A299214 Number of representations of integers by cyclotomic binary forms.

0, 0, 8, 16, 8, 0, 24, 4, 16, 8, 8, 12, 40, 0, 0, 40, 16, 4, 24, 8, 24, 0, 0, 0, 24, 8, 12, 24, 8, 0, 32, 8, 0, 8, 0, 16, 32, 0, 24, 8, 8, 0, 32, 0, 8, 0, 0, 12, 40, 12, 0, 32, 8, 0, 8, 0, 32, 8, 0, 0, 48, 0, 24, 40, 16, 0, 24, 8, 0, 0, 0, 4, 48, 8, 12, 24

Offset

1,3

Comments

a(m) is the number of solutions of the equation Phi_n(x,y) = m with n >= 3 and max{|x|,|y|} >= 2. Here the binary form Phi_n(x,y) is the homogeneous version of the cyclotomic polynomial phi_n(t).

One can prove that a(m) is always a multiple of 4.

Links

Michel Waldschmidt, Table of n, a(n) for n = 1..1000

Étienne Fouvry, Claude Levesque, Michel Waldschmidt, Representation of integers by cyclotomic binary forms, arXiv:1712.09019 [math.NT], 2017.

Maple

x := 'x'; y := 'y':
with(numtheory): for n from 3 to 1000 do
F[n] := expand(y^phi(n)*cyclotomic(n, x/y))  od:
g := 0:
for m from 1 to 1000 do
   for n from 3 to 60 do  # For the bounds see the reference.
      for x from -60 to 60 do
         for y from -60 to 60 do
            if F[n] = m and  max(abs(x), abs(y)) > 1
                then g := g+1 fi:
         od:
      od:
   od: a[m] := g: print(m, a[m]): g := 0
od:

Mathematica

For[n = 3, n <= 100, n++, F[n] = Expand[y^EulerPhi[n] Cyclotomic[n, x/y]]]; g = 0; For[m = 1, m <= 100, m++, For[n = 3, n <= 60, n++, For[x = -60, x <= 60, x++, For[y = -60, y <= 60, y++, If[F[n] == m && Max[Abs[x], Abs[y] ] > 1, g = g+1]]]]; a[m] = g; Print[m, " ", a[m]]; g = 0];
Array[a, 100] (* Jean-François Alcover, Dec 01 2018, from Maple *)

Prog

(Julia)
using Nemo
function countA296095(n)
    if n < 3 return 0 end
    R, x = PolynomialRing(ZZ, "x")
    K = Int(floor(5.383*log(n)^1.161)) # Bounds from
    M = Int(floor(2*sqrt(n/3)))        # Fouvry & Levesque & Waldschmidt
    N = QQ(n); count = 0
    for k in 3:K
        e = Int(eulerphi(ZZ(k)))
        c = cyclotomic(k, x)
        for m in 1:M, j in 0:M if max(j, m) > 1
            N == m^e*subst(c, QQ(j, m)) && (count += 1)
    end end end
    4*count
end
A299214list(upto) = [countA296095(n) for n in 1:upto]
print(A299214list(76)) # Peter Luschny, Feb 25 2018

Crossrefs

The sequence of indices m with a(m) != 0 is A296095.

The sequence of indices m with a(m) = 0 is A293654.

Sequence in context: A073925 A053321 A335772 * A174256 A037239 A205869

Adjacent sequences: A299211 A299212 A299213 * A299215 A299216 A299217

Keyword

nonn

Author

Michel Waldschmidt, Feb 16 2018

Status

approved