preliminary Results

Lemma 1 (see [2, page 43, Entry 27(v)]). If 𝛼 and 𝛽 are such that the modulus of each exponential argument is less than 1 and 𝛼𝛽=𝜋², then 𝑒^𝛼/24𝜒(𝑒^−𝛼)=𝑒^𝛽/24𝜒𝑒^−𝛽.(18)

Lemma 2 (see [15, page 241, Lemma 2.3]). Let 𝑋∶=𝑞^1/12(𝜒(𝑞)/𝜒(𝑞³)) and 𝑌∶=𝑞^1/6(𝜒(𝑞²)/𝜒(𝑞⁶)); then 𝑋𝑌×¹²+(𝑋𝑌𝑋𝑌)¹²₁₀+𝑋𝑌⁶+𝑋𝑌⁶+71(𝑋𝑌+)₂(𝑋𝑌)⁻¹⁰+16(𝑋𝑌)⁶+(𝑋𝑌)⁻⁶=(𝑋𝑌)₁₂+(𝑋𝑌)⁻²+200(𝑋+𝑌)(₄𝑋𝑌)⁻¹²+25(𝑋𝑌)⁸+(𝑋𝑌)⁻⁸+(𝑋𝑌)⁻⁴+550.(19)

Lemma 3 (see [15, page 241, Lemma 2.8]). Let 𝑋∶=𝑞^1/6(𝜒(𝑞)/𝜒(𝑞⁵)) and 𝑌∶=𝑞^1/3(𝜒(𝑞²)/𝜒(𝑞¹⁰)); then 𝑋𝑌³+𝑌𝑋³(𝑋𝑌)⁵+19𝑋𝑌+(𝑋𝑌)₋₁+(𝑋𝑌)⁻⁵+8(𝑋𝑌)³+(𝑋𝑌)⁻³+𝑋𝑌⁶+𝑋𝑌⁶+=52(𝑋𝑌(𝑋)⁶𝑌+)(₂𝑋𝑌)⁻⁶+13(𝑋𝑌)⁴+(𝑋𝑌)⁻⁴+(𝑋𝑌)⁻²+82.(20)

Lemma 4 (see [15, page 252, Lemma 2.13]). Let 𝑋∶=𝑞^1/4(𝜒(𝑞)/𝜒(𝑞⁷)) and 𝑌∶=𝑞^1/2(𝜒(𝑞²)/𝜒(𝑞¹⁴)); then 𝑋𝑌+¹²8+𝑋𝑋𝑌¹²+16𝑋𝑌¹⁰+𝑋𝑌¹⁰(𝑋𝑌)²+(𝑋𝑌)⁻²𝑌⁸+𝑌𝑋⁸7(𝑋𝑌)⁴+𝑋+(𝑋𝑌)⁻⁴−19𝑌⁶+𝑌𝑋⁶×(𝑋𝑌)¹⁰+(𝑋𝑌)⁻¹⁰+−8312((𝑋𝑋𝑌𝑌))⁶₂+(𝑋Y)⁻⁶+𝑋+(𝑋𝑌)⁻²𝑌⁴+𝑌𝑋⁴16(𝑋𝑌)⁸−288(𝑋𝑌)₄+(𝑋𝑌)⁻⁸+𝑋+(𝑋𝑌)⁻⁴+352𝑌²+𝑌𝑋²296(𝑋𝑌)²−256(𝑋𝑌)₆+(𝑋𝑌)⁻²−8(𝑋𝑌)₁₀+(𝑋𝑌)⁻⁶=(𝑋𝑌)₁₂+(𝑋𝑌)⁻¹⁰+1746+496(𝑋𝑌+)(₄𝑋𝑌)⁻¹²+145(𝑋𝑌)⁸+(𝑋𝑌)⁻⁸+(𝑋𝑌)⁻⁴.(21)

Lemma 5 (see [2, page 231, Entry 5(xii)]). Let 𝑃={16𝛼𝛽(1−𝛼)(1−𝛽)}^1/8 and 𝑄=(𝛽(1−𝛽)/𝛼(1−𝛼))^1/4; then 𝑄+𝑄⁻¹+2√2𝑃−𝑃⁻¹=0,(22) where 𝛽 has degree 3 over 𝛼.

Lemma 6 (see [2, page 282, Entry 13(xiv)]). Let 𝑃={16𝛼𝛽(1−𝛼)(1−𝛽)}^1/12 and 𝑄=(𝛽(1−𝛽)/𝛼(1−𝛼))^1/8; then 𝑄+𝑄⁻¹+2𝑃−𝑃⁻¹=0,(23) where 𝛽 has degree 5 over 𝛼.

Lemma 7 (see [2, page 315, Entry 13(xiv)]). Let 𝑃={16𝛼𝛽(1−𝛼)(1−𝛽)}^1/8 and 𝑄=(𝛽(1−𝛽)/𝛼(1−𝛼))^1/6; then 𝑄+𝑄⁻¹+7=2√2𝑃+𝑃⁻¹=0,(24) where 𝛽 has degree 7 over 𝛼.

Lemma 8 (see [5, page 378, Entry 41]). Let 𝑃=2^1/6{𝛼𝛽(1−𝛼)(1−𝛽)}^1/24 and 𝑄=(𝛽(1−𝛽)/𝛼(1−𝛼))^1/24; then 𝑄⁷+−𝑄8⁻⁷𝑃+1₋₆3𝑄⁵+𝑄⁻⁵+52𝑄³+𝑄⁻³+78𝑄+𝑄⁻¹+𝑃⁶=0,(25) where 𝛽 has degree 13 over 𝛼.

