%I #26 Feb 15 2021 03:44:15
%S 9,18,23,37,51
%N Numbers k such that S(24*(3*k+1)) !== 8*(3*k+1) (mod 24*(3*k+1)) where S(j) := Sum_{a=0..j-1, b=0..j-1} (a+b*i)^j and i is the imaginary unit; i.e., A230309(3*k+1) != 8*(3*k+1).
%C In most cases S(24*(3*k+1)) == 8*(3*k+1) (mod 24*(3*k+1)).
%t fu[n_] := fu[n] = Mod[Sum[PowerMod[i + j I, n, n], {i, 0, n - 1}, {j, 0, n - 1}], n]; Select[Range[50], ! fu[24*(3 # +1)] == 8*(3 # +1) &]
%Y Cf. A230308, A230309, A230310, A232057.
%K nonn,more,hard
%O 1,1
%A _José María Grau Ribas_, Nov 17 2013