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, 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

Author

Lei Zhou, Feb 13 2012

Keywords

• nonn
• tabf

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.

Examples

			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;
		

Crossrefs

Cf. A206292, A194712, A206225, A000010, A032447.

Programs

Mathematica

t = Select[Range[400], EulerPhi[#] <= 40 &]; SortBy[t, Cyclotomic[#, -2] &]