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A328331
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a(n) is the least k such that the average number of unitary divisors of {1..k} is >= n.
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6
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1, 6, 35, 190, 1015, 5304, 27417, 142142, 736782, 3816852, 19774690, 102446730, 530743749, 2749606626, 14244797910
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OFFSET
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1,2
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COMMENTS
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LINKS
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FORMULA
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Lim_{n->oo} a(n+1)/a(n) = exp(zeta(2)) = exp(Pi^2/6) = 5.180668... (since A064608(n) ~ n*log(n)/zeta(2)).
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EXAMPLE
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a(2) = 6 because the average number of unitary divisors of {1..6} is A064608(6)/6 = 13/6 > 2.
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MATHEMATICA
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seq={}; s = 0; k = 1; Do[While[s += 2^PrimeNu[k]; s < k*n, k++]; AppendTo[seq, k]; k++, {n, 1, 10}]; seq
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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STATUS
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approved
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