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Ricardo Andrade
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Alexey Ustinov
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Let me startI propose to collect here open problems from one example. It is known that (see J.W. Porter, On a theorem of Heilbronn) $$H(b)=\dfrac{1}{\varphi(b)}\sum\limits_{1\le a\le b\atop(a,b)=1}\ell(a/b)= \dfrac{2\log 2}{\zeta(2)}\cdot\log b+C_P-1+O_\varepsilon(b^{-1/6+\varepsilon}),$$ where $\ell(a/b)$ is a lengththe theory of standart continued fraction expansion and $C_P$ is Porter's constantfractions. Averaging over numerators and denominators one can prove asymptotic formula for the varianceAny types of continued fractions are welcome. Let $$E(R)=\dfrac{2}{R(R+1)}\sum\limits_{b\le R}\sum\limits_{a\le b}\ell(a/b) $$ and $${D}(R)=\dfrac{2}{R(R+1)}\sum\limits_{b\le R}\sum\limits_{a\le b}\left( \ell(a/b)-E(R)\right)^2. $$ Then (see D. Hensley, The Number of Steps in the Euclidean Algorithm) $${D}(R)=D_1\cdot\log R+o(\log R). $$ But if denominator is fixed then for the variance only right oder bound is known (see Bykovskii V.A., Estimate for dispersion of lengths of continued fractions): $$\dfrac{1}{b}\sum\limits_{a=1}^{b}\left(\ell\left(\dfrac{a}{b}\right)- \dfrac{2\log2}{\zeta(2)}\log b\right)^2\ll\log b.$$

Conjecture: $$\dfrac{1}{\varphi(b)}\sum\limits_{1\le a\le b\atop(a,b)=1}(\ell(a/b)-H(b))^2=D_1\log b+o(\log b).$$

Let me start from one example. It is known that (see J.W. Porter, On a theorem of Heilbronn) $$H(b)=\dfrac{1}{\varphi(b)}\sum\limits_{1\le a\le b\atop(a,b)=1}\ell(a/b)= \dfrac{2\log 2}{\zeta(2)}\cdot\log b+C_P-1+O_\varepsilon(b^{-1/6+\varepsilon}),$$ where $\ell(a/b)$ is a length of standart continued fraction expansion and $C_P$ is Porter's constant. Averaging over numerators and denominators one can prove asymptotic formula for the variance. Let $$E(R)=\dfrac{2}{R(R+1)}\sum\limits_{b\le R}\sum\limits_{a\le b}\ell(a/b) $$ and $${D}(R)=\dfrac{2}{R(R+1)}\sum\limits_{b\le R}\sum\limits_{a\le b}\left( \ell(a/b)-E(R)\right)^2. $$ Then (see D. Hensley, The Number of Steps in the Euclidean Algorithm) $${D}(R)=D_1\cdot\log R+o(\log R). $$ But if denominator is fixed then for the variance only right oder bound is known (see Bykovskii V.A., Estimate for dispersion of lengths of continued fractions): $$\dfrac{1}{b}\sum\limits_{a=1}^{b}\left(\ell\left(\dfrac{a}{b}\right)- \dfrac{2\log2}{\zeta(2)}\log b\right)^2\ll\log b.$$

Conjecture: $$\dfrac{1}{\varphi(b)}\sum\limits_{1\le a\le b\atop(a,b)=1}(\ell(a/b)-H(b))^2=D_1\log b+o(\log b).$$

I propose to collect here open problems from the theory of continued fractions. Any types of continued fractions are welcome.

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Alexey Ustinov
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Let me start from one example. It is known that (see J.W. Porter, On a theorem of Heilbronn) $$H(b)=\dfrac{1}{\varphi(b)}\sum\limits_{1\le a\le b\atop(a,b)=1}\ell(a/b)= \dfrac{2\log 2}{\zeta(2)}\cdot\log b+C_P-1+O_\varepsilon(d^{-1/6+\varepsilon}),$$$$H(b)=\dfrac{1}{\varphi(b)}\sum\limits_{1\le a\le b\atop(a,b)=1}\ell(a/b)= \dfrac{2\log 2}{\zeta(2)}\cdot\log b+C_P-1+O_\varepsilon(b^{-1/6+\varepsilon}),$$ where $\ell(c/d)$$\ell(a/b)$ is a length of standart continued fraction expansion and $C_P$ is Porter's constant. Averaging over numerators and denominators one can prove asymptotic formula for the variance. Let $$E(R)=\dfrac{2}{R(R+1)}\sum\limits_{b\le R}\sum\limits_{a\le b}\ell(a/b) $$ and $${D}(R)=\dfrac{2}{R(R+1)}\sum\limits_{b\le R}\sum\limits_{a\le b}\left( \ell(a/b)-E(R)\right)^2. $$ Then (see D. Hensley, The Number of Steps in the Euclidean Algorithm) $${D}(R)=D_1\cdot\log R+o(\log R). $$ But if denominator is fixed then for the variance only right oder bound is known (see Bykovskii V.A., Estimate for dispersion of lengths of continued fractions): $$\dfrac{1}{b}\sum\limits_{a=1}^{b}\left(\ell\left(\dfrac{a}{b}\right)- \dfrac{2\log2}{\zeta(2)}\log b\right)^2\ll\log b.$$

