A256696 R(k), the minimal alternating binary representation of k, concatenated for k = 0, 1, 2,....

0, 1, 2, 4, -1, 4, 8, -4, 1, 8, -2, 8, -1, 8, 16, -8, 1, 16, -8, 2, 16, -8, 4, -1, 16, -4, 16, -4, 1, 16, -2, 16, -1, 16, 32, -16, 1, 32, -16, 2, 32, -16, 4, -1, 32, -16, 4, 32, -16, 8, -4, 1, 32, -16, 8, -2, 32, -16, 8, -1, 32, -8, 32, -8, 1, 32, -8, 2, 32

Offset: 0

Author

Clark Kimberling, Apr 09 2015

Keywords

• easy
• sign
• base

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) = 2^n, the sum R(n) is the minimal alternating binary representation of n.

A055975 = trace of n, for n >= 1.

A091072 gives the numbers having positive trace.

A091067 gives the numbers having negative trace.

A072339 = number of terms in R(n).

A073122 = sum of absolute values of the terms in R(n).

Examples

			R(0) = 0
R(1) = 1
R(2) = 2
R(3) = 4 - 1
R(4) = 4
R(9) = 8 - 4 + 1
R(11) = 16 - 8 + 4 - 1
		

References

• D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, 1981, Vol. 2 (2nd ed.), p. 196, Exercise 27.

Links

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

Crossrefs

Cf. A000079, A256655 (Fibonacci based), A055975, A091072, A091067, A072339, A073122, A256701, A256702.

Programs

Mathematica

z = 100; b[n_] := 2^n; bb = Table[b[n], {n, 0, 40}];
s[n_] := Table[b[n + 1], {k, 1, b[n]}];
h[0] = {1}; h[n_] := Join[h[n - 1], s[n - 1]];
g = h[10]; r[0] = {0};
r[n_] := If[MemberQ[bb, n], {n}, Join[{g[[n]]}, -r[g[[n]] - n]]]
u = Flatten[Table[r[n], {n, 0, z}]]