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A308190 Number of steps to reach 5 when iterating x -> A111234(x) starting at x=n.

%I A308190 #17 Jun 15 2019 07:24:55
%S A308190 0,1,3,2,2,4,4,3,4,3,3,5,6,5,5,4,5,5,5,4,5,4,4,6,8,7,7,6,4,6,4,5,7,6,
%T A308190 6,6,7,6,6,5,6,6,6,5,4,5,5,7,10,9,6,8,6,8,8,7,6,5,5,7,6,5,7,6,5,8,8,7,
%U A308190 8,7,7,7,6,8,8,7,8,7,7,6,6,7,7,7,8,7,5,6,7,5,5,6,7,6,6,8,12
%N A308190 Number of steps to reach 5 when iterating x -> A111234(x) starting at x=n.
%C A308190 It is easy to show that every number n >= 5 eventually reaches 5. This was conjectured by Ali Sada. For A111234 sends a composite n > 5 to a smaller number, and sends a prime > 5 to a smaller number in two steps. Furthermore no number >= 5 can reach a number less than 5. So all numbers >= 5 eventually reach 5.
%D A308190 Ali Sada, Email to _N. J. A. Sloane_, Jun 13 2019.
%H A308190 Chai Wah Wu, <a href="/A308190/b308190.txt">Table of n, a(n) for n = 5..10000</a>
%t A308190 a[n_] := Module[{c = 0, x = n, y}, While[x != 5, y = Min[FactorInteger[x][[All, 1]]]; x = y + Quotient[x, y]; c++]; c];
%t A308190 Table[a[n], {n, 5, 100}] (* _Jean-François Alcover_, Jun 15 2019, from Python *)
%o A308190 (Python)
%o A308190 from sympy import factorint
%o A308190 def A308190(n):
%o A308190     c, x = 0, n
%o A308190     while x != 5:
%o A308190         y = min(factorint(x))
%o A308190         x = y + x//y
%o A308190         c += 1
%o A308190     return c # _Chai Wah Wu_, Jun 14 2019
%Y A308190 Cf. A111234, A308191, A308192, A308193, A308194.
%K A308190 nonn
%O A308190 5,3
%A A308190 _N. J. A. Sloane_, Jun 14 2019