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A256544 Number of ways to write n as the sum of three unordered elements of the set {floor(T(x)/3): x = 1,2,3,...}, where T(x) denotes the triangular number x*(x+1)/2.

%I A256544 #7 Apr 01 2015 15:49:12
%S A256544 1,1,2,3,3,4,4,5,4,6,5,6,6,6,6,8,6,8,7,9,7,9,8,8,9,9,9,10,9,9,11,9,12,
%T A256544 10,10,9,14,10,11,11,13,9,14,10,12,15,11,13,12,14,12,12,13,15,14,14,
%U A256544 11,16,11,17,14,14,14,16,13,16,15,17,12,15,17,15,17,15,14,20,13,15,19,14,18,16,21,12,19,15,16,22,18,15,18,14,21,19,18,18,17,19,18,17,18
%N A256544 Number of ways to write n as the sum of three unordered elements of the set {floor(T(x)/3): x = 1,2,3,...}, where T(x) denotes the triangular number x*(x+1)/2.
%C A256544 Conjecture: For any positive integer m, every nonnegative integer n can be written as floor(T(x)/m) + floor(T(y)/m) + floor(T(z)/m) with x,y,z nonnegative integers.
%C A256544 In the case m = 1, this is a well-known result in number theory.
%H A256544 Zhi-Wei Sun, <a href="/A256544/b256544.txt">Table of n, a(n) for n = 0..10000</a>
%e A256544 a(4) = 3 since 4 = floor(T(1)/3) + floor(T(2)/3) + floor(T(4)/3) = floor(T(1)/3) + floor(T(3)/3) + floor(T(3)/3) = floor(T(2)/3) + floor(T(2)/3) + floor(T(3)/3).
%t A256544 S[n_]:=Union[Table[Floor[x*(x+1)/6], {x, 0, (Sqrt[24n+21]-1)/2}]]
%t A256544 L[n_]:=Length[S[n]]
%t A256544 Do[r=0;Do[If[Part[S[n],x]>n/3,Goto[cc]];Do[If[Part[S[n],x]+2*Part[S[n],y]>n,Goto[bb]];
%t A256544 If[MemberQ[S[n], n-Part[S[n],x]-Part[S[n],y]]==True,r=r+1];
%t A256544 Continue,{y,x,L[n]}];Label[bb];Continue,{x,1,L[n]}];Label[cc];Print[n," ",r];Continue, {n,0,100}]
%Y A256544 Cf. A000217.
%K A256544 nonn
%O A256544 0,3
%A A256544 _Zhi-Wei Sun_, Apr 01 2015