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%I #4 Apr 14 2015 16:44:25
%S 1,1,1,1,2,2,1,2,2,2,1,2,2,2,3,1,2,2,2,3,3,1,2,2,2,3,3,2,1,2,2,2,3,3,
%T 2,3,1,2,2,2,3,3,2,3,3,1,2,2,2,3,3,2,3,3,3,1,2,2,2,3,3,2,3,3,3,2,1,2,
%U 2,2,3,3,2,3,3,3,2,3,1,2,2,2,3,3,2,3
%N Number of terms in the enhanced triangular-number representation of n.
%C See A256909 for definitions. The least n for which R(n) has 5 terms is given by R(7259) = 7140 + 105 + 10 + 3 + 1.
%H Clark Kimberling, <a href="/A256911/b256911.txt">Table of n, a(n) for n = 0..1000</a>
%e R(4) = 3 + 1, so a(4) = 2.
%e R(119) = 105 + 10 + 3 + 1, so a(119) = 4.
%t b[n_] := n (n + 1)/2; bb = Insert[Table[b[n], {n, 0, 200}], 2, 3]
%t s[n_] := Table[b[n], {k, 1, n + 1}];
%t h[1] = {0, 1, 2}; h[n_] := Join[h[n - 1], s[n]];
%t g = h[200]; r[0] = {0};
%t r[n_] := If[MemberQ[bb, n], {n}, Join[{g[[n]]}, r[n - g[[n]]]]];
%t t = Table[r[n], {n, 0, 120}] (*A256909 before concatenation*)
%t Flatten[t] (*A256909*)
%t Table[Last[r[n]], {n, 0, 120}] (*A256910*)
%t Table[Length[r[n]], {n, 0, 120}] (*A256911*)
%Y Cf. A000217, A256909 (representations) A256910 (trace).
%K nonn,easy
%O 0,5
%A _Clark Kimberling_, Apr 14 2015
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