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.

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

Offset: 0

Author

David James Sycamore, Oct 10 2019

Keywords

• nonn
• look

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.

Examples

			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.
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 0..10000
• Michael De Vlieger, Log log scatterplot of a(n), n = 0..2^20, showing a(n) = 0 instead as 1/2 for visibility.

Crossrefs

Cf. A132666, A025480.

Programs

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);

Mathematica

nn = 1000; c[] := 0; a[0] = a[1] = a[2] = 0; c[0] = 1; Do[If[# == 0, k = a[a[n - 1] - 1], k = #] &[c[a[n - 1]]]; Set[{a[n], c[a[n - 1]]}, {k, n - 1}], {n, 3, nn}]; Array[a, nn + 1, 0] (* _Michael De Vlieger, Jun 25 2025 *)

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) = A132666(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.