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A352451
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2^k appears in the binary expansion of a(n) iff 2^k appears in the binary expansion of n and k+1 divides n.
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5
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0, 1, 2, 1, 0, 1, 6, 1, 8, 1, 2, 1, 12, 1, 2, 5, 0, 1, 2, 1, 16, 5, 2, 1, 8, 17, 2, 1, 8, 1, 22, 1, 0, 1, 2, 1, 36, 1, 2, 5, 8, 1, 34, 1, 8, 5, 2, 1, 32, 1, 18, 1, 0, 1, 38, 17, 8, 1, 2, 1, 60, 1, 2, 5, 0, 1, 2, 1, 0, 5, 66, 1, 8, 1, 2, 1, 8, 65, 6, 1, 16, 1
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OFFSET
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0,3
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COMMENTS
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The idea is to keep the 1's in the binary expansion of a number whose positions are related in some way to that number.
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LINKS
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FORMULA
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a(n) <= n.
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EXAMPLE
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For n = 42:
- 42 = 2^5 + 2^3 + 2^1,
- 5+1 divides 42,
- 3+1 does not divide 42,
- 1+1 divides 42,
- so a(42) = 2^5 + 2^1 = 34.
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PROG
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(PARI) a(n) = { my (v=0, m=n, k); while (m, m-=2^k=valuation(m, 2); if (n%(k+1)==0, v+=2^k)); v }
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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