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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. 1

%I

%S 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 15,15,15,10,15,24,22,26,18,18,18,11,13,18,29,28,27,21,21,21,15,13,19,

%U 17,25,31,23,24,24,24,16,18,20,21

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

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

%e 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 (Python 3)

%o def seqr(n):

%o sequence = [n]

%o i = 1

%o while True:

%o if n - i > 0 and n - i not in sequence:

%o n -= i

%o sequence.append(n)

%o elif n + i not in sequence:

%o n += i

%o sequence.append(n)

%o else:

%o break

%o i += 1

%o return len(sequence)

%o print([seqr(n) for n in range(1000)])

%Y Iteration rule nearly identical to A005132.

%Y A334219 is essentially the same sequence.

%K nonn

%O 0,1

%A _Kevin J. Gomez_, May 25 2019

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Last modified October 24 20:35 EDT 2021. Contains 348233 sequences. (Running on oeis4.)