For Lemmas 9 to 15, we set 𝑃∶=(25𝑄6𝛼∶𝛽=𝛾𝛿(1𝛼−𝛿(𝛼1)−(1𝛼−)(𝛽1)−(1𝛿−)𝛾)(1−𝛿))^1/48,𝛽𝛾(1−𝛽)(1−𝛾)^1/48𝑅∶=𝛾𝛿(1−𝛾)(1−𝛿),𝛼𝛽(1−𝛼)(1−𝛽)^1/48𝑇∶=𝛽𝛿(1−𝛽)(1−𝛿),𝛼𝛾(1−𝛼)(1−𝛾)^1/48.(26)

Lemma 9 (see [5, page 381, Entry 50]). If 𝛼, 𝛽, 𝛾, and 𝛿 have degrees 1, 5, 7, and 35, respectively, then 𝑅⁴+𝑅⁻⁴−𝑄⁶+𝑄⁻⁶+5𝑄⁴+𝑄⁻⁴−10𝑄²+𝑄⁻²+15=0.(27)

Lemma 10 (see [5, page 381, Entry 51]). If 𝛼,𝛽,𝛾, and 𝛿 have degrees 1, 13, 3, and 39, respectively, then 𝑄⁴+𝑄⁻⁴−3𝑄²+𝑄⁻²−𝑇²+𝑇⁻²+3=0.(28)

Lemma 11 (see [5, page 381, Entry 52]). If 𝛼,𝛽,𝛾, and 𝛿 have degrees 1, 13, 5, and 65, respectively, then 𝑄⁶+𝑄⁻⁶−5𝑄+𝑄⁻¹²𝑇+𝑇⁻¹²−𝑇⁴+𝑇⁻⁴=0.(29)

Lemma 12 (see [16, page 277, Lemma 3.1]). If 𝛼,𝛽,𝛾, and 𝛿 have degrees 1, 3, 7, and 21, respectively, then 𝑅²+𝑅⁻²−𝑄⁴−𝑄⁻⁴+3=0.(30)

Lemma 13 (see [15, page 243, Theorem 2.5]). If 𝛼, 𝛽, 𝛾, and 𝛿 have degrees 1, 2, 3, and 6, respectively, then 𝑅⁴+𝑅⁻⁴+𝑃²−2𝑃⁻²=0.(31)

Lemma 14 (see [15, page 248, Theorem 2.10]). If 𝛼, 𝛽, 𝛾 and 𝛿 have degrees 1, 2, 5, and 10, respectively, then 𝑅⁶+𝑅⁻⁶+5𝑅²+𝑅⁻²=4𝑃⁻⁴−𝑃⁴.(32)

Lemma 15 (see [15, page 252, Theorem 2.12]). If 𝛼, 𝛽, 𝛾 and 𝛿 have degrees 1, 2, 7, and 14, respectively, then 2√2𝑇³+𝑇⁻³=4𝑃⁻¹−2𝑃+4𝑃³−𝑃⁵.(33)

3. Properties of 𝐼_𝑘,𝑛

In this section, we study some properties of 𝐼_𝑘,𝑛. We also establish a relation connecting 𝐼_𝑘,𝑛 and Ramanujan's class invariants 𝐺_𝑛.

Theorem 16. For all positive real numbers 𝑘 and 𝑛, one has (i)𝐼_𝑘,1=1,(ii)𝐼_𝑘,1𝑛=1𝐼_𝑘,𝑛,(iii)𝐼_𝑘,𝑛=𝐼_𝑛,𝑘.(34)

Proof . Using the definition of 𝐼_𝑘,𝑛 and Lemma 1, we easily arrive at (i). Replacing 𝑛 by 1/𝑛 in 𝐼_𝑘,𝑛 and using Lemma 1, we find that 𝐼_𝑘,𝑛𝐼_𝑘,1/𝑛=1, which completes the proof of (ii). To prove (iii), we use Lemma 1 in the definition of 𝐼_𝑘,𝑛 to arrive at (𝐼_𝑘,𝑛/𝐼_𝑛,𝑘)=1.

Remark 17. By using the definitions of 𝜒(𝑞) and 𝐼_𝑘,𝑛, it can be seen that 𝐼_𝑘,𝑛 has positive real value less than 1 and that the values of 𝐼_𝑘,𝑛 decrease as 𝑛 increases when 𝑘>1. Thus, by Theorem 16(i), 𝐼_𝑘,𝑛<1 for all 𝑛>1 if 𝑘>1.

Theorem 18. For all positive real numbers k, m, and n, one has 𝐼_{𝑘,𝑛/𝑚}=𝐼_{𝑚𝑘,𝑛}𝐼^{−1𝑛𝑘,𝑚}.(35)

Proof. Using the definition of 𝐼_𝑘,𝑛, we obtain 𝐼_{𝑚𝑘,𝑛}𝐼_{𝑛𝑘,𝑚}=𝜒𝑒^{√−𝜋𝑛/𝑚𝑘}𝑒^{√𝜋(√𝑚/𝑛𝑘−𝑛/𝑚𝑘)/24}𝜒𝑒^{√−𝜋𝑚/𝑛𝑘}.(36) Using Lemma 1 in the denominator of the right-hand side of (36) and simplifying, we complete the proof.

