A094778 "Dropping time" in juggler sequence problem starting at 2n+1 (number of steps to reach a lower number than starting value).

0, 5, 4, 2, 5, 2, 2, 2, 2, 2, 2, 2, 5, 2, 2, 2, 4, 4, 15, 5, 2, 4, 4, 2, 5, 2, 4, 4, 2, 5, 2, 2, 2, 2, 7, 2, 4, 5, 10, 2, 7, 2, 4, 7, 8, 2, 2, 5, 5, 8, 5, 13, 8, 2, 7, 8, 13, 10, 4, 2, 4, 2, 4, 4, 8, 4, 4, 2, 5, 2, 2, 2, 2, 2, 2, 4, 2, 5, 5, 2, 2, 24, 10, 2, 4, 2, 26, 5, 2, 2, 4, 10, 2, 5, 2, 4, 70, 4, 5, 5, 5

Offset

0,2

Comments

If the starting value is even then of course the next step in the trajectory is smaller.

Links

Table of n, a(n) for n=0..100.

Example

9 -> 27 -> 140 -> 11 -> 36 -> 6, taking 5 steps, so a(4) = 5.

Crossrefs

Cf. A007320, A094683.

Sequence in context: A225063 A309442 A213205 * A260849 A246746 A180131

Adjacent sequences: A094775 A094776 A094777 * A094779 A094780 A094781

Keyword

easy,nonn

Author

Jason Earls, Jun 10 2004

Status

approved