This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A380103 #8 Jan 25 2025 23:02:05 %S A380103 657,2439,7657,41839,231469 %N A380103 Minimal conductors c of cyclic cubic number fields K with elementary bicyclic 3-class group Cl_3(K)=(3,3) and second 3-class group M=Gal(F_3^2(K)/K) of assigned coclass cc(M)=0,1,2,3,... %C A380103 The coclass cc(M) for one of the fields K with conductor c = a(n) is n-1, and for each field K with conductor c < a(n), the coclass cc(M) is less than n-1. Among the 3-groups M of coclass cc(M)=1, we distinguish the abelian 3-group A=(3,3) by formally putting cc(A)=0, in accordance with the FORMULA. This is a significant difference to quadratic fields, which are firstly uniquely determined by their discriminant, and secondly cannot have an abelian second 3-class group. %H A380103 D. C. Mayer, <a href="http://arxiv.org/abs/1403.3833">The distribution of second p-class groups on coclass graphs</a>, arXiv:1403.3833 [math.NT], 2014; J. Théor. Nombres Bordeaux 25 (2013), 401-456. %H A380103 Daniel Constantin Mayer, <a href="/A380103/a380103.m.txt">Magma program "CyclCoClass.m" with endless loop</a> %F A380103 According to Theorem 3.12 on page 435 of "The distribution of second p-class groups on coclass graphs", the coclass of the non-abelian 3-group M is given by cc(M)+1=log_3(h_3(E_2)), where h_3(E_2) is the second largest 3-class number among the four unramified cyclic cubic extensions E_1,..,E_4 of the cyclic cubic field K, and log_3 denotes the logarithm with respect to the basis 3. An exception is the abelian 3-group A=(3,3) with correct cc(A)=1, where the FORMULA yields cc(A)=0. %e A380103 We have M abelian for c=657=9*73 (two fields in a doublet), cc(M)=1 for c=2439=9*271 (two fields in a doublet), cc(M)=2 for c=7657=13*19*31 (three fields in a quartet), cc(M)=3 for c=41839=7*43*139 (two fields in a quartet), cc(M)=4 for c=231469=7*43*769 (four fields in a quartet). If the conductor c has two prime divisors, then cc(M)=1. For cc(M) > 1, exactly three prime divisors of the conductor c are required. %o A380103 (Magma) // See Links section. %Y A380103 Analog of A379524 for real quadratic fields. %K A380103 nonn,hard,more %O A380103 1,1 %A A380103 _Daniel Constantin Mayer_, Jan 15 2025