Corollary 19. For all positive real numbers k and n, one has 𝐼_𝑘²_,𝑛=𝐼_{𝑛𝑘,𝑛}𝐼_{𝑘,𝑛/𝑘}.(37)

Proof. Setting 𝑘=𝑛 in Theorem 18 and simplifying using Theorem 16(ii), we obtain 𝐼_𝑘²_,𝑚=𝐼_{𝑚𝑘,𝑘}𝐼_{𝑘,𝑚/𝑘}.(38) Replacing 𝑚 by 𝑛, we complete the proof.

Theorem 20. Let k, a, b, c, and d be positive real numbers such that ab=cd. Then 𝐼_𝑎,𝑏𝐼_{𝑘𝑐,𝑘𝑑}=𝐼_{𝑘𝑎,𝑘𝑏}𝐼_𝑐,𝑑.(39)

Proof. From the definition of 𝐼_𝑘,𝑛, we deduce that, for positive real numbers 𝑘,𝑎,𝑏,𝑐, and 𝑑, 𝐼_{𝑘𝑎,𝑘𝑏}𝐼^{−1𝑎,𝑏}=𝜒𝑒^{√−𝜋𝑎𝑏}𝐼_{𝑘𝑐,𝑘𝑑}𝑒^{√𝜋(𝑘√𝑎𝑏−𝑎𝑏)/24}𝜒𝑒^{√−𝑘𝜋𝑎𝑏}𝐼^{−1𝑐,𝑑}=𝜒𝑒^{√−𝜋𝑐𝑑}𝑒^{√𝜋(𝑘√𝑐𝑑−𝑐𝑑)/24}𝜒𝑒^{√−𝑘𝜋𝑐𝑑}.(40) Now the result follows readily from (40), and the hypothesis that 𝑎𝑏=𝑐𝑑.

Corollary 21. For any positive real numbers 𝑛 and 𝑝, we have 𝐼_{𝑛𝑝,𝑛𝑝}=𝐼_{𝑛𝑝2,𝑛}𝐼_𝑝,𝑝.(41)

Proof. The result follows immediately from Theorem 20 with 𝑎=𝑝²,𝑏=1,𝑐=𝑑=𝑝, and 𝑘=𝑛.

Now, we give some relations connecting the parameter 𝐼_𝑘,𝑛 and Ramanujan's class invariants 𝐺_𝑛.

Theorem 22. Let 𝑘 and 𝑛 be any positive real numbers. Then (i)𝐼_𝑘,𝑛=𝐺_𝑛/𝑘𝐺^−1𝑛𝑘,(ii)𝐺_1/𝑛=𝐺_𝑛.(42)

Proof. Proof of (i) follows easily from the definitions of 𝐼_𝑘,𝑛 and 𝐺_𝑛 from (10) and (4), respectively. To prove (ii), we set 𝑘=1 in part (i) and use Theorem 16(i) and (iii).

4. General Theorems and Explicit Evaluations of 𝐼_𝑘,𝑛

In this section, we prove some general theorems for the explicit evaluations of 𝐼_𝑘,𝑛 and find its explicit values.

Theorem 23. One has 𝐼𝐼^3,𝑛_3,4𝑛¹²+𝐼^3,𝑛𝐼_3,4𝑛⁻¹²+𝐼^3,𝑛𝐼_3,4𝑛⁶+𝐼^3,𝑛×𝐼_3,𝑛𝐼_3,4𝑛⁻⁶+71𝐼𝐼_{3,4𝑛3,𝑛}¹⁰+𝐼_3,𝑛𝐼_3,4𝑛⁻¹⁰+16𝐼_3,𝑛𝐼_3,4𝑛⁶+(𝐼_3,𝑛𝐼_3,4𝑛)⁻⁶=𝐼𝐼_3,4𝑛²+𝐼_3,𝑛𝐼_3,4𝑛⁻²_3,𝑛+200𝐼𝐼^3,4𝑛¹²_3,𝑛+𝐼_3,𝑛𝐼_3,4𝑛⁻¹²+25𝐼_3,𝑛𝐼_3,4𝑛⁸+𝐼_3,n𝐼_3,4𝑛⁻⁸𝐼_3,4𝑛⁴+𝐼_3,𝑛𝐼_3,4𝑛⁻⁴+550.(43)

Proof. The proof follows easily from the definition of 𝐼_𝑘,𝑛 and Lemma 2.

Corollary 24. One has ((iii))𝐼𝐼_3,23,4=−44+27√3−3458−264√3^1/12,=⎛⎜⎜⎜⎜⎝−2+3√6−−6+3√62⎞⎟⎟⎟⎟⎠^1/2(iii)𝐼_3,1/2,(iv)𝐼_3,1/4=−44+27√3+3458−264√3^1/12,=⎛⎜⎜⎜⎜⎝−6+3√6+−2+3√62⎞⎟⎟⎟⎟⎠^1/2.(44)

Proof. Setting 𝑛=1/2 in Theorem 23 and using Theorem 16(ii), we obtain 𝐼^243,2+𝐼^−243,2+176𝐼^123,2+𝐼^−123,2−1002=0.(45) Equivalently, 𝐵²+176𝐵−1004=0,(46) where 𝐵=𝐼^123,2+𝐼^−123,2.(47) Solving (46) and using the fact in Remark 17, we obtain 𝐵=54√5−88.(48) Employing (48) in (47), solving the resulting equation for 𝐼_3,2, and noting that 𝐼_3,2<1, we arrive at 𝐼_3,2=−44+27√3−3458−264√3^1/12.(49) This completes the proof of (i).
Again setting 𝑛=1 in Theorem 23 and using Theorem 16(i), we obtain +𝐼16^63,4+𝐼𝐼^−63,4_63,4I^103,4+𝐼^−103,4=25𝐼+_83,4𝐼^−63,4+71𝐼^23,4+𝐼^−23,4+𝐼^−83,4+200𝐼^43,4+𝐼^−43,4+550.(50) Equivalently, 𝐷²+1²𝐷⁴+4𝐷²−50=0,(51) where 𝐷=𝐼^23,4+𝐼^−23,4.(52) Since the first factor of (51) is nonzero, solving the second factor, we deduce that 𝐷=−2+3√6^1/2.(53) Employing (53) in (52), solving the resulting equation, and using the fact that 𝐼_3,4<1, we obtain 𝐼_3,4=⎛⎜⎜⎜⎜⎝−2+3√6−−6+3√62⎞⎟⎟⎟⎟⎠^1/2.(54) This completes the proof of (ii).
Now (iii) and (iv) follow from (i) and (ii), respectively, and Theorem 16(ii).

