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A232056
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).
1
9, 18, 23, 37, 51
OFFSET
1,1
COMMENTS
In most cases S(24*(3*k+1)) == 8*(3*k+1) (mod 24*(3*k+1)).
MATHEMATICA
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) &]
CROSSREFS
KEYWORD
nonn,more,hard
AUTHOR
STATUS
approved