%I #18 Aug 08 2020 14:36:31
%S 1,1,1,1,1,1,1,2,2,1,1,2,1,1,2,2,1,2,1,2,2,1,1,3,3,1,3,2,1,2,1,2,2,1,
%T 3,3,1,1,2,3,1,2,1,2,4,1,1,3,4,3,2,2,1,4,3,3,2,1,1,4,1,1,4,3,3,2,1,2,
%U 2,3,1,3,1,1,5,2,4,2,1,4,3,1,1,4,3,1,2,3,1,4,4,2,2,1,3,5,1,3,4,5
%N Let f(n) = lcd(n)lpf(n), where lcd(n) is the largest common difference between consecutive divisors of n (ordered by size) and lpf(n) is the largest prime factor of n. Sequence gives number of iterations for f(n) to reach 0 or 1.
%H Michael De Vlieger, <a href="/A075661/b075661.txt">Table of n, a(n) for n = 1..10000</a>
%H Jason Earls, <a href="https://www.semanticscholar.org/paper/SmarandacheiterationsofthefirstkindonandEarls/3685dfcee58bf0d2e15a15198e4005ecbadc0c8a">Smarandache iterations of the first kind on functions involving divisors and prime factors</a>, in Smarandache Notions Journal (2004), Vol. 14.1, pp 259264.
%e a(24)=3 because 24 > 9 > 3 > 1.
%t Array[1 + Length@ NestWhileList[Function[n, Max@ Differences@ #  SelectFirst[Reverse@ #, PrimeQ] &@ Divisors[n]], #, # > 0 &] &, 100] (* _Michael De Vlieger_, Mar 28 2018 *)
%Y Cf. A074348.
%K nonn
%O 1,8
%A _Jason Earls_, Sep 23 2002