Theorem 25. One has 𝐼𝐼^5,𝑛_5,4𝑛⁶+𝐼^5,𝑛𝐼_5,4𝑛⁻⁶+𝐼^5,𝑛𝐼_5,4𝑛³+𝐼^5,𝑛×𝐼_5,𝑛𝐼_5,4𝑛⁻³+19𝐼𝐼_{5,4𝑛5,𝑛}⁵+𝐼_5,𝑛𝐼_5,4𝑛⁻⁵+8𝐼_5,𝑛𝐼_5,4𝑛³+𝐼_5,𝑛𝐼_5,4𝑛⁻³=𝐼𝐼_5,4𝑛+𝐼_5,𝑛𝐼_5,4𝑛_5,𝑛+52𝐼𝐼^5,4𝑛_5,𝑛⁶+𝐼_5,𝑛𝐼_5,4𝑛⁻⁶+13𝐼_5,𝑛𝐼_5,4𝑛⁴+𝐼_5,𝑛𝐼_5,4𝑛⁻⁴𝐼_5,4𝑛²+𝐼_5,𝑛𝐼_5,4𝑛⁻²+82.(55)

Proof. The proof follows from Lemma 3 and the definition of 𝐼_𝑘,𝑛.

Corollary 26. One has ((iii))𝐼𝐼_5,25,4=−14+5√10−445−140√10^1/6,=11+5√5^1/4−−4+11+5√5(iii)𝐼_5,1/22,(iv)𝐼_5,1/4=−14+5√10+445−140√10^1/6,=11+5√5^1/4+−4+11+5√52.(56)

Proof. Setting 𝑛=1/2 in Theorem 25 and using Theorem 16(ii), we obtain 𝐶²+56𝐶−216=0,(57) where 𝐶=𝐼^65,2+𝐼^−65,2.(58) Solving (57) and noting the fact in Remark 17, we obtain 𝐶=−28+10√10.(59) Employing (59) in (58), solving the resulting equation, and noting that 𝐼_5,2<1, we obtain 𝐼_5,2=−14+5√10−445−140√10^1/6.(60) This completes the proof of (i).
Again, setting 𝑛=1 in Theorem 25 and using Theorem 16(i), we obtain 𝐵⁸−22𝐵⁴−4=0,(61) where 𝐵=𝐼_5,4+𝐼^−14,5.(62) Solving (61), we obtain 𝐵=11+5√5^1/4.(63) Using (63) in (62), solving the resulting equation, and noting that 𝐼_5,4<1, we arrive at 𝐼_5,4=11+5√5^1/4−−4+11+5√52.(64) This completes the proof of (ii).
Now (iii) and (iv) follow from (i) and (ii), respectively, and Theorem 16(ii).

Theorem 27. One has 𝐼𝐼^7,𝑛_4,7𝑛¹²+𝐼^7,𝑛𝐼_4,7𝑛⁻¹²+16𝐼^7,𝑛𝐼_4,7𝑛¹⁰+𝐼^7,𝑛𝐼_4,7𝑛⁻¹⁰𝐼^7,𝑛+8𝐼_7,𝑛𝐼_7,4𝑛²+𝐼_7,𝑛𝐼_7,4𝑛⁻²𝐼_4,7𝑛⁸+𝐼^7,𝑛×7𝐼_7,𝑛𝐼_4,7𝑛⁻⁸+𝐼𝐼_7,4𝑛⁴+𝐼_7,𝑛𝐼_7,4𝑛⁻⁴−19_7,𝑛𝐼_4,7𝑛⁶+𝐼^7,𝑛×𝐼_7,𝑛𝐼_4,7𝑛⁻⁶𝐼_7,4𝑛¹⁰+𝐼_7,𝑛𝐼_7,4𝑛⁻¹⁰+−8312𝐼𝐼_7,𝑛7,𝑛𝐼_7,4𝑛⁶+𝐼_7,𝑛𝐼_7,4𝑛⁻⁶+𝐼𝐼_7,4𝑛²+𝐼_7,𝑛𝐼_7,4𝑛⁻²_7,𝑛𝐼_4,7𝑛⁴+𝐼^7,𝑛×16𝐼_7,𝑛𝐼_4,7𝑛⁻⁴−288𝐼_7,𝑛𝐼_7,4𝑛⁸+𝐼_7,𝑛𝐼_7,4𝑛⁻⁸+𝐼𝐼_7,4𝑛⁴+𝐼_7,𝑛𝐼_7,4𝑛⁻⁴+352_7,𝑛𝐼_4,7𝑛²+𝐼^7,𝑛×296𝐼_7,𝑛𝐼_4,7𝑛⁻²−256𝐼_7,𝑛𝐼_7,4𝑛²+𝐼_7,𝑛𝐼_7,4𝑛⁻²−8𝐼_7,𝑛𝐼_7,4𝑛⁶+𝐼_7,𝑛𝐼_7,4𝑛⁻⁶=𝐼𝐼_7,4𝑛¹⁰+𝐼_7,𝑛𝐼_7,4𝑛⁻¹⁰+1746_7,𝑛+145𝐼𝐼^7,4𝑛_7,𝑛¹²+𝐼_7,𝑛𝐼_7,4𝑛⁻¹²+496𝐼_7,𝑛𝐼_7,4𝑛⁸+𝐼_7,𝑛𝐼_7,4𝑛⁻⁸𝐼_7,4𝑛⁴+𝐼_7,𝑛𝐼_7,4𝑛⁻⁴.(65)

