cp's OEIS Frontend

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.

Also the positions in A071542 where new records appear, record values appearing in the ascending order, i.e., as A001477 (because A071542 is a monotone and surjective function).

What is s = lim sup a(n)/(n log_2(n))? A counting argument suggests s >= 1/2, and in any case s <= 1.

Essentially also the partial sums of A086876. - Antti Karttunen, Nov 10 2012 (per personal mail from Carl R. White, Nov 02 2012)

For all n, the following holds: A213708(n) <= A179016(n) <= A173601(n). This sequence gives the distance of the node n in the infinite trunk of beanstalk (A179016(n)) from the greater edge of the A086876(n) wide window which it at that point must pass through.

The increasing steps seem to be quite constrained in their magnitude, compared to the decreasing steps. (This depends on how the "tendrils",i.e. the finite side-trees on the other side of the infinite trunk grow and reach their tops).

For all n, the following holds: A213708(n) <= A179016(n) <= A173601(n). This sequence gives the distance of the node n in the infinite trunk of beanstalk (A179016(n)) from the lesser edge of the A086876(n) wide window which it at that point must pass through.

The increasing steps seem to be quite constrained in their magnitude, compared to the decreasing steps. (This depends on how the "tendrils", i.e. the finite side-trees on the other side of the infinite trunk grow and reach their tops).

Number of integers k which require exactly n steps to reach 0, when starting from k and iterating the map: x -> x - (number of runs in binary representation of x).

a(n) tells from how many starting values one can end to 0 in n steps, with the iterative process described in A219652 (if going around in 0->0 loop is disallowed).

a(n) tells from how many starting values one can end to 0 in n steps, with the iterative process described in A219642 (if going around in 0->0 loop is disallowed).