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A241665
Number of iterations of A241663 needed to reach either 0 or 1.
2
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, 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, 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
OFFSET
1,7
COMMENTS
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
LINKS
EXAMPLE
A241663(11)=7, A241663(7)=3, A241663(3)=0. Thus, a(11)=3.
PROG
(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.
(PARI)
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
A241665(n) = { my(s=(1==n)); while(n>1, n = A241663(n); s++); (s); }; \\ Antti Karttunen, Oct 01 2018
CROSSREFS
Sequence in context: A101491 A276949 A205794 * A175307 A324825 A316557
KEYWORD
nonn
AUTHOR
Colin Defant, Apr 26 2014
EXTENSIONS
More terms from Alois P. Heinz, Apr 30 2014
Terms a(88) .. a(105) from Antti Karttunen, Oct 01 2018
STATUS
approved