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A241665 Number of iterations of A241663 needed to reach either 0 or 1. 2

%I #21 Oct 01 2018 21:06:52

%S 1,1,1,1,1,1,2,1,1,1,3,1,2,1,1,1,3,1,2,1,1,1,3,1,2,1,1,1,3,1,2,1,1,1,

%T 2,1,2,1,1,1,3,1,2,1,1,1,3,1,2,1,1,1,3,1,3,1,1,1,4,1,2,1,1,1,2,1,2,1,

%U 1,1,3,1,2,1,1,1,2,1,2,1,1,1,3,1,3,1,1,1,4,1,2,1,1,1,2,1,2,1,1,1,3,1,2,1,1

%N Number of iterations of A241663 needed to reach either 0 or 1.

%C It might be more natural to define the initial terms as a(0) = a(1) = 0 for the sake of recurrence. - _Antti Karttunen_, Oct 01 2018

%H Antti Karttunen, <a href="/A241665/b241665.txt">Table of n, a(n) for n = 1..65537</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="/A241665/a241665.py.txt">Python program</a>

%e A241663(11)=7, A241663(7)=3, A241663(3)=0. Thus, a(11)=3.

%o (Python) See Defant link. Enter m=4, as well as starting and ending values of n. The third string of numbers will be this sequence.

%o (PARI)

%o A241663(n) = {my(f = factor(n)); prod(i=1, #f~, if ((f[i, 1] == 2) || (f[i, 1] == 3), 0, f[i, 1]^(f[i, 2]-1)*(f[i, 1]-4))); } \\ From A241663

%o A241665(n) = { my(s=(1==n)); while(n>1, n = A241663(n); s++); (s); }; \\ _Antti Karttunen_, Oct 01 2018

%Y Cf. A241663, A241668.

%K nonn

%O 1,7

%A _Colin Defant_, Apr 26 2014

%E More terms from _Alois P. Heinz_, Apr 30 2014

%E Terms a(88) .. a(105) from _Antti Karttunen_, Oct 01 2018

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Last modified March 28 04:13 EDT 2024. Contains 371235 sequences. (Running on oeis4.)