Corollary 28. One has (i)𝐼_7,2=2^−1/4×−⎛⎜⎜⎝−8−5√2+163+116√2−4+−8−5√2+163+116√2²⎞⎟⎟⎠^1/4(ii)𝐼_7,4,=3−6+5√2−−58+45√2(iii)𝐼_7,1/2√2,×⎛⎜⎜⎝−8−=5√2^−1/4,+2+163+116√2−4+−8−5√2+163+116√2²⎞⎟⎟⎠^1/4(iv)𝐼_7,1/4=3−6+5√2+−58+45√2√2.(66)

Proof. Setting 𝑛=1/2 and simplifying using Theorem 16(ii), we obtain −𝐼96^247,2+𝐼𝐼^−247,2_127,2+32𝐼^207,2+𝐼^−207,2−40𝐼^167,2+𝐼^−167,2+𝐼^−127,2−192𝐼^87,2+𝐼^−87,2+64𝐼^47,2+𝐼^−47,2+462=0.(67) Equivalently, 𝐴²−4𝐴⁴+32𝐴³−42𝐴²−128𝐴−191=0,(68) where 𝐴=𝐼^47,2+𝐼^−47,2.(69) By using the fact in Remark 17, it is seen that the first factor of (68) is nonzero, and so from the second factor, we deduce that 𝐴=−8−5√2+163+116√2.(70) Combining (69) and (70) and noting that 𝐼_7,2<1, we obtain 𝐼_7,2=2^−1/4⎛⎜⎜⎝−8−5√−2+163+116√2−4+−8−5√2+163+116√2²⎞⎟⎟⎠^1/4.(71) This completes the proof of (i).

To prove (ii), setting 𝑛=1 and simplifying using Theorem 16(i), we arrive at 𝐸²𝐸²−4𝐸⁴+108𝐸²−1134=0,(72)

where 𝐸=𝐼^27,4+𝐼^−27,4.(73)

Using the fact in Remark 17 it is seen that the first two factors of (72) are nonzero, and so solving the third factor, we obtain 𝐸=3−6+5√2.(74)

Combining (73) and (74) and noting that 𝐼_7,4<1, we deduce that 𝐼_7,4=3−6+5√2−−58+45√2√2.(75) So the proof of (ii) is complete.

Now (iii) and (iv) follow from (i) and (ii), respectively, and Theorem 16(ii).

Theorem 29. One has 𝐼_7,𝑛𝐼_7,25𝑛²+𝐼_7,𝑛𝐼_7,25𝑛⁻²−𝐼𝐼^7,25𝑛_7,𝑛³+𝐼𝐼^7,25𝑛_7,𝑛⁻³+5𝐼_7,25𝑛𝐼_7,𝑛²+𝐼^7,25𝑛−10𝐼_7,25𝑛𝐼_7,𝑛⁻²𝐼_7,𝑛+𝐼𝐼^7,25𝑛_7,𝑛+15=0.(76)

Proof. Using (5) in Lemma 9, we find that 𝑄=⎷𝑞𝜒𝑞⁵𝜒𝑞⁷𝜒(𝑞)𝜒𝑞³⁵,𝑅=⎷𝑞^3/2𝜒(𝑞)𝜒𝑞⁵𝜒𝑞⁷𝜒𝑞³⁵.(77) Setting 𝑞=𝑒^{√−𝜋𝑛/7} and using the definition of 𝐼_𝑘,𝑛 in (77), we get 𝑄=𝐼^7,25𝑛𝐼_7,𝑛^1/2,𝑅=𝐼_7,𝑛𝐼_7,25𝑛^1/2.(78) Employing (78) in (27), we complete the proof.

Corollary 30. One has (i)𝐼_7,5=ℎ−√−36+ℎ²(ii)𝐼_7,1/5√6,=ℎ+√−36+ℎ²(iii)𝐼_7,25√6,=𝑑−√−144+𝑑²(iv)𝐼_7,1/2512,=𝑑+√−144+𝑑²12,(79) where ℎ=5+(62−6√105)^1/3+(62+6√105)^1/3 and 𝑑=12+2^2/3(135−15√21)^1/3+2^2/3(135+15√21)^1/3.

