Entry 160 dealt with discriminants d with class number 1. Going higher, there are only three fundamental d=4p with class number 2, namely p=5,13,17. Given the Kronecker symbol (dm), we propose
K(k5)=√2π2√20(1+√52)3/4(20∏m=1[Γ(m20)](−20m))1/4K(k13)=√2π2√52(3+√132)3/4(52∏m=1[Γ(m52)](−52m))1/4K(k37)=√2π2√148(6+√37)3/4(148∏m=1[Γ(m148)](−148m))1/4
Going to class number 4, the red exponent will now be 1/8.
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