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Equivalently, satisfies the property: A000005(a(n)) = a(n)-a(n-1). The first differences a(n)-a(n-1) are given in A259935.

V. S. Guba (2015) proved that such an infinite sequence exists. Numerical evidence suggests that it may also be unique -- is it? All terms below 10^10 are defined uniquely.

If the current definition does not uniquely define the sequence, the "lexicographically earliest" condition may be added to make the sequence well-defined.

From Vladimir Shevelev, Jul 21 2015: (Start)

If a(k), a(k+1), a(k+2) is an arithmetic progression, then a(k+1) is in A175304.

Indeed, by the definition of this sequence, a(n)-a(n-1) = d(a(n)), for all n>=1, where d(n) = A000005(n). Hence, have a(k+1) - a(k) = a(k+2) - a(k+1) = d(a(k+1)) = d(a(k+2)). So a(k+1) + d(a(k+2)) = a(k+2) and a(k+1) + d(a(k+1)) = a(k+2).

Therefore, d(a(k+1) + d(a(k+1))) = d(a(k+2))= d(a(k+1)), i.e., a(k+1) is in A175304. Thus, if there are infinitely many pairs of the same consecutive terms of A259935, then A175304 is infinite (see there my conjecture). (End)

From Antti Karttunen, Nov 27 2015: (Start)

If multiple apparently infinite branches would occur at some point of computing, then even if the "lexicographically earliest" condition were then added to the definition, it would not help us much (when computing the sequence), as we would still not know which of the said branches were truly infinite. [See also Max Alekseyev's latter Jul 9 2015 posting on SeqFan-list, where he notes the same thing.] Note that many of the derived sequences tacitly assume that the uniqueness-conjecture is true. See also comments at A262693 and A262896.

One sufficient (but not a necessary) condition for the uniqueness of this sequence is that the sequence A262509 has infinite number of terms. Please see further comments there.

The graph of sequence exhibits two markedly different slopes, depending on whether it is on the "fast lane" of A049820 (even numbers) or the "slow lane" [odd numbers, for example when traversing the 1356 odd terms from 123871 to 113569 at range a(9859) .. a(8504)]. See A263086/A263085 for the "average cumulative speed difference" between the lanes. In general, slow and fast lane stay separate, except when they terminate into one of the squares (A262514) that work as "exchange ramps", forcing the parity (and thus the speed) to change. In average, the odd squares are slightly better than the even squares in attracting lanes going towards smaller numbers (compare A263253 to A263252). The cumulative effect of this bias is that the odd terms are much rarer in this sequence than the even terms (compare A263278 to A262516).

(End)

Starting offset is zero, because a(0) = 0 is a special case in this sequence.

Numbers where A155043 takes a unique value. (Those values are given by A262508.)

Numbers n such that there does not exist any other number h from which one could reach zero in exactly the same number of steps as from n by repeatedly applying the map where k is replaced by k - A000005(k) = A049820(k). Thus in the tree where zero is the root and parent-child relation is given by A049820(child) = parent, all the numbers > n+t (where t is a small value depending on n) have n as their common ancestor. As it is guaranteed that there is at least one infinite path in such a tree, any n in this sequence can be neither a leaf nor any other vertex in a finite side-tree, as then at least one node in the infinite part would have the same distance to the root, thus it must be that n itself is in the infinite part and thus has an infinite number of descendant vertices. Also, for the same reason, the tree cannot branch to two infinite parts from any ancestor of n (which are nodes nearer to the root of the tree, zero).

From the above it follows that if this sequence is infinite, then A259934 is guaranteed to be the only infinite sequence which starts with a(0)=0 and satisfies the condition A049820(a(k)) = a(k-1) for all k>=1, where A049820(n) = n-d(n) and d(n) is the number of divisors of n (A000005). This is a sufficient condition for the uniqueness of A259934, although not necessary. See e.g. A179016 which is the unique infinite solution to a similar problem, even though A086876 contains no ones after its two initial terms.

Is it possible for any even terms to occur after zero? If not, then apart from zero, this would be a subsequence of A262517.

Number of perfect squares (A000290) encountered before zero is reached when starting from k = n and repeatedly applying the map that replaces k by k - d(k), where d(k) is the number of divisors of k (A000005). This count includes n itself if it is a square, but excludes the final zero.

Also number of times the parity (of numbers encountered) changes until zero is reached when iterating A049820. This count includes also the last parity change 1 - d(1) -> 0 if coming to zero through 1.

There is a lower bound for this sequence that grows without limit if and only if either (1) A259934 is indeed the unique sequence (satisfying its given condition) and it contains an infinite number of squares (see A262514), or (2) more generally, if each one of all (hypothetically multiple) infinite branches of the tree (defined by parent-child relation A049820(child) = parent) contains an infinite number of squares. See also comments in A262509.

Terms were computed from the data file of A259934 provided by Max Alekseyev, which contained terms up to A259934(397420670) = 9999999974.

Odd terms of A259934 appear in separate runs, where the first term of each run is always an odd square, like 121 and 113569 are. See A262514 for the explanation.

Numbers n for which A259934(n) is a perfect square. Indexing begins from zero, because a(0) = 0 is a special case.

Terms were computed from the data file of A259934 provided by Max Alekseyev, which contained terms up to A259934(397420670) = 9999999974.

Every other term is even and every other odd. See comments at A262514.

a(31) > 10^8.

All terms are squares (A262688 gives the square roots).

The even and odd terms alternate.

Some of the terms are shared with A262687, but none with A262514.