Proof. Setting 𝑛=1/5 in Theorem 29 and simplifying using Theorem 16(ii), we obtain 𝐼^67,5+𝐼^−67,5−5𝐼^47,5+𝐼^−47,5+10𝐼^27,5+𝐼^−27,5−17=0.(80) Equivalently, 𝐻³−5𝐻²+7𝐻−7=0,(81) where 𝐻=𝐼^27,5+𝐼^−27,5.(82) Solving (81) and noting the fact in Remark 17, we obtain 𝐻=5+62−6√105^1/3+62+6√105^1/33(83) Combining (82) and (83) and noting that 𝐼_7,5<1, we deduce that 𝐼_7,5=ℎ−√−36+ℎ²√6,(84) where ℎ=5+62−6√105^1/3+62+6√105^1/3.(85) This completes the proof of (i).
Again setting 𝑛=1 and simplifying using Theorem 16(i), we arrive at 𝑈³−6𝑈²+7𝑈−3=0,(86) where 𝑈=𝐼_7,25+𝐼^−17,25.(87) Solving (86) and using Remark 17, we get 𝑈=6+159−√21/2^1/3+159+√21/2^1/33.(88) Combining (87) and (88) and noting that 𝐼_7,25<1, we obtain 𝐼_7,25=𝑑−√−144+𝑑²12,(89) where 𝑑=12+2^2/3135−15√21^1/3+2^2/3135+15√21^1/3.(90) This completes the proof of (ii). Now (ii) and (iv) easily follow from (i) and (ii), respectively, and Theorem 16(ii).

Theorem 31. One has 𝐼𝐼^13,𝑛_13,9𝑛²+𝐼^13,𝑛𝐼_13,9𝑛⁻²−3𝐼^13,𝑛−𝐼_13,𝑛𝐼_13,9𝑛+𝐼𝐼^13,𝑛_13,9𝑛⁻¹𝐼_13,9𝑛+𝐼_13,𝑛𝐼_13,9𝑛⁻¹+3=0.(91)

Proof. Proceeding as in the proof of Theorem 29, using (5) in Lemma 10, setting 𝑞∶=𝑒^{√−𝜋𝑛/13}, and using the definition of 𝐼_𝑘,𝑛, we arrive at 𝑄=𝐼^13,9𝑛𝐼_13,𝑛^1/2,𝑇=𝐼_13,𝑛𝐼_13,9𝑛^1/2.(92) Employing (92) in (28), we complete the proof.

Corollary 32. One has (i)𝐼_13,3=1+√13−−2+2√13(ii)𝐼_13,94,=2+√3−3+4√3(iii)𝐼_13,1/32,=1+√13+−2+2√13(iv)𝐼_13,1/94,=2+√3+3+4√32.(93)

Proof. Setting 𝑛=1/3 in Theorem 31 and simplifying using Theorem 16(ii), we obtain 𝑉²−3𝑉−1=0,(94) where 𝑉=𝐼^213,3+𝐼^−213,3.(95) Solving (94) and using Remark 17, we get 𝑉=3+√132.(96) Combining (95) and (96) and noting that 𝐼_13,3<1, we obtain 𝐼_13,3=1+√13−−2+2√134.(97) So we complete the proof of (i).
Again setting 𝑛=1 and using Theorem 16(i), we obtain 𝐽²−4𝐽+1=0,(98) where 𝐽=𝐼_13,9+𝐼^−113,9.(99) Solving (98) and using Remark 17, we get 𝐽=2+√3.(100) Combing (99) and (100) and noting that 𝐼_13,9<1, we deduce that 𝐼_13,9=2+√3−3+4√32.(101) So the proofs of (ii) is complete. Now the proof of (iii) and (iv) follow from (i) and (ii), respectively, and Theorem 16(ii).

Theorem 33. One has 𝐼𝐼^13,25𝑛_13,𝑛³𝐼𝐼^13,25𝑛_13,𝑛⁻³−5𝐼^13,25𝑛𝐼_13,𝑛+𝐼𝐼^13,25𝑛_13,𝑛⁻¹×𝐼+2_13,25𝑛−𝐼𝐼_13,𝑛+𝐼_13,25𝑛𝐼_13,𝑛⁻¹+2_13,25𝑛𝐼_13,𝑛²+𝐼_13,25𝑛𝐼_13,𝑛⁻²=0.(102)

Proof. Using (5) in Lemma 11, setting 𝑞∶=𝑒^{√−𝜋𝑛/13}, and using the definition of 𝐼_𝑘,𝑛, we arrive at 𝑄=𝐼^13,25𝑛𝐼_13,𝑛^1/2,𝑇=𝐼_13,𝑛𝐼_13,25𝑛^1/2.(103) Employing (103) in (29), we complete the proof.

Corollary 34. One has (i)𝐼_13,5=3+√65−58+6√65(ii)𝐼_13,252,=𝑐−√−36+𝑐²(iii)𝐼_13,1/56,=3+√65+58+6√65(iv)𝐼_13,1/252,=𝑐+√−36+𝑐²6,(104) where 𝑐=6+(1080−15√39)^1/3+(1080+15√39)^1/3.

Proof. Setting 𝑛=1/5 and simplifying using Theorem 16(ii), we arrive at 𝐿³−23𝐿−42=0,(105) where 𝐿=𝐼^213,5+𝐼^−213,5.(106) Solving (105) and using the fact in Remark 17, we obtain 𝐿=3+√652.(107) Employing (107) in (106), solving the resulting equation, and noting that 𝐼_13,5<1, we obtain 𝐼_13,5=3+√65−58+6√652.(108) This completes the proof of (i).
To prove (ii), setting 𝑛=1 and simplifying using Theorem 16(i), we arrive at 𝐴³−6𝐴²−23𝐴−18=0,(109) where 𝐴=𝐼_13,25+𝐼^−113,25.(110) Solving (109) and using the fact in Remark 17, we obtain 𝐴=6+1080−15√39^1/3+1080+15√39^1/33.(111) Employing (111) and (110), solving the resulting equation, and noting that 𝐼_13,25<1, we obtain 𝐼_13,25=𝑐−√−36+𝑐²6,(112) where 𝑐=6+(1080−15√39)^1/3+(1080+15√39)^1/3.
This completes the proof of (ii). Now the proofs of (iii) and (iv) follow from (i) and (ii), respectively, and Theorem 16(ii).

