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A306536
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The smallest integer k such that floor((2*n)^k/k) is an odd number.
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0
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12, 3, 5, 5, 3, 7, 10, 3, 5, 11, 3, 7, 5, 3, 17, 12, 3, 5, 5, 3, 11, 10, 3, 5, 7, 3, 7, 5, 3, 11, 11, 3, 5, 5, 3, 13, 10, 3, 5, 7, 3, 10, 5, 3, 17, 7, 3, 5, 5, 3, 11, 10, 3, 5, 7, 3, 10, 5, 3, 7, 7, 3, 5, 5, 3, 15, 7, 3, 5, 12, 3, 10, 5, 3, 7
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OFFSET
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1,1
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COMMENTS
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For n > 0, there are infinitely many numbers k such that floor(n^k/k) is an odd number. For odd number n, it's obvious because k can be n^j for any j >= 0. For n = 2, k can be 3*2^j for any j >= 1.
For even number n > 2, k can be A090368(n/2)^j for any j >= 1. Proof: if odd number k makes n^k == 1 (mod k), then n^k = 1 + k*(odd number t), so floor(n^k/k) = t is an odd number. A090368(n/2)^j (j>0) is such k.
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LINKS
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FORMULA
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a(2^j) <= 12. (This is because floor((2*2^j)^12/12) = floor(2^(12j+10)/3) = (2^(12j+10) - 1)/3 is an odd integer for all j >= 0. - Jianing Song, Feb 24 2019)
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PROG
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(PARI) a(n) = {k=1; while((2*n)^k\k%2==0, k++); k; }
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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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