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%I #12 Feb 13 2016 23:17:20
%S 0,1,1,2,2,3,3,3,3,4,4,4,4,4,4,5,5,6,6,6,7,7,7,7,8,8,8,8,8,8,9,9,9,9,
%T 10,10,10,10,10,11,11,11,11,11,11,12,12,12,12,12,12,13,13,13,13,13,13,
%U 14,14,14,14,14,14,15,15,15,15,15,16,16,16,16,16,16,16,17,17,17,17,17,17,17,17,18
%N Number of iterations of A268395 needed to reach zero: a(0) = 0, for n >= 1, a(n) = 1 + a(A268395(n)).
%C Set k = n, take the k-th Xfactorial A048631(k) and find what is the maximum exponent h so that polynomial (X+1)^h still divides it (over GF(2), this can be computed with A268389, maybe also more directly). Set this h as a new value of k, and repeat. a(n) tells how many steps are needed before zero is reached.
%C Note that so far it is just an empirical observation that A268672(n) > 0 for all n > 0.
%H Antti Karttunen, <a href="/A268708/b268708.txt">Table of n, a(n) for n = 0..65537</a>
%F a(0) = 0, for n >= 1, a(n) = 1 + a(A268395(n)).
%o (Scheme, with memoization-macro definec)
%o (definec (A268708 n) (if (zero? n) 0 (+ 1 (A268708 (A268395 n)))))
%Y Cf. A048631, A268389, A268395, A268672.
%Y Cf. also A268709, A268710.
%K nonn
%O 0,4
%A _Antti Karttunen_, Feb 11 2016
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