Theorem 35. One has 𝐼_7,𝑛𝐼_7,9𝑛+𝐼_7,𝑛𝐼_7,9𝑛⁻¹−𝐼𝐼^7,9𝑛_7,𝑛²+𝐼𝐼^7,9𝑛_7,𝑛⁻²+3=0.(113)

Proof. Using (5) in Lemma 12, setting 𝑞∶=𝑒^{√−𝜋𝑛/7}, and using the definition of 𝐼_𝑘,𝑛, we arrive at 𝑄=𝐼^7,9𝑛𝐼_7,𝑛^1/2,𝑅=𝐼_7,𝑛𝐼_7,9𝑛^1/2.(114) Employing (114) in (30), we complete the proof.

Corollary 36. One has (i)𝐼_7,3=5−√212^1/4(ii)𝐼_7,9,=1+√21−6+2√21(iii)𝐼_7,1/34,=5+√212^1/4(iv)𝐼_7,1/9,=1+√21+6+2√214.(115)

Proof. Setting 𝑛=1/3 and simplifying using Theorem 16(ii), we arrive at 𝐼^47,3+𝐼^−47,3−5=0.(116) Solving (116) and noting the fact in Remark 17, we obtain 𝐼_7,3=5−√212^1/4.(117) This completes the proof of (i).
To prove (ii), setting 𝑛=1 and simplifying using Theorem 16(i), we arrive at 𝐷²−𝐷−5=0,(118) where 𝐷=𝐼_7,9+𝐼^−17,9.(119) Solving (118) and using the fact in Remark 17, we obtain 𝐷=1+√212.(120) Employing (120) in (119), solving the resulting equation, and noting that 𝐼_7,9<1, we deduce that 𝐼_7,9=1+√21−6+2√214.(121) This completes the proof of (ii). Now the proofs of (iii) and (iv) follow from (i) and (ii), respectively, and Theorem 16(ii).

5. General Theorems and Explicit Evaluations of 𝐺_𝑛𝑘𝐺_𝑛/𝑘

In this section we evaluate some explicit values of the product 𝐺_𝑛𝑘𝐺_𝑛/𝑘 by establishing some general theorems and employing the values of 𝐼_𝑘,𝑛 obtained in Section 4. We recall from Theorem 22(ii) that 𝐺_1/𝑛=𝐺_𝑛 for ready references in this section.

Theorem 37. One has (i)𝐼_3,𝑛𝐼_3,4𝑛²+𝐼_3,𝑛𝐼_3,4𝑛⁻²+𝐺_𝑛/3𝐺_4𝑛/3𝐺_9𝑛/3𝐺_36𝑛/3⁻¹(−ii)2𝐼𝐺_{𝑛/363,𝑛}𝐺_4𝑛/3𝐺_9𝑛/3𝐺_36𝑛/3=0,+𝐼^−63,𝑛+2√2𝐺_𝑛/3𝐺_3𝑛⁻³−𝐺_𝑛/3𝐺_3𝑛³=0.(122)

Proof. To prove (i), using (5) in Lemma 13, setting 𝑞∶=𝑒^{√−𝜋𝑛/3}, and employing the definitions of 𝐼_𝑘,𝑛 and 𝐺_𝑛, we obtain 𝑅=⎷𝑞^1/4𝜒(𝑞)𝜒𝑞²𝜒𝑞³𝜒𝑞⁶=𝐼^3,𝑛𝑃=𝐺_𝑛/3𝐼_3,4𝑛^1/2,𝐺_4𝑛/3𝐺_9𝑛/3𝐺_36𝑛/3^−1/2.(123) Employing (123) in (31), we complete the proof. (ii) follows similarly from Lemma 5 and the definition of 𝐼_𝑘,𝑛 and 𝐺_𝑛 with 𝑞∶=𝑒^{√−𝜋𝑛/3}.

Corollary 38. One has (i)𝐺₆𝐺_3/2=1+√3(ii)𝐺₁₂2,𝐺_4/3=2^−13/6(iii)𝐺₃₉6+3√6+−2+3√6,𝐺_13/3=2^1/63+√13^1/3.(124)

Proof. Setting 𝑛=1/2 in Theorem 37(i) and simplifying using Theorem 16(ii) and the result 𝐺_1/𝑛=𝐺_𝑛, we obtain 2𝐺₆𝐺_3/2²−𝐺₆𝐺_3/2⁻²−2=0.(125) Solving (125) and noting that 𝐺₆𝐺_3/2>1, we complete the proof of (i).
To prove (ii), setting 𝑛=1 in Theorem 37(i); using Theorem 16(i), and noting that 𝐺_1/𝑛=𝐺_𝑛, we obtain 𝐼^23,4+𝐼^−23,4+𝐺²³𝐺_4/3𝐺₁₂⁻¹−2𝐺²³𝐺_4/3𝐺₁₂=0.(126) Employing (53) in (126), solving the resulting equation, and noting that 𝐺²³𝐺_4/3𝐺₁₂>1, we obtain 𝐺²³𝐺_4/3𝐺₁₂=6+3√6+−2+3√64.(127) Using the value 𝐺₃=2^1/12 from [5, p. 189] in (127), we complete the proof of (ii).
To prove (iii), setting 𝑛=13 in Theorem 37(ii), we obtain 𝐼^63,13+𝐼^−63,13+2√2𝐺_3/13𝐺₃₉⁻²−𝐺_3/13𝐺₃₉²=0.(128) Cubing (96) and then employing in (128) and solving the resulting equation, we complete the proof.

