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1. If we denote $G(s) = F(2s)$, a trivial coordinate change, then \eqref{1} becomes $\zeta(s)=G(s/2)+G((s+1)/2)$. This can be interpreted as the statement that $\zeta(s)$ is the image of $G(s)$ under the linear operator $T$ that sends an analytic function $g(s)$ to the function $T(g):=g\left(\frac{s}{2}\right) + g\left(\frac{s+1}{2}\right)$.

   This linear operator has some very nice properties. Notably, it is "Riemann-hypothesis preserving", in the following sense: if $g(s)$ is a complex polynomial with all its zeros on the line $\operatorname{Re}(s)=1/2$, then $T(f)$ is also such a polynomial. (This is a reformulation, via a simple coordinate change, of the statement that the operator that maps a polynomial $h(z)$ to $h(z-i/4)+h(z+i/4)$ is "hyperbolicity preserving", that is, sends polynomials with only real zeros to polynomials with only real zeros -- a special case of a more general result that I think is due to Polya).

   This hyperbolicity/RH preservation property also extends to certain entire functions of genus 0 or 1 (usually assumed to satisfy the functional equation $f(1-s)=f(s)$). See section 10.23 of Titschmarsh's book "The Theory of the Riemann Zeta Function", 2nd ed. This was used in some of Polya's failed (though still very interesting) attacks on the Riemann hypothesis. I've seen it used in some other places, for example in the paper "The Riemann hypothesis for certain integrals of Eisenstein series" by Lagarias and Suzuki.

   Now, $\zeta(s)$ is not an entire function, and doesn't exactly satisfy $\zeta(1-s)=\zeta(s)$, so the relevance of these properties of the linear operator $T$ to RH-type questions isn't entirely clear. But the equation $\zeta(s)=F(s)+F(s+1)$ is still suggestive of the possibility that by studying $F(s)$ and its zeros we could learn something about $\zeta(s)$ and its zeros, using these sorts of Polya-style ideas.

2. The same linear operator $T$ is also the transfer operator for the doubling map $x\mapsto 2x \bmod 1$ from ergodic theory.

3. The Bernoulli polynomials $B_n(x)$ (which are themselves related to $\zeta(s)$ in all sorts of ways) are the eigenfunctions of this transfer operator, satisfying the equation $$ T [B_n] = \frac{1}{2^n} B_n, $$ and they also have the symmetry $B_n(1-x)=B_n(x)$.

1. By rewriting \eqref{1} as $F(s) = \zeta(s) - F(s+1)$, one sees that this gives a way to analytically continue $F(s)$ to the region $\operatorname{Re}(s)>0$, then inductively to the region $\operatorname{Re}(s)>-1$, $\operatorname{Re}(s)>-2$, etc, using the formulas \begin{align*} F(s) &= \zeta(s) - F(s+1) \\ &= \zeta(s) - \zeta(s+1) + F(s+2) \\ &= \zeta(s) - \zeta(s+1) + \zeta(s+2) - F(s+3) = \ldots \\ &= \zeta(s) - \zeta(s+1) + \zeta(s+2) - \zeta(s+3) + \ldots + (-1)^k \zeta(s+k) + (-1)^{k+1} F(s+k+1) \end{align*} This shows that $F(s)$ can be analytically continued to a meromorphic function on $\mathbb{C}$, with poles at $s=1, 0, -1, -2, \ldots$. The pole at $s=n$ for integer $n\le 1$ is a simple pole with residue $(-1)^n$.

2. $F(s)$ has the special values \begin{align*} F(2) &= 1, \\ F(3) &= -1 + \zeta(2), \\ F(4) &= 1 - \zeta(2) + \zeta(3), \\ &\ \ \vdots \\ F(n) &= (-1)^n + \sum_{k=2}^{n-1} (-1)^{n+k+1} \zeta(k). \end{align*} (Proof: the evaluation of $F(2)$ is a trivial telescoping series, and the other ones follow from it by induction using \eqref{1}.)

