%I
%S 1,2,3,4,6,5,12,8,10,7,9,18,14,30,20,24,16,15,11,22,42,13,28,36,21,26,
%T 17,40,48,32,60,34,19,27,54,38,66,44,25,50,33,23,46,70,78,52,90,56,72,
%U 45,84,39,35,29,58,31,62,102,68,80,96,64,120
%N This irregular table contains indices j, k, l,... in each row such that the values Phi(j,m) < Phi(k,m)< Phi(l,m)< ... of cyclotomic polynomials Phi(.,.) are sorted given any constant integer argument m >= 2.
%C Based on A002202 "Values taken by totient function phi(m)", A000010 can only take certain even numbers. So for the worst case, the largest Phi(k,m) with degree d (even positive integer) will be (1k^(d+1))/(1k) (or smaller)and the smallest Phi(k,m) with degree d+2 will be (1+k^(d+3))/(1+k) (or larger).
%C (1+k^(d+3))/(1+k)(1k^(d+1))/(1k)=(k/(k^21))*(2+k^d*(k^3(k^2+k+1)))
%C k^3>k^2+k+1 when k>=2.
%C This means that this sequence can be segmented to sets in which Cyclotomic(k,m) shares the same degree of Polynomial and it can be generated in this way.
%e For those k's that make A000010(k) = 1
%e Phi(1,m) = 1m
%e Phi(2,m) = 1m
%e Phi(1,m) < Phi(2,m)
%e So, a(1) = 1, a(2) = 2;
%e For those k's (k > 2) that make A000010(k) = 2
%e Phi(3,m) = 1  m + m^2
%e Phi(4,m) = 1 + m^2
%e Phi(6,m) = 1 + m + m^2
%e Obviously when integer m > 1, Phi(3,m) < Phi(4,m) < Phi(6,m)
%e So a(3)=3, a(4)=4, and a(5)=6
%e For those k's that make A000010(k) = 4
%e Phi(5,m) = 1  m + m^2  m^3 + m^4
%e Phi(8,m) = 1 + m^4
%e Phi(10,m) = 1 + m + m^2 + m^3 + m^4
%e Phi(12,m) = 1  m^2 + m^4
%e Obviously when integer m > 1, Phi(5,m) < Phi(12,m) < Phi(8,m) < Phi(10,m),
%e So a(6) = 5, a(7) = 12, a(8) = 8, and a(9) = 10.
%e The table starts
%e 1,2;
%e 3,4,6;
%e 5,12,8,10;
%t t = Select[Range[400], EulerPhi[#] <= 40 &]; SortBy[t, Cyclotomic[#, 2] &]
%Y Cf. A206292, A194712, A206225, A000010, A032447.
%K nonn,tabf
%O 1,2
%A _Lei Zhou_, Feb 13 2012
