|
%I #21 Mar 20 2018 16:15:37
%S 0,1,1,1,2,1,3,1,2,1,3,1,4,1,2,1,3,1,4,1,3,1,4,1,3,1,3,1,4,1,5,1,3,1,
%T 3,1,4,1,4,1,5,1,6,1,3,1,4,1,4,1,3,1,4,1,4,1,4,1,5,1,6,1,3,1,4,1,5,1,
%U 4,1,5,1,6,1,3,1,4,1,5,1,4,1,5,1,4,1,4
%N Number of iterations of A058026 needed to reach either 0 or 1.
%C I conjecture that, for n>3 and n odd, we have a(n)>=log((49/15)n)/log(7).
%H Reinhard Zumkeller, <a href="/A241664/b241664.txt">Table of n, a(n) for n = 1..10000</a>
%H C. Defant, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL18/Defant/defant5.html">On Arithmetic Functions Related to Iterates of the Schemmel Totient Functions</a>, J. Int. Seq. 18 (2015) # 15.2.1
%H Colin Defant, <a href="/A241664/a241664_1.py.txt">Python program for A241664</a>
%e A058026(7)=5, A058026(5)=3, A058026(3)=1. As it takes 3 iterations to reach 1, a(7)=3.
%o The link above provides a Python program. Enter m=2 as well as starting and ending values of n. The third string of numbers will be this sequence.
%o (Haskell)
%o a241664 n = fst $ until ((<= 1) . snd)
%o (\(u, v) -> (u + 1, a058026 v)) (0, n)
%o -- _Reinhard Zumkeller_, May 10 2014
%K nonn,look
%O 1,5
%A _Colin Defant_, Apr 26 2014
%E a(1) corrected by _Reinhard Zumkeller_, May 10 2014
|
