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A344274
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a(n) is the least k such that the average number of infinitary divisors of {1..k} is >= n.
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2
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1, 6, 24, 105, 385, 1554, 6063, 23688, 92610, 362112, 1416360, 5539296, 21663378, 84725487, 331362185, 1295952084, 5068450464, 19822658688
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OFFSET
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1,2
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LINKS
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FORMULA
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Lim_{n->oo} a(n+1)/a(n) = exp(1/(2*c)) = 3.9109891037..., where c is A327576.
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EXAMPLE
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a(2) = 6 since the average number of infinitary divisors of {1..6} is A327573(6)/6 = 13/6 > 2.
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MATHEMATICA
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f[p_, e_] := 2^DigitCount[e, 2, 1]; idivnum[1] = 1; idivnum[n_] := Times @@ (f @@@ FactorInteger[n]); seq={}; s = 0; k = 1; Do[While[s = s + idivnum[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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