%I #16 Mar 09 2018 04:44:13
%S 5,8,9,10,11,16,17,18,20,25,26,29,32,34,36,40,41,45,50,53,55,58,64,65,
%T 68,72,74,81,82,85,89,90,98,100,101,104,106,113,116,122,125,128,130,
%U 136,137,144,145,146,149,153,160,162,164,170,173,176,178,180,185
%N Integers represented by a cyclotomic binary form Phi{k}(x,y) with positive integers x and y where max(x, y) >= 2 and the index k is not prime.
%C A cyclotomic binary form is a homogeneous polynomial in two variables of the form p(x, y) = y^phi(k)*Phi(k, x/y) where Phi(k, z) is a cyclotomic polynomial of index k and phi is Euler's totient function. An integer m is represented by p if p(x,y) = m has an integer solution.
%C m is in this sequence if and only if m is in A296095 but not in A300332. This means m can be represented by a cyclotomic binary form but not as m = Sum_{j in 0:p-1} x^j*y^(p-j-1) with p prime.
%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.
%e 1037 is in this sequence because 1037 = f(26,19) = f(29,14) with f(x,y) = y^2 + x^2 are the only representations of 1037 by a cyclotomic binary form (which has index 4).
%e 1031 is not in this sequence because 1031 = f(5,2) where f(x,y) = x^4 + y*x^3 + y^2*x^2 + y^3*x + y^4 (which has index 5).
%o (Julia)
%o using Nemo
%o function isA300331(n)
%o R, z = PolynomialRing(ZZ, "z")
%o N = QQ(n)
%o # Bounds from Fouvry & Levesque & Waldschmidt
%o logn = log(n)^1.161
%o K = Int(floor(5.383*logn))
%o M = Int(floor(2*(n/3)^(1/2)))
%o r = false
%o k = 2
%o while k <= K
%o if k == 7
%o K = Int(ceil(4.864*logn))
%o M = Int(ceil(2*(n/11)^(1/4)))
%o end
%o e = Int(eulerphi(ZZ(k)))
%o c = cyclotomic(k, z)
%o for y in 2:M, x in 1:y
%o if N == y^e*subst(c, QQ(x, y))
%o isprime(ZZ(k)) && return false
%o r = true
%o end
%o end
%o k += 1
%o end
%o return r
%o end
%o A300331list(upto) = [n for n in 1:upto if isA300331(n)]
%o println(A300331list(185))
%Y Cf. A296095, A293654, A299930, A300332.
%K nonn
%O 1,1
%A _Peter Luschny_, Mar 06 2018