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Numbers whose unitary divisors have a mean unitary abundancy index that is larger than 2.
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223092870, 281291010, 300690390, 6469693230, 6915878970, 8254436190, 8720021310, 9146807670, 9592993410, 10407767370, 10485364890, 10555815270, 11125544430, 11532931410, 11797675890, 11823922110, 12095513430, 12328305990, 12598876290, 12929686770, 13162479330
COMMENTS
The least odd term is A070826(43) = 5.154... * 10^74, and the least term that is coprime to 6 is Product_{k=3..219} prime(k) = 1.0459... * 10^571. The least nonsquarefree ( A013929) term is a(613) = 148802944290 = 2 * 3 * 5 * 7 * 11 * 13 * 17 *19 * 23^2 * 29.
EXAMPLE
223092870 is a term since A374783(223092870)/ A374784(223092870) = 666225/330752 = 2.014... > 2.
MATHEMATICA
f[p_, e_] := 1 + 1/(2*p^e); r[1] = 1; r[n_] := Times @@ f @@@ FactorInteger[n]; Select[Range[4*10^8], s[#] > 2 &]
PROG
(PARI) is(n) = {my(f = factor(n)); prod(i = 1, #f~, 1 + 1/(2*f[i, 1]^f[i, 2])) > 2; }
Decimal expansion of the asymptotic mean of the ratio between the number of squarefree divisors and the number of cubefree divisors.
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8, 5, 6, 2, 0, 0, 5, 0, 7, 9, 3, 7, 4, 7, 7, 1, 4, 9, 3, 9, 7, 2, 8, 1, 0, 8, 9, 5, 9, 5, 1, 6, 0, 4, 0, 4, 9, 8, 8, 4, 9, 0, 3, 1, 5, 8, 4, 1, 3, 2, 7, 1, 3, 1, 8, 5, 9, 6, 9, 5, 5, 8, 0, 3, 4, 0, 3, 8, 6, 6, 0, 8, 9, 6, 0, 1, 1, 9, 5, 9, 2, 1, 0, 5, 5, 5, 3, 0, 9, 0, 7, 8, 0, 9, 2, 3, 1, 4, 3, 4, 9, 2, 7, 3, 9
COMMENTS
For a positive integer k the ratio between the number of squarefree divisors and the number of cubefree divisors is r(k) = A034444(k)/ A073184(k). r(k) <= 1 with equality if and only if k is squarefree ( A005117). The asymptotic second raw moment is <r(k)^2> = Product_{p prime} (1 - 5/(9*p^2)) = 0.76780883634140395932... and the asymptotic standard deviation is 0.29730736888962774256... .
FORMULA
Equals Product_{p prime} (1 - 1/(3*p^2)).
In general, the asymptotic mean of the ratio between the number of k-free divisors and the number of (k-1)-free divisors, for k >= 3, is Product_{p prime} (1 - 1/(k*p^2)).
EXAMPLE
0.85620050793747714939728108959516040498849031584132...
MATHEMATICA
$MaxExtraPrecision = 1000; m = 1000; c = LinearRecurrence[{0, 1/3}, {0, -(2/3)}, m]; RealDigits[Exp[NSum[Indexed[c, n] * PrimeZetaP[n]/n, {n, 2, m}, NSumTerms -> m, WorkingPrecision -> m]], 10, 105][[1]]
PROG
(PARI) prodeulerrat(1 - 1/(3*p^2))
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