Conjecture: $$\dfrac{1}{\varphi(b)}\sum\limits_{1\le a\le b\atop(a,b)=1}(\ell(a/b)-H(b))^2=D_1\log b+o(\log b).$$

Let me start from one example. It is known that (see J.W. Porter, On a theorem of Heilbronn) $$H(b)=\dfrac{1}{\varphi(b)}\sum\limits_{1\le a\le b\atop(a,b)=1}\ell(a/b)= \dfrac{2\log 2}{\zeta(2)}\cdot\log b+C_P-1+O_\varepsilon(d^{-1/6+\varepsilon}),$$ where $\ell(c/d)$ is a length of standart continued fraction expansion and $C_P$ is Porter's constant. Averaging over numerators and denominators one can prove asymptotic formula for the variance. Let $$E(R)=\dfrac{2}{R(R+1)}\sum\limits_{b\le R}\sum\limits_{a\le b}\ell(a/b) $$ and $${D}(R)=\dfrac{2}{R(R+1)}\sum\limits_{b\le R}\sum\limits_{a\le b}\left( \ell(a/b)-E(R)\right)^2. $$ Then (see D. Hensley, The Number of Steps in the Euclidean Algorithm) $${D}(R)=D_1\cdot\log R+o(\log R). $$ But if denominator is fixed then for the variance only right oder bound is known (see Bykovskii V.A., Estimate for dispersion of lengths of continued fractions): $$\dfrac{1}{b}\sum\limits_{a=1}^{b}\left(\ell\left(\dfrac{a}{b}\right)- \dfrac{2\log2}{\zeta(2)}\log b\right)^2\ll\log b.$$

Conjecture: $$\dfrac{1}{\varphi(b)}\sum\limits_{1\le a\le b\atop(a,b)=1}(\ell(a/b)-H(b))^2=D_1\log b+o(\log b).$$

Let me start from one example. It is known that (see J.W. Porter, On a theorem of Heilbronn) $$H(b)=\dfrac{1}{\varphi(b)}\sum\limits_{1\le a\le b\atop(a,b)=1}\ell(a/b)= \dfrac{2\log 2}{\zeta(2)}\cdot\log b+C_P-1+O_\varepsilon(b^{-1/6+\varepsilon}),$$ where $\ell(a/b)$ is a length of standart continued fraction expansion and $C_P$ is Porter's constant. Averaging over numerators and denominators one can prove asymptotic formula for the variance. Let $$E(R)=\dfrac{2}{R(R+1)}\sum\limits_{b\le R}\sum\limits_{a\le b}\ell(a/b) $$ and $${D}(R)=\dfrac{2}{R(R+1)}\sum\limits_{b\le R}\sum\limits_{a\le b}\left( \ell(a/b)-E(R)\right)^2. $$ Then (see D. Hensley, The Number of Steps in the Euclidean Algorithm) $${D}(R)=D_1\cdot\log R+o(\log R). $$ But if denominator is fixed then for the variance only right oder bound is known (see Bykovskii V.A., Estimate for dispersion of lengths of continued fractions): $$\dfrac{1}{b}\sum\limits_{a=1}^{b}\left(\ell\left(\dfrac{a}{b}\right)- \dfrac{2\log2}{\zeta(2)}\log b\right)^2\ll\log b.$$

Conjecture: $$\dfrac{1}{\varphi(b)}\sum\limits_{1\le a\le b\atop(a,b)=1}(\ell(a/b)-H(b))^2=D_1\log b+o(\log b).$$

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Alexey Ustinov
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