3. $F(s)$ has no zeros in the region $\operatorname{Re}(s)>2$.

   Proof: if $\operatorname{Re}(s)>2$ then it is easily checked that the first term $\frac{1}{2}$ in the series defining $F(s)$ dominates the sum of the remaining terms.

4. Numerically using Mathematica I found some zeros of $F(s)$, at these points: \begin{align*} Z_1 &\approx 0.901294 + 14.11648 i, \\ Z_2 & \approx 0.85788022 + 21.0356764 i, \\ Z_3 & \approx 0.8389893 + 24.982853 i, \\ Z_4 & \approx 0.821207 + 30.4765 i, \\ Z_5 & \approx 0.812678 + 32.8777 i, \\ Z_6 & \approx 0.0987755 + 1.27788 i, \\ Z_7 & \approx -1.47031 + 1.65906 i. \end{align*}

5. The functional equation for $\zeta(s)$ can be reformulated in terms of an auxiliary function related to $F(s)$. More precisely, define \begin{align*} q(s) &= \pi^{-s/2} \Gamma\left(\frac{s}{2}\right), \\ H(s) &= q(s) F(s) - q(1-s) F(2-s). \end{align*} Then $H(s)$ is a meromorphic function. With this notation, the functional equation for $\zeta(s)$ $$q(s)\zeta(s) = q(1-s)\zeta(1-s), \label{2} \tag{2}$$ can be rewritten as $$ q(s) F(s) + q(s) F(s+1) = q(1-s) F(1-s) + q(1-s) F(2-s), $$ or, rearranging terms, $$ q(s) F(s) - q(1-s) F(2-s) = q(1-s) F(1-s) - q(s) F(s+1), $$ This is simply the statement that $$ H(1-s) = H(s). \label{3} \tag{3} $$ So we see that \eqref{2} and \eqref{3} are equivalent.

6. $F(s)$ has the Mellin transform representation $$ F(s) = \frac{1}{\Gamma(s-1)} \int_0^{\infty} \left(- e^x \log(1-e^{-x}) - 1\right) x^{s-2} dx \qquad (\operatorname{Re}(s)>2). $$ (Proof: expand the integrand in a series in powers of $e^{-x}$ and integrate termwise.)

1. If we denote $G(s) = F(2s)$, a trivial coordinate change, then \eqref{1} becomes $\zeta(s)=G(s/2)+G((s+1)/2)$. This can be interpreted as the statement that $\zeta(s)$ is the image of $G(s)$ under the linear operator $T$ that sends an analytic function $g(s)$ to the function $T(g):=g\left(\frac{s}{2}\right) + g\left(\frac{s+1}{2}\right)$.

   This linear operator has some very nice properties. Notably, it is "Riemann-hypothesis preserving", in the following sense: if $g(s)$ is a complex polynomial with all its zeros on the line $\operatorname{Re}(s)=1/2$, then $T(f)$ is also such a polynomial. (This is a reformulation, via a simple coordinate change, of the statement that the operator that maps a polynomial $h(z)$ to $h(z-i/4)+h(z+i/4)$ is "hyperbolicity preserving", that is, sends polynomials with only real zeros to polynomials with only real zeros -- a special case of a more general result that I think is due to Polya).

   This hyperbolicity/RH preservation property also extends to certain entire functions of genus 0 or 1 (usually assumed to satisfy the functional equation $f(1-s)=f(s)$). See section 10.23 of Titschmarsh's book "The Theory of the Riemann Zeta Function", 2nd ed. This was used in some of Polya's failed (though still very interesting) attacks on the Riemann hypothesis. I've seen it used in some other places, for example in the paper "The Riemann hypothesis for certain integrals of Eisenstein series" by Lagarias and Suzuki.

