A383730 a(0) = 0, a(n) = a(n-1) + A002260(n) * (-1)^(n-1) if not already in the sequence, otherwise a(n) = a(n-1) - A002260(n) * (-1)^(n-1).

0, 1, 2, 4, 3, 5, 8, 9, 7, 10, 6, 5, 7, 4, 8, 13, 12, 14, 11, 15, 20, 26, 25, 27, 24, 28, 23, 29, 22, 21, 19, 16, 20, 15, 21, 14, 22, 21, 23, 20, 24, 19, 25, 32, 40, 49, 48, 50, 47, 51, 46, 52, 45, 53, 44, 54, 55, 57, 60, 64, 59, 65, 58, 66, 75, 85, 74, 73, 71

Offset: 0

Author

Markel Zubia, May 06 2025

Keywords

• sign
• look
• hear

Comments

This sequence is a bidirectional form of Recamán's sequence.

Another way to define the sequence: starting at 0, take steps of size 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, ... alternating left and right while avoiding repeated values (negative values are allowed).

The sequence is unbounded either above or below.

Conjecture: the sequence is unbounded both above and below.

Conjecture: each integer appears finitely often.

It exhibits a mix of chaotic and periodic behavior, including long plateaus and sudden large jumps.

Around term 25000, the sequence settles near -3000 in a visually fractal structure. After ~1.85 million terms, it appears to settle again near -500000. Astonishingly, after ~66.7 million steps, it jumps sharply from around -500000 to +4.51 million.

Examples

			a(1) = 0 + 1.
a(2) = 0 + 1 + 1 = 2, since 0 + 1 - 1 = 0 already appears in the sequence, as a(0) = 0.
a(6) = 0 + 1 + 1 + 2 - 1 + 2 + 3 = 8.
		

Links

• Markel Zubia, Table of n, a(n) for n = 0..9999
• Markel Zubia, Plot of the first 100k terms
• Markel Zubia, Close-up of the fractal-like pattern
• Markel Zubia, Plot of the first 10M terms
• Markel Zubia, Plot of the first 1B terms

Crossrefs

Cf. A005132.

Other bidirectional extensions of Recamán's sequence: A063733, A079053, A064288, A064289, A064387, A064388, A064389, A228474.

Cf. A002260.

Programs

Python

def a(n):
    t, k, curr = 1, 1, 0
    seen = set()
    for i in range(n):
        seen.add(curr)
        step = (-1)**i * t
        if curr + step not in seen:
            curr = curr + step
        else:
            curr = curr - step
        t += 1
        if t > k:
            t, k = 1, k + 1
    return curr