A308419 Stopping time for Recamán-like iteration of each n: a(0) = n, a(k) = a(k-1) - k if positive and not already in the sequence, a(k) = a(k-1) + k if not already in the sequence, otherwise stop.

24, 24, 13, 21, 3, 3, 3, 15, 6, 6, 6, 15, 12, 9, 9, 9, 16, 20, 15, 12, 12, 12, 8, 10, 12, 20, 15, 15, 15, 10, 15, 24, 22, 26, 18, 18, 18, 11, 13, 18, 29, 28, 27, 21, 21, 21, 15, 13, 19, 17, 25, 31, 23, 24, 24, 24, 16, 18, 20, 21

Offset: 0

Author

Kevin J. Gomez, May 25 2019

Keywords

• nonn

Comments

a(0) is the index of the first repeated value in Recamán's sequence (A005132).

a(n) appears to grow like sqrt(2n).

Examples

			For n = 8, the Recamán-like sequence generated is 8, 7, 5, 2, 6, 1; the sequence halts after a(8) = 6 terms since 1 - 6 = -5 is negative and 1 + 6 = 7 is already in the sequence.
		

Crossrefs

Iteration rule nearly identical to A005132.

A334219 is essentially the same sequence.

Programs

Python

def seqr(n):
    sequence = [n]
    i = 1
    while True:
        if n - i > 0 and n - i not in sequence:
            n -= i
            sequence.append(n)
        elif n + i not in sequence:
            n += i
            sequence.append(n)
        else:
            break
        i += 1
    return len(sequence)
print([seqr(n) for n in range(1000)])

Formula

a(n) >= ceiling((sqrt(1 + 8*n)-1)/2). - Markel Zubia, May 03 2025