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A255974 R(k), the minimal alternating triangular-number representation of k, concatenated for k = 0, 1, 2,....

%I A255974 #4 Apr 13 2015 09:38:49
%S A255974 0,1,3,-1,3,6,-3,1,6,-1,6,10,-3,10,-3,1,10,-1,10,15,-6,3,-1,15,-3,15,
%T A255974 -3,1,15,-1,15,21,-6,1,21,-6,3,-1,21,-3,21,-3,1,21,-1,21,28,-6,28,-6,
%U A255974 1,28,-6,3,-1,28,-3,28,-3,1,28,-1,28,36,-10,3,36,-6,36
%N A255974 R(k), the minimal alternating triangular-number representation of k, concatenated for k = 0, 1, 2,....
%C A255974 Suppose that b = (b(0), b(1), ... ) is an increasing sequence of positive integers satisfying b(0) = 1 and b(n+1) <= 2*b(n) for n >= 0.  Let B(n) be the least b(m) >= n.  Let R(0) = 1, and for n > 0, let R(n) = B(n) - R(B(n) - n).  The resulting sum of the form R(n) = B(n) - B(m(1)) + B(m(2)) - ... + ((-1)^k)*B(k) is the minimal alternating b-representation of n.  The sum B(n) + B(m(2)) + ... is the positive part of R(n), and the sum B(m(1)) + B(m(3)) + ... , the nonpositive part of R(n).  The number ((-1)^k)*B(k) is the trace of n.  If b(n) = n(n+1)/2, the n-th triangular number, then the sum R(n) is the minimal alternating triangular-number representation of n.
%H A255974 Clark Kimberling, <a href="/A255974/b255974.txt">Table of n, a(n) for n = 0..1000</a>
%e A255974 R(0) = 0
%e A255974 R(1) = 1
%e A255974 R(2) = 3 - 1
%e A255974 R(3) = 3
%e A255974 R(4) = 6 - 3 + 1
%e A255974 R(5) = 6 - 1
%e A255974 R(8) = 10 - 3 + 1
%e A255974 R(11) = 15 - 6 + 3 - 1
%t A255974 b[n_] := n (n + 1)/2; bb = Table[b[n], {n, 0, 1000}];
%t A255974 s[n_] := Table[b[n], {k, 1, n}];
%t A255974 h[1] = {1}; h[n_] := Join[h[n - 1], s[n]];
%t A255974 g = h[100]; r[0] = {0};
%t A255974 r[n_] := If[MemberQ[bb, n], {n}, Join[{g[[n]]}, -r[g[[n]] - n]]]
%t A255974 t = Table[r[n], {n, 0, 120}]  (* A255974 actual representations *)
%t A255974 Flatten[t]  (* A255974 sequence *)
%Y A255974 Cf. A000217, A256655 (Fibonacci based), A256696 (binary), A256789 (squares).
%K A255974 easy,sign
%O A255974 0,3
%A A255974 _Clark Kimberling_, Apr 11 2015