   Now, $\zeta(s)$ is not an entire function, and doesn't exactly satisfy $\zeta(1-s)=\zeta(s)$, so the relevance of these properties of the linear operator $T$ to RH-type questions isn't entirely clear. But the equation $\zeta(s)=F(s)+F(s+1)$ is still suggestive of the possibility that by studying $F(s)$ and its zeros we could learn something about $\zeta(s)$ and its zeros, using these sorts of Polya-style ideas.

2. The same linear operator $T$ (or rather, $\frac{1}{2}T$) is also the transfer operator for the doubling map $x\mapsto 2x \bmod 1$ from ergodic theory.

3. The Bernoulli polynomials $B_n(x)$ (which are themselves related to $\zeta(s)$ in all sorts of ways) are the eigenfunctions of this transfer operator, satisfying the equation $$ T [B_n] = \frac{1}{2^{n-1}} B_n, $$ and they also have the symmetry $B_n(1-x)=B_n(x)$.

1. By rewriting \eqref{1} as $F(s) = \zeta(s) - F(s+1)$, one sees that this gives a way to analytically continue $F(s)$ to the region $\operatorname{Re}(s)>0$, then inductively to the region $\operatorname{Re}(s)>-1$, $\operatorname{Re}(s)>-2$, etc, using the formulas \begin{align*} F(s) &= \zeta(s) - F(s+1) \\ &= \zeta(s) - \zeta(s+1) + F(s+2) \\ &= \zeta(s) - \zeta(s+1) + \zeta(s+2) - F(s+3) = \ldots \\ &= \zeta(s) - \zeta(s+1) + \zeta(s+2) - \zeta(s+3) + \ldots + (-1)^k \zeta(s+k) + (-1)^{k+1} F(s+k+1) \end{align*} This shows that $F(s)$ can be analytically continued to a meromorphic function on $\mathbb{C}$, with poles at $s=1, 0, -1, -2, \ldots$. The pole at $s=n$ for integer $n\le 1$ is a simple pole with residue $(-1)^{n-1}$.

2. $F(s)$ has the special values \begin{align*} F(2) &= 1, \\ F(3) &= -1 + \zeta(2), \\ F(4) &= 1 - \zeta(2) + \zeta(3), \\ &\ \ \vdots \\ F(n) &= (-1)^n + \sum_{k=2}^{n-1} (-1)^{n+k+1} \zeta(k). \end{align*} (Proof: the evaluation of $F(2)$ is a trivial telescoping series, and the other ones follow from it by induction using \eqref{1}.)

3. $F(s)$ has no zeros in the region $\operatorname{Re}(s)>2$.

   Proof: if $\operatorname{Re}(s)>2$ then it is easily checked that the first term $\frac{1}{2}$ in the series defining $F(s)$ dominates the sum of the remaining terms.

4. Numerically using Mathematica I found some zeros of $F(s)$, at these points: \begin{align*} Z_1 &\approx 0.901294 + 14.11648 i, \\ Z_2 & \approx 0.85788022 + 21.0356764 i, \\ Z_3 & \approx 0.8389893 + 24.982853 i, \\ Z_4 & \approx 0.821207 + 30.4765 i, \\ Z_5 & \approx 0.812678 + 32.8777 i, \\ Z_6 & \approx 0.0987755 + 1.27788 i, \\ Z_7 & \approx -1.47031 + 1.65906 i. \end{align*}

5. The functional equation for $\zeta(s)$ can be reformulated in terms of an auxiliary function related to $F(s)$. More precisely, define \begin{align*} q(s) &= \pi^{-s/2} \Gamma\left(\frac{s}{2}\right), \\ H(s) &= q(s) F(s) - q(1-s) F(2-s). \end{align*} Then $H(s)$ is a meromorphic function. With this notation, the functional equation for $\zeta(s)$ $$q(s)\zeta(s) = q(1-s)\zeta(1-s), \label{2} \tag{2}$$ can be rewritten as $$ q(s) F(s) + q(s) F(s+1) = q(1-s) F(1-s) + q(1-s) F(2-s), $$ or, rearranging terms, $$ q(s) F(s) - q(1-s) F(2-s) = q(1-s) F(1-s) - q(s) F(s+1), $$ This is simply the statement that $$ H(1-s) = H(s). \label{3} \tag{3} $$ So we see that \eqref{2} and \eqref{3} are equivalent.

