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%I #31 Feb 19 2019 10:24:05
%S 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,
%T 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,
%U 8,0,0,48,0,24,40,16,0,24,8,0,0,0,4,48,8,12,24
%N Number of representations of integers by cyclotomic binary forms.
%C 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).
%C One can prove that a(m) is always a multiple of 4.
%H Michel Waldschmidt, <a href="/A299214/b299214.txt">Table of n, a(n) for n = 1..1000</a>
%H Étienne Fouvry, Claude Levesque, Michel Waldschmidt, <a href="https://arxiv.org/abs/1712.09019">Representation of integers by cyclotomic binary forms</a>, arXiv:1712.09019 [math.NT], 2017.
%p x := 'x'; y := 'y':
%p with(numtheory): for n from 3 to 1000 do
%p F[n] := expand(y^phi(n)*cyclotomic(n, x/y)) od:
%p g := 0:
%p for m from 1 to 1000 do
%p for n from 3 to 60 do # For the bounds see the reference.
%p for x from -60 to 60 do
%p for y from -60 to 60 do
%p if F[n] = m and max(abs(x), abs(y)) > 1
%p then g := g+1 fi:
%p od:
%p od:
%p od: a[m] := g: print(m, a[m]): g := 0
%p od:
%t 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];
%t Array[a, 100] (* _Jean-François Alcover_, Dec 01 2018, from Maple *)
%o (Julia)
%o using Nemo
%o function countA296095(n)
%o if n < 3 return 0 end
%o R, x = PolynomialRing(ZZ, "x")
%o K = Int(floor(5.383*log(n)^1.161)) # Bounds from
%o M = Int(floor(2*sqrt(n/3))) # Fouvry & Levesque & Waldschmidt
%o N = QQ(n); count = 0
%o for k in 3:K
%o e = Int(eulerphi(ZZ(k)))
%o c = cyclotomic(k, x)
%o for m in 1:M, j in 0:M if max(j, m) > 1
%o N == m^e*subst(c, QQ(j,m)) && (count += 1)
%o end end end
%o 4*count
%o end
%o A299214list(upto) = [countA296095(n) for n in 1:upto]
%o print(A299214list(76)) # _Peter Luschny_, Feb 25 2018
%Y The sequence of indices m with a(m) != 0 is A296095.
%Y The sequence of indices m with a(m) = 0 is A293654.
%K nonn
%O 1,3
%A _Michel Waldschmidt_, Feb 16 2018
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