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A252735
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a(1) = 0; for n > 1: a(2n) = a(n), a(2n+1) = 1 + a(A064989(n)).
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9
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0, 0, 1, 0, 2, 1, 3, 0, 1, 2, 4, 1, 5, 3, 2, 0, 6, 1, 7, 2, 3, 4, 8, 1, 2, 5, 1, 3, 9, 2, 10, 0, 4, 6, 3, 1, 11, 7, 5, 2, 12, 3, 13, 4, 2, 8, 14, 1, 3, 2, 6, 5, 15, 1, 4, 3, 7, 9, 16, 2, 17, 10, 3, 0, 5, 4, 18, 6, 8, 3, 19, 1, 20, 11, 2, 7, 4, 5, 21, 2, 1, 12, 22, 3, 6, 13, 9, 4, 23, 2, 5, 8, 10, 14, 7, 1, 24, 3, 4, 2, 25, 6, 26, 5, 3, 15, 27, 1
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OFFSET
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1,5
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COMMENTS
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Consider the binary tree illustrated in A005940: If we start from any n, computing successive iterations of A252463 until 1 is reached (i.e., we are traversing level by level towards the root of the tree, starting from that vertex of the tree where n is located at), a(n) gives the number of odd numbers > 1 encountered on the path (i.e., excluding the final 1 from the count but including the starting n if it was odd).
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LINKS
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FORMULA
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a(1) = 0; for n > 1: a(2n) = a(n), a(2n+1) = 1 + a(A064989(n)).
Other identities:
For all n >= 2:
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MATHEMATICA
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PROG
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(Scheme, with memoization-macro definec) (definec (A252735 n) (cond ((= 1 n) 0) ((odd? n) (+ 1 (A252735 (A064989 n)))) (else (A252735 (/ n 2)))))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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