A343554 a(1) = 0. For n >= 1, if there exists i < j < n such that a(i) = a(j) = a(n), take the largest such i and set a(n+1) = n-i; otherwise a(n+1) = 0.

0, 0, 0, 2, 0, 3, 0, 4, 0, 4, 0, 4, 4, 3, 0, 6, 0, 6, 0, 4, 8, 0, 5, 0, 5, 0, 4, 14, 0, 5, 7, 0, 6, 17, 0, 6, 18, 0, 6, 6, 4, 21, 0, 8, 0, 7, 0, 4, 21, 0, 5, 26, 0, 6, 15, 0, 6, 17, 0, 6, 6, 4, 21, 21, 15, 0, 10, 0, 9, 0, 4, 23, 0, 5, 44, 0, 6, 17, 44, 0, 7, 50

Offset

1,4

Comments

A variant of "Van Eck's" sequence A181391, but "looking back" for the second to last appearance of the value of a(n) instead of the last one. If a(n) has appeared just once before, a(n+1) is also 0.

Links

Allan Teitelman, Table of n, a(n) for n = 1..10000

Allan Teitelman, C Code

Maple

b:= proc() [0$3] end:
a:= proc(n) option remember; local t; t:= (h->
     `if`(h=0, 0, n-1-h))(`if`(n=1, 0, b(a(n-1))[1]));
      b(t):= [b(t)[2..3][], n]; t
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Apr 19 2021

Mathematica

a[1]=0; a[n_]:=a[n]=If[Length[s=Position[Array[a, n-1], a[n-1]]]>2, First@@{s[[-1]]-s[[-3]]}, 0]; Array[a, 100] (* Giorgos Kalogeropoulos, Apr 24 2021 *)

Prog

(C) see link above
(Python)
def aupton(terms):
  n, alst, locs2 = 1, [0], dict()
  while n < terms:
    an = alst[-1]
    if an in locs2:
      if len(locs2[an]) > 1:
        i = min(locs2[an]); locs2[an].discard(i); anp1 = n - i
      else: anp1 = 0
    else: locs2[an] = set(); anp1 = 0
    alst.append(anp1); locs2[an].add(n); n += 1
  return alst
print(aupton(82)) # Michael S. Branicky, Apr 24 2021

Crossrefs

Cf. A181391.

Sequence in context: A338569 A343757 A108760 * A137304 A234581 A234582

Adjacent sequences: A343551 A343552 A343553 * A343555 A343556 A343557

Keyword

easy,nonn

Author

Allan Teitelman, Apr 19 2021

Status

approved