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%I #2 Mar 30 2012 17:35:24
%S 1,2,2,3,3,3,6,6,2,2,5,5,5,5,5,6,6,6,6,6,6,7,11,8,12,2,6,3,12,20,4,4,
%T 12,4,4,4,15,15,6,6,6,6,6,6,15,10,10,10,10,10,10,10,10,10,10,11,11,11,
%U 11,11,11,11,11,11,11,11,18,18,6,6,18,18,6,6,18,18,6,6,13,11,18,8,20,15,6
%N Table with T(n,k) = the number of ways to represent k as the sum of a square and a cube modulo n.
%C The top left corner is T(1,0).
%C It appears that this table does not contain any 0's.
%C It appears that row n is constant iff n is squarefree, and no prime divisor of n is == 1 (mod 6). It is not hard to show that such rows are constant, since the cubes are equi-distributed in such moduli.
%e The 6 ways to represent 0 (mod 4) are 0^2+0^3, 0^2+2^3, 1^2+3^3, 2^2+0^3, 2^2+2^3, and 3^2+3^3.
%o (PARI) al(n)=local(v);v=vector(n);for(i=0,n-1,for(j=0,n-1,v[(i^2+j^3)%n+1]++));v
%Y Cf. A022549, A002476, A045309.
%K nonn,tabl
%O 1,2
%A _Franklin T. Adams-Watters_, Dec 03 2009
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