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%I #14 Oct 16 2019 08:50:05
%S 2,3,4,7,8,11,12,17,30,39,52,59
%N Records in A317993.
%C Companion sequence of A328420.
%C It seems that this sequence is infinite (i.e., A317993 is unbounded), but for each n, to really construct a number k such that A317993(k) > a(n) seems impossible.
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Multiplicative_group_of_integers_modulo_n ">Multiplicative group of integers modulo n</a>
%e Let (Z/mZ)* be the multiplicative group of integers modulo m. We have (Z/mZ)* = (Z/104Z)* has 8 solutions, namely m = 104, 105, 112, 140, 144, 156, 180, 210; for all k' < 104, (Z/mZ)* = (Z/k'Z)* has fewer than 8 solutions. So A317993(104) = 8 is a term.
%o (PARI) b(n) = if(abs(n)==1||abs(n)==2, 2, my(i=0, k=eulerphi(n), N=floor(exp(Euler)*k*log(log(k^2))+2.5*k/log(log(k^2)))); for(j=k+1, N, if(znstar(j)[2]==znstar(n)[2], i++)); i)
%o my(t=0); for(k=1, 20000, if(b(k)>t, print1(b(k), ", "); t=b(k))) \\ Warning: program runs for about 30 min
%Y Cf. A317993, A328420.
%K nonn,hard,more
%O 1,1
%A _Jianing Song_, Oct 14 2019
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