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A206710
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.
1
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, 17, 40, 48, 32, 60, 34, 19, 27, 54, 38, 66, 44, 25, 50, 33, 23, 46, 70, 78, 52, 90, 56, 72, 45, 84, 39, 35, 29, 58, 31, 62, 102, 68, 80, 96, 64, 120
OFFSET
1,2
COMMENTS
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 (1-k^(d+1))/(1-k) (or smaller)and the smallest Phi(k,m) with degree d+2 will be (1+k^(d+3))/(1+k) (or larger).
(1+k^(d+3))/(1+k)-(1-k^(d+1))/(1-k)=(k/(k^2-1))*(2+k^d*(k^3-(k^2+k+1)))
k^3>k^2+k+1 when k>=2.
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.
EXAMPLE
For those k's that make A000010(k) = 1
Phi(1,-m) = -1-m
Phi(2,-m) = 1-m
Phi(1,-m) < Phi(2,-m)
So, a(1) = 1, a(2) = 2;
For those k's (k > 2) that make A000010(k) = 2
Phi(3,-m) = 1 - m + m^2
Phi(4,-m) = 1 + m^2
Phi(6,-m) = 1 + m + m^2
Obviously when integer m > 1, Phi(3,m) < Phi(4,m) < Phi(6,m)
So a(3)=3, a(4)=4, and a(5)=6
For those k's that make A000010(k) = 4
Phi(5,-m) = 1 - m + m^2 - m^3 + m^4
Phi(8,-m) = 1 + m^4
Phi(10,-m) = 1 + m + m^2 + m^3 + m^4
Phi(12,-m) = 1 - m^2 + m^4
Obviously when integer m > 1, Phi(5,m) < Phi(12,m) < Phi(8,m) < Phi(10,m),
So a(6) = 5, a(7) = 12, a(8) = 8, and a(9) = 10.
The table starts
1,2;
3,4,6;
5,12,8,10;
MATHEMATICA
t = Select[Range[400], EulerPhi[#] <= 40 &]; SortBy[t, Cyclotomic[#, -2] &]
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Lei Zhou, Feb 13 2012
STATUS
approved