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

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, -3, 1, 15, -1, 15, 21, -6, 1, 21, -6, 3, -1, 21, -3, 21, -3, 1, 21, -1, 21, 28, -6, 28, -6, 1, 28, -6, 3, -1, 28, -3, 28, -3, 1, 28, -1, 28, 36, -10, 3, 36, -6, 36

Offset: 0

Author

Clark Kimberling, Apr 11 2015

Keywords

• easy
• sign

Comments

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.

Examples

			R(0) = 0
R(1) = 1
R(2) = 3 - 1
R(3) = 3
R(4) = 6 - 3 + 1
R(5) = 6 - 1
R(8) = 10 - 3 + 1
R(11) = 15 - 6 + 3 - 1
		

Links

• Clark Kimberling, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A000217, A256655 (Fibonacci based), A256696 (binary), A256789 (squares).

Programs

Mathematica

b[n_] := n (n + 1)/2; bb = Table[b[n], {n, 0, 1000}];
s[n_] := Table[b[n], {k, 1, n}];
h[1] = {1}; h[n_] := Join[h[n - 1], s[n]];
g = h[100]; r[0] = {0};
r[n_] := If[MemberQ[bb, n], {n}, Join[{g[[n]]}, -r[g[[n]] - n]]]
t = Table[r[n], {n, 0, 120}]  (* A255974 actual representations *)
Flatten[t]  (* A255974 sequence *)