A096303 Number of iterations of n -> n + (number of 1's in binary representation of n) needed for the trajectory of n to join the trajectory of A010062.

0, 0, 0, 1, 0, 4, 0, 3, 2, 0, 1, 0, 2, 0, 1, 1, 0, 2, 0, 1, 6, 0, 2, 5, 0, 4, 1, 0, 3, 2, 0, 3, 2, 1, 1, 0, 5, 0, 2, 4, 0, 3, 1, 0, 2, 7, 0, 7, 1, 6, 1, 0, 5, 3, 0, 2, 4, 2, 1, 0, 3, 1, 6, 0, 0, 2, 0, 1, 5, 0, 2, 4, 0, 3, 1, 0, 2, 5, 0, 5, 1, 4, 1, 0, 3, 10, 0, 2, 2, 9, 1, 0, 1, 8, 1, 0, 8, 0, 7, 7, 0, 6, 6, 6, 0

Offset

1,6

Comments

Conjecture: For any positive integer starting value n, iterations of n -> n + (number of 1's in binary representation of n) will eventually join A010062.

Links

Rémy Sigrist, Table of n, a(n) for n = 1..10000

Index entries for Colombian or self numbers and related sequences

Example

a(6)=4 because the trajectory for 1 (sequence A010062) starts
1->2->3->5->7->10->12->14->17->19->22->25...
and the trajectory for 6 starts
6->8->9->11->14->17->19->22->25->28->31->36...
so the sequence beginning with 6 joins A010062 after 4 steps.

Prog

(PARI) a(n) = { my (o=1); for (k=0, oo, while (o<n, o=o+hammingweight(o)); if (o==n, return (k), n=n+hammingweight(n))) } \\ Rémy Sigrist, Apr 05 2020

Crossrefs

Cf. A010062.

For records see A229743, A229744.

Sequence in context: A086165 A301408 A227290 * A262695 A021252 A348991

Adjacent sequences: A096300 A096301 A096302 * A096304 A096305 A096306

Keyword

base,nonn

Author

Jason Earls, Jun 25 2004

Status

approved