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%I #26 Nov 05 2012 14:01:41
%S 0,1,1,0,1,0,1,1,0,1,1,0,1,0,1,1,0,0,1,1,1,1,0,1,1,0,0,1,1,1,0,1,1,0,
%T 1,0,0,0,1,0,1,1,0,0,1,1,1,0,1,1,0,1,0,0,0,0,1,1,0,1,1,1,1,1,1,0,0,1,
%U 1,0,1,1,0,0,1,1,1,0,1,1,0,1,0,0,0,0,1,1
%N Sequence A179016 reduced modulo 2.
%C It holds for all n>=1 that a(n) = A179016(n)-A213723(A179016(n-1)) meaning that a(n) = 1 when the next node upwards in the infinite trunk of beanstalk sequence (A179016) is the larger of the two possible branches from A179016(n), and 0 when it is the smaller of the said branches. That is, this sequence tells whether A179016 proceeds "left" or "right" at each step.
%C If we were able to find the values of this sequence "a priori" (without needing the value of A179016 at the same point and taking modulo 2 from it), then A179016 could be computed in a more straightforward "bottom-up manner", as then we would have enough information to find the correct path in the binary tree of beanstalk up to the infinity.
%H Antti Karttunen, <a href="/A213729/b213729.txt">Table of n, a(n) for n = 0..2493</a>
%F a(n) = A000035(A179016(n)).
%o (Scheme): (define (A213729 n) (A000035 (A179016 n)))
%o ;; Alternative definition:
%o (define (A213729v2 n) (if (zero? n) n (- (A179016 n) (A213723 (A179016 (-1+ n))))))
%Y a(n) = A000035(A179016(n)). Binary complement of A213728. Cf. A213730, A213733. Run lengths: A218545.
%K nonn
%O 0
%A _Antti Karttunen_, Nov 01 2012
%E Offset changed from 1 to 0 by _Antti Karttunen_, Nov 05 2012
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