%I
%S 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,
%T 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,
%U 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
%N "Dropping time" in juggler sequence problem starting at 2n+1 (number of steps to reach a lower number than starting value).
%C If the starting value is even then of course the next step in the trajectory is smaller.
%e 9 > 27 > 140 > 11 > 36 > 6, taking 5 steps, so a(4) = 5.
%Y Cf. A007320, A094683.
%K easy,nonn
%O 0,2
%A _Jason Earls_, Jun 10 2004
