A361869 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.
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, 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, 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
Offset: 0
Keywords
Examples
a(5) = 2 because x_0 = 5 > x_1 = A003415(5) = 1 > x_2 = A003415(1) = 0, but x_3 = A003415(0) = 0. a(6) = 3 because x_0 = 6 > x_1 = A003415(6) = 5 > ... > x_3 = 0 but x_4 = 0.
Links
- Robert Israel, Table of n, a(n) for n = 0..10000
Programs
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Maple
ader:= proc(n) local t; n * add(t[2]/t[1], t = ifactors(n)[2]) end proc: f:= proc(n) option remember; local t; t:= ader(n); if t < n then procname(t)+1 else 0 fi end proc: map(f, [$0..1000]);
Comments