The case p=7 is the next case to consider since the next odd, negative discriminant with class number 1 is -7. The corresponding quadratic form is . See  for a treatment of the classical theory of binary quadratic forms. We havewhere Also, Equation (7.1) with (7.2), and (7.4) are contained in Entry 5(i) of Chapter 21 of Ramanujan's second notebook , [1, p. 467,]. Equation (7.3) is analogous to (6.8) but does not appear in Ramanujan's notebooks. We expect a result like (7.3) to hold since the class number . A result of Legendre's [11, Equation (1), p. 97,] gives an equivalent formulation in terms of L-series: We define the following functions (analogous to the cubic case): and for j=1,2,3 In a later version of this paper we will examine these functions under the action of the congruence subgroup, .
The proof of the following is analogous to the cubic case [7, Lemma 2.1, p. 36,]:From (7.9)--(7.12) we easily find that
For j=1,2,3 we defineand
From (3.14), (7.1) we haveThroughout this section q and r are related by If we write the as functions of r then by using the multidimensional analogue [##
of (3.3) it can be shown thatfor j=1, 2, 3.
By (7.14), (7.19) we have
As noted in Section 4.3 the main problem of constructing a 7-th order iteration is to find a relationship between and . In view of (7.22) and (7.23), we would like to find a relationship between and . Instead we get each of the in terms of , , . This is achieved by solving a certain cubic and taking 7-th roots. We now give some details of this construction. LetNow we define the following functions: Then by using the theory of modular functions it can be shown that
Also, Our functions and are related. We find that Hence are the roots of the following cubic equation Now observe that Similarly we find that Thus after solving a certain cubic, and taking 7-th roots we may obtain each of the in terms of , , and . From (3.18) we have
Finally, to construct an septic iteration, we need some initial values. Two obvious candidates are , . From (3.10) we haveWe need to find the values of the for r=1. Fortunately, by (7.21), these coincide with the values of the . So when r=1, and . Thus we can in this case solve equations (7.27) and (7.33) to obtain Thus via (7.34) we find that , , are roots of the cubic Then after solving this cubic we can find the initial values using (7.36). To obtain the septic iteration we need only consider in our identities and equations to obtain the following theorem.
Define sequences , , , and by where and are the roots of the cubic equation where are the roots of the cubic where and where where and Then converges septically to .