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Plotting records (from A071383 and A071385) suggests that an upper bound of the form $$ \log r_2(n)\ll\log^\alpha n $$ is plausible, perhaps for $\alpha$ around $0.6$ or $0.7$. Polynomial bounds seem too large, while polylog bounds seem too small.

Edit: See the answer by GH from MOthe answer by GH from MO which shows that $$ \leq\frac{(\log 2+o(1))\log n}{\log\log n} $$ should be used in place of $\ll\log^\alpha n.$

Plotting records (from A071383 and A071385) suggests that an upper bound of the form $$ \log r_2(n)\ll\log^\alpha n $$ is plausible, perhaps for $\alpha$ around $0.6$ or $0.7$. Polynomial bounds seem too large, while polylog bounds seem too small.

Edit: See the answer by GH from MO which shows that $$ \leq\frac{(\log 2+o(1))\log n}{\log\log n} $$ should be used in place of $\ll\log^\alpha n.$

Plotting records (from A071383 and A071385) suggests that an upper bound of the form $$ \log r_2(n)\ll\log^\alpha n $$ is plausible, perhaps for $\alpha$ around $0.6$ or $0.7$. Polynomial bounds seem too large, while polylog bounds seem too small.

Edit: See the answer by GH from MO which shows that $$ \leq\frac{(\log 2+o(1))\log n}{\log\log n} $$ should be used in place of $\ll\log^\alpha n.$

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GH from MO
GH from MO
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Plotting records (from A071383 and A071385) suggests that an upper bound of the form $$ \log r_2(n)\ll\log^\alpha n $$ is plausible, perhaps for $\alpha$ around $0.6$ or $0.7$. Polynomial bounds seem too large, while polylog bounds seem too small.

Edit: See the answer by GH from MO which shows that $$ \frac{2\log n}{\log\log n} $$$$ \leq\frac{(\log 2+o(1))\log n}{\log\log n} $$ should be used in place of $\log^\alpha n.$$\ll\log^\alpha n.$

Plotting records (from A071383 and A071385) suggests that an upper bound of the form $$ \log r_2(n)\ll\log^\alpha n $$ is plausible, perhaps for $\alpha$ around $0.6$ or $0.7$. Polynomial bounds seem too large, while polylog bounds seem too small.

Edit: See the answer by GH from MO which shows that $$ \frac{2\log n}{\log\log n} $$ should be used in place of $\log^\alpha n.$

Plotting records (from A071383 and A071385) suggests that an upper bound of the form $$ \log r_2(n)\ll\log^\alpha n $$ is plausible, perhaps for $\alpha$ around $0.6$ or $0.7$. Polynomial bounds seem too large, while polylog bounds seem too small.

Edit: See the answer by GH from MO which shows that $$ \leq\frac{(\log 2+o(1))\log n}{\log\log n} $$ should be used in place of $\ll\log^\alpha n.$

GH
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Charles
Charles
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Plotting records (from A071383 and A071385) suggests that an upper bound of the form $$ \log r_2(n)\ll\log^\alpha n $$ is plausible, perhaps for $\alpha$ around $0.6$ or $0.7$. Polynomial bounds seem too large, while polylog bounds seem too small.

Edit: See the answer by GH from MO which shows that $$ \frac{2\log n}{\log\log n} $$ should be used in place of $\log^\alpha n.$

Plotting records (from A071383 and A071385) suggests that an upper bound of the form $$ \log r_2(n)\ll\log^\alpha n $$ is plausible, perhaps for $\alpha$ around $0.6$ or $0.7$. Polynomial bounds seem too large, while polylog bounds seem too small.

Plotting records (from A071383 and A071385) suggests that an upper bound of the form $$ \log r_2(n)\ll\log^\alpha n $$ is plausible, perhaps for $\alpha$ around $0.6$ or $0.7$. Polynomial bounds seem too large, while polylog bounds seem too small.

Edit: See the answer by GH from MO which shows that $$ \frac{2\log n}{\log\log n} $$ should be used in place of $\log^\alpha n.$

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Charles
Charles
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