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 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. 1
 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 (list; graph; refs; listen; history; text; internal format)
 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

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Last modified June 8 18:04 EDT 2023. Contains 363165 sequences. (Running on oeis4.)