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A306754
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The bottom entry in the difference table of the positions of the ones in the binary representation of n.
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1
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0, 0, 1, 1, 2, 2, 1, 0, 3, 3, 2, 1, 1, -1, 0, 0, 4, 4, 3, 2, 2, 0, 1, 1, 1, -2, -1, -2, 0, 1, 0, 0, 5, 5, 4, 3, 3, 1, 2, 2, 2, -1, 0, -1, 1, 2, 1, 1, 1, -3, -2, -4, -1, -1, -2, -3, 0, 2, 1, 3, 0, -1, 0, 0, 6, 6, 5, 4, 4, 2, 3, 3, 3, 0, 1, 0, 2, 3, 2, 2, 2, -2
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OFFSET
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0,5
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COMMENTS
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By convention, a(0) = 0.
For any n > 0:
- let (b_1, b_2, ..., b_h) be the positions of the ones in the binary representation of n,
- h = A000120(n) and 0 <= b_1 < b_2 < ... < b_h,
- n = Sum_{k = 1..h} 2^b_k,
- a(n) is the unique value remaining after taking successively the first differences of (b_1, ..., b_h) h-1 times.
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LINKS
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FORMULA
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a(2^k) = k for any k >= 0.
a(2^k-1) = [k=2].
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EXAMPLE
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For n = 59:
- the binary representation of 59 is "111011",
- so h = 5 and b_1 = 0, b_2 = 1, b_3 = 3, b_4 = 4, b_5 = 5,
- the corresponding difference table is:
0 1 3 4 5
1 2 1 1
1 -1 0
-2 1
3
- hence a(59) = 3.
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PROG
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(PARI) a(n) = { my (h=hammingweight(n), o=0, v=0); forstep (k=h-1, 0, -1, my (w=valuation(n, 2)); o += w; v += (-1)^k * binomial(h-1, k) * o; o++; n\=2^(1+w)); v };
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CROSSREFS
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See A306607 for a similar sequence.
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KEYWORD
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sign,base
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AUTHOR
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STATUS
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approved
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