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
Keywords
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.
Programs
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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
Comments