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

Author

Jason Earls, Jun 10 2004

Keywords

• easy
• nonn

Comments

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

Examples

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

Crossrefs

Cf. A007320, A094683.