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 A320046 #17 May 12 2024 18:11:33 %S A320046 2,2,6,6,10,6,18,12,18,10,22,12,26,18,30,30,34,18,54,30,42,22,46,24, %T A320046 50,26,54,42,58,30,62,32,66,34,90,42,74,54,90,60,82,42,98,66,90,46,94, %U A320046 60,98,50,102,90,106,54,150,84,114,58,118,60,122,62,126,102,130,66,134,102,138,90 %N A320046 Largest k such that (Z/kZ)* is isomorphic to (Z/nZ)*. %C A320046 All terms are even because (Z/kZ)* is isomorphic to (Z/(2k)Z)* for odd k. Most terms are congruent to 2 modulo 4. Among the first 10000 terms there are 8980 ones congruent to 2 modulo 4. %H A320046 Jianing Song, <a href="/A320046/b320046.txt">Table of n, a(n) for n = 1..10000</a> %H A320046 Wikipedia, <a href="http://en.wikipedia.org/wiki/Multiplicative_group_of_integers_modulo_n">Multiplicative group of integers modulo n</a> %F A320046 n <= a(n) <= A028476(n). %e A320046 The solutions to (Z/kZ)* = C_6 are k = 7, 9, 14 and 18, so a(7) = a(9) = a(14) = a(18) = 18. %e A320046 The solutions to (Z/kZ)* = C_2 X C_20 are k = 55, 75, 100, 110 and 150, so a(55) = a(75) = a(100) = a(110) = a(150) = 150. %e A320046 The solutions to (Z/kZ)* = C_2 X C_12 are k = 35, 39, 45, 52, 70, 78 and 90, so a(35) = a(39) = a(45) = a(52) = a(70) = a(78) = a(90) = 90. %o A320046 (PARI) a(n) = if(abs(n)==1||abs(n)==2, 2, my(i=0, search_max = A057635(eulerphi(n))); for(j=eulerphi(n)+1, search_max, if(znstar(j)[2]==znstar(n)[2], i=j)); i) \\ search_max is the largest k such that phi(k) = phi(n). See A057635 for its program %Y A320046 Cf. A028476, A057635, A320045. %K A320046 nonn %O A320046 1,1 %A A320046 _Jianing Song_, Oct 04 2018