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 2020-10-11, 18:11 #9 Till     "Tilman Neumann" Jan 2016 Germany 1B216 Posts For completeness I'ld like to add that it's more than counting. For any n>2, we can obtain the values of A004215 (natural numbers representable by 4 squares but no less) < 2^n from the set of quadratic residues mod 2^n as follows (pseudocode): Code: computeA004215Mod2PowN(int n) { Set input = {quadratic residues modulo 2^n}; // the set of quadratic residues modulo 2^n Set output = new Set(); for (qr in input) { output.add(2^n - qr); } output.remove(2^n); if (n is odd) { output.remove(2^(n-1)); } else { output.remove(2^(n-1) + 2^(n-2)); } return output; }