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A058032
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Largest m such that 2^n / primorial(m) >= 1.
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0
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1, 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 16, 17, 17, 17, 17, 17, 17, 18, 18, 18, 18
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OFFSET
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0,3
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COMMENTS
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Primorial order of powers of 2.
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LINKS
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FORMULA
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a(n)=Max{s| q(s)=Sign[Floor[2^n/A002110(n)]]=1}
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EXAMPLE
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for n=1 and 2, when 2 and 4 is divided by 2 gives quotient = 1 or 2, but when divided by 6 q<1, so the largest suitable primorial is the first; thus a(1)=a(2)=1. n=11, 2^11=2048. The largest primorial P, such that 2048/P > 1 is P=210, the 4th = A002110(4). So a(11)=4.
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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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