A208547 Phi(k,m) with squarefree k values in sorted order for any integer m > 1.

1, 2, 6, 3, 10, 5, 14, 7, 15, 30, 22, 11, 21, 26, 42, 13, 34, 17, 38, 19, 33, 66, 46, 23, 35, 39, 78, 70, 58, 29, 62, 31, 51, 102, 57, 74, 114, 37, 55, 82, 110, 41, 86, 43, 69, 138, 94, 47, 65, 210, 130, 105, 106, 53, 87, 174, 118, 59, 77, 93, 122, 186, 154

Offset: 1

Author

Lei Zhou, Feb 28 2012

Keywords

• nonn
• easy

Comments

Phi(k,m) denotes cyclotomic polynomial numbers Cyclotomic(k, m).

When k = Product(p_i^j_i), i = 1, 2,..., and p_i are prime factors of k, then Phi(k, m) = Phi(Product(p_i), m^(Product(p_i^(j_i-1)))).

For this reason, number space of Phi(k, m) is still traversed with Phi(k, m) terms with only squarefree k values.

This sequence sorts the Phi(k, m) value along k-axis for all squarefree k values.

Examples

			For those squarefree numbers 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 squarefree numbers that make A000010(k) = 2
Phi(3,m) = 1 + m + m^2
Phi(6,m) = 1 - m + m^2
Obviously when integer m > 1, Phi(6,m) < Phi(3,m)
So a(3)=6, a(4)=3 (noting that Phi(6,m) > Phi(2,m) when m > 2, and Phi(6,2) = Phi(2,2))
For those squarefree numbers that make A000010(k) = 4
Phi(5,m) = 1 + m + m^2 + m^3 + m^4
Phi(10,m) = 1 - m + m^2 - m^3 + m^4
Obviously when integer m > 1, Phi(10,m) < Phi(5,m),
So a(5) = 10, and a(6) = 5 (noting Phi(10,m) - Phi(3,m) = m((m^2 + m + 2)(m - 2) + 2) >= 4 > 0 when m >= 2).
		

Crossrefs

Cf. A005117, A206225.

Programs

Mathematica

phiinv[n_, pl_] := Module[{i, p, e, pe, val}, If[pl == {}, Return[If[n == 1, {1}, {}]]]; val = {}; p = Last[pl]; For[e = 0; pe = 1, e == 0 || Mod[n, (p - 1) pe/p] == 0, e++; pe *= p, val = Join[val, pe*phiinv[If[e == 0, n, n*p/pe/(p - 1)], Drop[pl, -1]]]]; Sort[val]]; phiinv[n_] := phiinv[n, Select[1 + Divisors[n], PrimeQ]]; eb = 60; t =  Select[Range[Max[Table[phiinv[n], {n, 1, eb}]]], ((EulerPhi[#] <= eb) && SquareFreeQ[#]) &]; SortBy[t, Cyclotomic[#, 2]&]