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A213710
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Number of steps to reach 0 when starting from 2^n and iterating the map x -> x - (number of 1's in binary representation of x): a(n) = A071542(2^n) = A218600(n)+1.
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14
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1, 2, 3, 5, 8, 13, 22, 39, 69, 123, 221, 400, 730, 1344, 2494, 4656, 8728, 16406, 30902, 58320, 110299, 209099, 397408, 757297, 1446946, 2771952, 5323983, 10250572, 19780123, 38243221, 74058514, 143592685, 278661809, 541110612, 1051158028, 2042539461, 3969857206
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OFFSET
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0,2
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COMMENTS
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Conjecture: A179016(a(n))= 2^n for all n apart from n=2. This is true if all powers of 2 except 2 itself occur in A179016 as in that case they must occur at positions given by this sequence.
This is easy to prove: It suffices to note that after 3 no integer of form (2^k)+1 can occur in A005187, thus for all k >= 2, A213725((2^k)+1) = 1 or equally: A213714((2^k)+1) = 0. - Antti Karttunen, Jun 12 2013
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LINKS
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FORMULA
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PROG
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(Scheme, two alternatives)
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CROSSREFS
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One more than A218600, which is the partial sums of A213709, thus the latter also gives the first differences of this sequence.
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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