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%I #16 Aug 06 2015 07:40:21
%S 0,0,0,0,6,0,0,0,5,6,0,0,0,0,5,0,4,5,0,6,4,0,55,0,0,0,4,0,141,5,0,0,
%T 140,4,1,5,0,0,54,6,0,4,2,0,145,55,0,0,3,0,6,0,2,4,0,0,1,141,0,5,0,0,
%U 3,0,2,140,0,4,4,1,4,5,0,0,1,0,2,54,5,6,3,0,3,4,4,2,0,0,4,145,0,55,139,0,1,0,0,3,53,0,3,6,0,0,3,2,14,4,0
%N Number of iterations of A234742 needed when started from n before a fixed point is reached.
%C The fixed points of A234742 are in A235035, thus the latter gives the zeros of this sequence.
%C It is not known whether the sequence is well-defined for all values. For example, does a(455) or a(1361) have a finite value? Cf. sequences A260735 and A260441.
%H Antti Karttunen, <a href="/A260712/b260712.txt">Table of n, a(n) for n = 1..454</a>
%F If A234742(n) = n, then a(n) = 0, otherwise a(n) = 1 + a(A234742(n)).
%F Other identities:
%F a(A235035(n)) = 0.
%F a(2n) = a(n).
%o (PARI)
%o allocatemem((2^30));
%o A234742(n) = factorback(subst(lift(factor(Mod(1, 2)*Pol(binary(n)))), x, 2)); \\ After _M. F. Hasler_'s Feb 18 2014 code.
%o A260712(n) = {my(prev=-1,i=-1); until((n==prev), prev = n; n = A234742(n); i++); return(i); };
%o for(n=1, 454, write("b260712.txt", n, " ", A260712(n)));
%o (Scheme, two alternatives, the first one using memoizing definec-macro)
%o (definec (A260712 n) (let ((next (A234742 n))) (if (= next n) 0 (+ 1 (A260712 next)))))
%o (define (A260712loop n) (let loop ((n (A234742 n)) (prev_n n) (i 0)) (if (= n prev_n) i (loop (A234742 n) n (+ 1 i)))))
%Y Cf. A235035 (gives the positions of zeros).
%Y Cf. A234742, A260735, A260735, A260441.
%Y Subsequences: A260713, A260716.
%K nonn
%O 1,5
%A _Antti Karttunen_, Aug 04 2015
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