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 A327883 a(0)=a(1)=0; thereafter a(n+1) is the index of the earliest as yet unused occurrence of a(n) if a(n) has occurred before; otherwise a(n+1) = a(a(n)-1). Once an occurrence of a(n) has been used it cannot be used again. 2
 0, 0, 0, 1, 0, 2, 0, 4, 1, 3, 0, 6, 2, 5, 0, 10, 3, 9, 1, 8, 4, 7, 0, 14, 5, 13, 2, 12, 6, 11, 0, 22, 7, 21, 4, 20, 8, 19, 1, 18, 9, 17, 3, 16, 10, 15, 0, 30, 11, 29, 6, 28, 12, 27, 2, 26, 13, 25, 5, 24, 14, 23, 0, 46, 15, 45, 10, 44, 16, 43, 3, 42, 17, 41, 9, 40, 18, 39, 1, 38, 19, 37, 8 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,6 COMMENTS For n > 2, a(n+1) is either the index or repeat of an earlier term (see Name). Hence terms are referred to here as being "index" or "repeat" respectively. Starting from a(3)=1, index and repeat terms occur alternatively throughout the sequence. a(n) < n for all n >= 0. For k >= 2, numbers 0,1,...,2^k-2 occur as terms in the interval between a(2^k-2) = 0 and a(2^(k+1)-2) = 0; see Formula (e.g., k = 2 --> 0,1,2 occur between a(2) and a(6); k = 3 --> 0,1,..,6 occur between a(6) and a(14)). Thus every integer >= 0 occurs infinitely many times in the sequence. There appears to be a proper copy subsequence given by a(4*m+2) = a(m-1); m >0 (noticed by Carl J Love). There may be other (independent) copies to be found (e.g. by selecting terms sequentially, one from each of the above mentioned intervals). A family of sequences similar to this one can be described using the same rule as above, with offset k >= 0 and initial terms a(k) = a(k+1) = k. Conjecture: Every such sequence contains a proper copy of itself as a(4*m+2+k) = a(m+k-1). Index terms > 0 are 1,2,4,3,6,5,10,9,8,7,14,13,... (see A132666). Repeat terms from a(4)=0 are 0,0,1,0,2,0,3,1,4,0,5,2,6,0,7,... (union of the nonnegative integers interleaved with the copy subsequence described above). For any n >= 0, a k >= 0 exists such that a^k(n)=0 (e.g., n=5 -> k=2; a^2(5)=0). Replacing a(a(n)-1) with a(a(n)+1) in the Name produces A025480 with 0 prepended. LINKS Rémy Sigrist, Table of n, a(n) for n = 0..10000 FORMULA Conjectured formulae: a(2^k-2)=0 = a(3*2^k-2) = 0; (k>=0). a(5*2^k-2) = 1, a(7*2^k-2) = 2, (k>=0). a(11*2^(2*k)-2) = a(9*2^(2*k+1)-2) = 3 (k>=0). a(11*2^(2*k+1)-2) = a(9*2^(2k)-2) = 4 (k>=0). a(2*k+1) = A32666(k); k >= 1. a(4*k) = k-1; k >= 1. 1st-level copy subsequence: a(k-1) = a(4*k+2), k >= 1. (m-th)-level (dependent) copy subsequence: a(k) = a(4^m*(k+2) - 2), m >= 1, k >= 0. EXAMPLE a(2) = 0 since a(1) = 0 was last seen as a(0); a(3) = 1 since a(2) = 0 was last seen as a(1); a(4) = 0 since a(3) = 1 has not been seen before, so a(4) = a(a(3)-1) = a(0) = 0; a(327883)=163736. MAPLE # Code by Carl J Love (via Mapleprimes). (This code is derived from Name. An alternative code (same author), based on a recursion deduced from empirical formulae (see above) produces identical output up to 100000 terms.) restart: a:= module() export pos:= table([0= 0]), nextpos:= table([0= 1]), used:= table(sparse, [0= 2]), ModuleApply:= proc(n::nonnegint) option remember; local p:= thisproc(n-1), r:= `if`(used[p] > 1, pos[p], thisproc(p-1)); (pos[r], used[r], nextpos[r]):= (nextpos[r], used[r]+1, n); r end proc; (ModuleApply(0), ModuleApply(1)):= (0, 0) end module : seq(a(k), k= 0..100); CROSSREFS Cf. A132666, A025480. Sequence in context: A305796 A347961 A020781 * A007432 A079124 A242071 Adjacent sequences:  A327880 A327881 A327882 * A327884 A327885 A327886 KEYWORD nonn AUTHOR David James Sycamore, Oct 10 2019 STATUS approved

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Last modified November 30 19:02 EST 2021. Contains 349424 sequences. (Running on oeis4.)