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<p class=MsoNormal>Let f = A005228, and let g = complement of f.<o:p></o:p></p>
<p class=MsoNormal><o:p> </o:p></p>
<p class=MsoNormal>The fact that f and g are complements leads to<o:p></o:p></p>
<p class=MsoNormal><o:p> </o:p></p>
<p class=MsoNormal> [1] #{k: f(k) <= n} + #{k: g(k)
<= n} = n<o:p></o:p></p>
<p class=MsoNormal><o:p> </o:p></p>
<p class=MsoNormal>And the fact that f is the running sum of g leads to<o:p></o:p></p>
<p class=MsoNormal><o:p> </o:p></p>
<p class=MsoNormal> [2] f(n) = 1 + SUM(k < n; g(k))<o:p></o:p></p>
<p class=MsoNormal><o:p> </o:p></p>
<p class=MsoNormal>In order to guess at the asymptotics of f, we handwave f and
g into real functions and freely translate [1] and [2] into similar real
relationships, getting:<o:p></o:p></p>
<p class=MsoNormal><o:p> </o:p></p>
<p class=MsoNormal> [3] f^-1(x) + g^-1(x) = x
where f^-1 is the inverse of f<o:p></o:p></p>
<p class=MsoNormal><o:p> </o:p></p>
<p class=MsoNormal> [4] f = g’<o:p></o:p></p>
<p class=MsoNormal><o:p> </o:p></p>
<p class=MsoNormal>This leads to<o:p></o:p></p>
<p class=MsoNormal><o:p> </o:p></p>
<p class=MsoNormal> [5] f^-1(x) + (f’)^-1(x) = x<o:p></o:p></p>
<p class=MsoNormal><o:p> </o:p></p>
<p class=MsoNormal>Of course, I have blithely ignored a lot of details, but one
might hope that a function satisfying [5] would give a good asymptotic to
A005228.<o:p></o:p></p>
<p class=MsoNormal><o:p> </o:p></p>
<p class=MsoNormal>I wish I could solve [5].<o:p></o:p></p>
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