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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.

%I A308419 #20 May 09 2025 00:37:55
%S A308419 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,
%T A308419 15,15,15,10,15,24,22,26,18,18,18,11,13,18,29,28,27,21,21,21,15,13,19,
%U A308419 17,25,31,23,24,24,24,16,18,20,21
%N 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.
%C A308419 a(0) is the index of the first repeated value in Recamán's sequence (A005132).
%C A308419 a(n) appears to grow like sqrt(2n).
%F A308419 a(n) >= ceiling((sqrt(1 + 8*n)-1)/2). - _Markel Zubia_, May 03 2025
%e A308419 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.
%o A308419 (Python)
%o A308419 def seqr(n):
%o A308419     sequence = [n]
%o A308419     i = 1
%o A308419     while True:
%o A308419         if n - i > 0 and n - i not in sequence:
%o A308419             n -= i
%o A308419             sequence.append(n)
%o A308419         elif n + i not in sequence:
%o A308419             n += i
%o A308419             sequence.append(n)
%o A308419         else:
%o A308419             break
%o A308419         i += 1
%o A308419     return len(sequence)
%o A308419 print([seqr(n) for n in range(1000)])
%Y A308419 Iteration rule nearly identical to A005132.
%Y A308419 A334219 is essentially the same sequence.
%K A308419 nonn
%O A308419 0,1
%A A308419 _Kevin J. Gomez_, May 25 2019