Theorem 39. One has (=i)4𝐼𝐺^5,𝑛𝐼_{5,4𝑛𝑛/5}³+𝐼_5,𝑛𝐼_5,4𝑛⁻³+5𝐼_5,𝑛𝐼_5,4𝑛+𝐼_5,𝑛𝐼_5,4𝑛⁻¹𝐺_4𝑛/5𝐺_25𝑛/5𝐺_100𝑛/5²(−ii)𝐺𝐼_{𝑛/535,𝑛}𝐺_4𝑛/5𝐺_25𝑛/5𝐺_100𝑛/5⁻²,+𝐼^−35,𝑛+2𝐺_𝑛/5𝐺_5𝑛⁻²−𝐺_𝑛/5𝐺_5𝑛²=0.(129)

Proof. Using (5) in Lemma 14, setting 𝑞∶=𝑒^{√−𝜋𝑛/5}, and employing the definitions of 𝐼_𝑘,𝑛 and 𝐺_𝑛, we obtain 𝑅=⎷𝑞^1/2𝜒(𝑞)𝜒𝑞²𝜒𝑞⁵𝜒𝑞¹⁰=𝐼^5,𝑛𝑃=𝐺_𝑛/5𝐼_5,4𝑛^1/2,𝐺_4𝑛/5𝐺_25𝑛/5𝐺_100𝑛/5^−1/2.(130) Employing (130) in (32), we complete the proof of (i). Similarly, (ii) follows from Lemma 6 and the definition of 𝐼_𝑘,𝑛 and 𝐺_𝑛 with 𝑞∶=𝑒^{√−𝜋𝑛/5}.

Corollary 40. One has (i)𝐺₁₀𝐺_5/2=3+√102^1/4(ii)𝐺₂₀,𝐺_5/4=12−311+5√5^1/4+11+5√5^3/4+𝔐,(131) where 𝔐 denotes 16+11+5√5(−3+11+5√5)².

Proof. Setting 𝑛=1/2 in Theorem 39(i) and simplifying using Theorem 16(ii) and the result 𝐺_1/𝑛=𝐺_𝑛, we obtain 4𝐺₁₀𝐺_5/2⁴−𝐺₁₀𝐺_5/2⁻⁴−12=0.(132) Solving (132) and noting that 𝐺₁₀𝐺_5/2>1, we complete the proof of (i).
For proof of (ii), setting 𝑛=1 in Theorem 39(ii) and simplifying using Theorem 18(i) and the result 𝐺_1/𝑛=𝐺_𝑛, we obtain 𝐼^35,4+𝐼^−35,4+2𝐺_4/5𝐺₂₀⁻²−𝐺_4/5𝐺₂₀²=0.(133) Employing the value of 𝐼_5,4+𝐼^−15,4 from (63) in (133), solving the resulting equation, and noting that 𝐺_4/5𝐺₂₀>1, we get 𝐺_4/5𝐺₂₀=12−311+5√5^1/4+11+5√5^3/4+𝔐.(134) So the proof is complete.

Theorem 41. One has 2√=24𝐺𝐼_2,𝑛𝑛/2𝐼_2,49𝑛^3/2+𝐼_2,𝑛𝐼_2,49𝑛^−3/2𝐺_4𝑛/2𝐺_47𝑛/2𝐺_196𝑛/2^1/2−2𝐺_𝑛/2𝐺_4𝑛/2𝐺_47𝑛/2𝐺_196𝑛/2^−1/2+4𝐺_𝑛/2𝐺_4𝑛/2𝐺_47𝑛/2𝐺_196𝑛/2^−3/2−𝐺_𝑛/2𝐺_4𝑛/2𝐺_47𝑛/2𝐺_196𝑛/2^−5/2.(135)

Proof. Using (5) in Lemma 15, setting 𝑞∶=𝑒^{√−𝜋𝑛/2}, and employing the definitions of 𝐼_𝑘,𝑛 and 𝐺_𝑛, we obtain, 𝑅=⎷𝑞^1/3𝜒(𝑞)𝜒𝑞⁷𝜒𝑞²𝜒𝑞¹⁴=𝐼^2,𝑛𝑃=𝐺_𝑛/2𝐼_2,49𝑛^1/2,𝐺_4𝑛/2𝐺_47𝑛/2𝐺_196𝑛/2^−1/2.(136) Employing (136) in (33), we complete the proof.

Corollary 42. One has 𝐺₁₄𝐺_7/2=2+4+8√24.(137)

Proof. Setting 𝑛=1/7 and simplifying using Theorem 16(ii) and the result 𝐺_1/𝑛=𝐺_𝑛, we get 4𝐺₁₄𝐺_7/2⁶−4√2𝐺₁₄𝐺_7/2⁵−2𝐺₁₄𝐺_7/2⁴+4𝐺₁₄𝐺_7/2²−1=0.(138) Solving (138) and noting that 𝐺₁₄𝐺_7/2>1, we complete the proof.

Theorem 43. One has 𝐼^47,𝑛+𝐼^−47,𝑛+7=2√2𝐺_7/𝑛𝐺_7𝑛⁻³+𝐺_7/𝑛𝐺_7𝑛³.(139)

Proof. Using (5) in Lemma 7, setting 𝑞∶=𝑒^{√−𝜋𝑛/7}, and employing definitions of 𝐼_𝑘,𝑛 and 𝐺_𝑛, we arrive at