%I #50 May 30 2023 07:45:30

%S 0,1,2,2,0,2,3,2,0,4,3,2,0,2,5,1,0,2,0,2,0,4,3,2,0,4,2,0,0,2,0,2,0,6,

%T 3,1,0,2,5,1,0,2,3,2,0,2,5,2,0,6,3,1,0,2,0,1,0,4,3,2,0,2,7,2,0,1,3,2,

%U 0,3,3,2,0,2,2,2,0,1,3,2,0,0,3,2,0,4,3,1,0,2,0,1,0,4,7,1,0,2,2,3

%N Let x_0, x_1, x_2, ... be the iterations of the arithmetic derivative A003415 starting with x_0 = n. a(n) is the greatest k such that x_0 > x_1 > ... > x_k.

%C a(n) is the number of iterations of A003415 starting at n until the sequence of iterates stops decreasing.

%C a(n) = 0 if and only if A003415(n) >= n.

%C First differs from A099307 at n=15, where a(15) = 1 while A099307(15) = 0.

%H Robert Israel, <a href="/A361869/b361869.txt">Table of n, a(n) for n = 0..10000</a>

%e a(5) = 2 because x_0 = 5 > x_1 = A003415(5) = 1 > x_2 = A003415(1) = 0, but x_3 = A003415(0) = 0.

%e a(6) = 3 because x_0 = 6 > x_1 = A003415(6) = 5 > ... > x_3 = 0 but x_4 = 0.

%p ader:= proc(n) local t;

%p n * add(t[2]/t[1], t = ifactors(n)[2])

%p end proc:

%p f:= proc(n) option remember; local t;

%p t:= ader(n);

%p if t < n then procname(t)+1 else 0 fi

%p end proc:

%p map(f, [$0..1000]);

%Y Cf. A003415, A099307.

%K nonn

%O 0,3

%A _Robert Israel_, May 28 2023