A206864 Prime numbers of the form Phi_k(m), where k > 2, |m| > 1, and Phi_k(m) is the k-th cyclotomic polynomial evaluated at m.

3, 5, 7, 11, 13, 17, 31, 37, 43, 61, 73, 101, 127, 151, 157, 197, 211, 241, 257, 307, 331, 401, 421, 463, 521, 547, 577, 601, 677, 683, 757, 1093, 1123, 1297, 1483, 1601, 1723, 2551, 2731, 2801, 2917, 2971, 3137, 3307, 3541, 3907, 4357, 4423, 4561, 4831, 5113

Offset: 1

Author

Lei Zhou, Feb 13 2012

Keywords

• nonn

Comments

These are the prime numbers picked from sequence A206942.

Choosing negative m does not generate more primes, so it does not need negative m part in the Mathematica program.

The provided mathematica program generate this sequence in six steps:

Step 1: Find the minimum m such that Phi(6, m) is greater than the search boundary maxdata, and adjust the search boundary to the next: ( maxdata = 5200; max = Ceiling[(1 + Sqrt[1 + 4*(maxdata - 1)])/2]; )

Step 2: Find the even number eulerbound such that 2^(eulerbound+1)-1 > maxdata: ( eulerbound = 2*Floor[(Log[2, maxdata])/2 + 0.5]; )

This is the maximum possible value of Phi(k, 2) when Phi(k, m) has a totient function value of eulerbound;

Step 3: Adjust (up) the eulerbound such that it is an element of A002202 and find the group of ks such that Phi(k, m) has the same totient function value eulerbound: ( 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]]; While[eulergroup = phiinv[eulerbound]; lu = Length[eulergroup]; lu == 0, eulerbound = eulerbound + 2]; )

Step 4: Make list of k values such that the totient function of Phi(k, m) smaller or equal to the chosen Euler boundary eulerbound, and sort it in the order of the Phi(k, 2): ( Select[Range[eulergroup[[Length[eulergroup]]]], EulerPhi[#] <= eulerbound &]; ap = SortBy[t, Cyclotomic[#, 2] &])

Step 5: Scan k in the set of ap, 1 < m <= max, for all appeared primes that are smaller than maxdata: (a = {}; Do[i = 2; While[i++; cc = Cyclotomic[ap[[i]], m]; cc <= maxdata, If[PrimeQ[cc], a = Append[a, cc]]], {m, 2, max}];)

Step 6: Remove duplicate and sort the set generated in the above step: ( Sort[DeleteDuplicates[a]] )

Through these steps, a mathematically abundant algorithm is presented to find all the terms up to an arbitrary bound, without requiring the user to determine any other search parameters.

Examples

			Prime 3 = Phi_6(2); so a(1) = 3;
Prime 5 = Phi_4(2), so a(2) = 5;
...
Prime 17 = Phi_8(2), so a(6)=17;
Primes 19 and 23 are not in A206942;
Prime 31 = Phi_5(2), so a(7)=31.
		

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
• Étienne Fouvry, Claude Levesque, Michel Waldschmidt, Representation of integers by cyclotomic binary forms, arXiv:1712.09019 [math.NT], 2017.

Crossrefs

Cf. A206942, A006511.

Programs

Julia

# Function isA206942 is defined in A206942.
L = [n for n in 1:5113 if isprime(ZZ(n)) && isA206942(n)]
println(L) # Peter Luschny, Feb 21 2018

Mathematica

maxdata = 5200; max = Ceiling[(1 + Sqrt[1 + 4*(maxdata - 1)])/2]; eulerbound = 2*Floor[(Log[2, maxdata])/2 + 0.5]; 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]]; While[eulergroup = phiinv[eulerbound]; lu = Length[eulergroup]; lu == 0, eulerbound = eulerbound + 2]; t = Select[Range[eulergroup[[Length[eulergroup]]]], EulerPhi[#] <= eulerbound &]; ap = SortBy[t, Cyclotomic[#, 2] &]; a = {}; Do[i = 2; While[i++; cc = Cyclotomic[ap[[i]], m]; cc <= maxdata, If[PrimeQ[cc], a = Append[a, cc]]], {m, 2, max}]; Sort[DeleteDuplicates[a]]
(* Alternatively: *)
isA206864[n_] := If[! PrimeQ[n], Return[False],
    K = Floor[5.383 Log[n]^1.161]; M = Floor[2 Sqrt[n/3]];
    For[k = 3, k <= K, k++, For[x = 2, x <= M, x++,
        If[n == Cyclotomic[k, x], Return[True]]]];
    Return[False]
]; Select[Range[1000], isA206864] (* Peter Luschny, Feb 21 2018 *)