A340612 a(0) = 0; for n > 0, if n appears in the sequence then a(n) = lastindex(n), where lastindex(n) is the index of the last appearance of n. Otherwise a(n) = a(n-1) - n if nonnegative and not already in the sequence, otherwise a(n) = a(n-1) + n.

0, 1, 3, 2, 6, 11, 4, 11, 19, 10, 9, 7, 19, 32, 18, 33, 17, 16, 14, 12, 32, 53, 31, 8, 32, 57, 83, 56, 28, 57, 27, 22, 24, 15, 49, 84, 48, 85, 47, 86, 46, 5, 47, 90, 134, 89, 40, 42, 36, 34, 84, 135, 187, 21, 75, 20, 27, 29, 87, 146, 206, 145, 207, 144, 80, 145, 79, 146, 78, 147, 77, 148, 76, 149

Offset

0,3

Comments

This sequences uses the same rules as Recamán's sequence A005132 if the value of n itself has not previously appeared in the sequence. However if n has previously appeared then a(n) = lastindex(n), where lastindex(n) is the sequence index of the last appearance of n.

The terms appear to be clustered in bands which are themselves composed of thinner bands. No values appear outside these groupings. See the linked image.

The smallest value not to have appeared after 1 million terms is 13. It is unknown if all terms eventually appear.

Links

Table of n, a(n) for n=0..73.

Scott R. Shannon, Image of the first 1 million terms.

Example

a(3) = 2, as a(2) = 3 = n, thus a(3) = 2.
a(5) = 11, as 5 has not previously appeared in the sequence, but 1 has, a(5) = a(4) + 5 = 6 + 5 = 11.
a(11) = 7, as a(7) = 11 = n, thus a(11) = 7.

Prog

(Python)
def aupton(nn):
  alst, index = [0], {0: 0} # data list, map of last occurrence
  for n in range(1, nn+1):
    if n in index:
      an = index[n]
    else:
      an = alst[-1] - n
      if an < 0 or an in index:
        an = alst[-1] + n
    alst.append(an)
    index[an] = n
  return alst
print(aupton(65)) # Michael S. Branicky, Jan 13 2021

Crossrefs

Cf. A005132, A340593, A339673, A336760, A336761.

Sequence in context: A331889 A369247 A367544 * A362373 A364929 A109876

Adjacent sequences: A340609 A340610 A340611 * A340613 A340614 A340615

Keyword

nonn

Author

Scott R. Shannon, Jan 13 2021

Status

approved