6. $F(s)$ has the Mellin transform representation $$ F(s) = \frac{1}{\Gamma(s-1)} \int_0^{\infty} \left(- e^x \log(1-e^{-x}) - 1\right) x^{s-2} dx \qquad (\operatorname{Re}(s)>2). $$ (Proof: expand the integrand in a series in powers of $e^{-x}$ and integrate termwise.)

Has anyone seen/studied this function before? It has the property that $$\zeta(s) = F(s)+F(s+1), \label{1}\tag{$*$} $$ where $\zeta(s)$ is the Riemann zeta function. This is trivial to check: $$ F(s)+F(s+1) = \sum_{n=1}^\infty \frac{n}{(n+1)n^s} + \sum_{n=1}^\infty \frac{1}{(n+1)n^s} = \sum_{n=1}^\infty \frac{n+1}{n+1} \cdot \frac{1}{n^{s-1}} = \zeta(s). $$ It also has several other properties which make it seem interesting. I'll start by giving some motivation and then describe the properties I noticed, and ask some related questions.

1. By rewriting \eqref{1} as $F(s) = \zeta(s) - F(s+1)$, one sees that this gives a way to analytically continue $F(s)$ to the region $\operatorname{Re}(s)>0$, then inductively to the region $\operatorname{Re}(s)>-1$, $\operatorname{Re}(s)>-2$, etc, using the formulas \begin{align*} F(s) &= \zeta(s) - F(s+1) \\ &= \zeta(s) - \zeta(s+1) + F(s+2) \\ &= \zeta(s) - \zeta(s+1) + \zeta(s+2) - F(s+3) = \ldots \\ &= \zeta(s) - \zeta(s+1) + \zeta(s+2) - \zeta(s+3) + \ldots + (-1)^k \zeta(s+k) + (-1)^{k+1} F(s+k+1) \end{align*} This shows that $F(s)$ can be analytically continued to a meromorphic function on $\mathbb{C}$, with poles at $s=1, 0, -1, -2, \ldots$. The pole at $s=n$ for integer $n\le 1$ is a simple pole with residue $(-1)^n$.

2. $F(s)$ has the special values \begin{align*} F(2) &= 1, \\ F(3) &= -1 + \zeta(2), \\ F(4) &= 1 - \zeta(2) + \zeta(3), \\ &\ \ \vdots \\ F(n) &= (-1)^n + \sum_{k=2}^{n-1} (-1)^{n+k+1} \zeta(k). \end{align*} (Proof: the evaluation of $F(2)$ is a trivial telescoping series, and the other ones follow from it by induction using \eqref{1}.)

3. $F(s)$ has no zeros in the region $\operatorname{Re}(s)>2$.

   Proof: if $\operatorname{Re}(s)>2$ then it is easily checked that the first term $\frac{1}{2}$ in the series defining $F(s)$ dominates the sum of the remaining terms.

4. Numerically using Mathematica I found some zeros of $F(s)$, at these points: \begin{align*} Z_1 &\approx 0.901294 + 14.11648 i, \\ Z_2 & \approx 0.85788022 + 21.0356764 i, \\ Z_3 & \approx 0.8389893 + 24.982853 i, \\ Z_4 & \approx 0.821207 + 30.4765 i, \\ Z_5 & \approx 0.812678 + 32.8777 i, \\ Z_6 & \approx 0.0987755 + 1.27788 i, \\ Z_7 & \approx -1.47031 + 1.65906 i. \end{align*}

5. The functional equation for $\zeta(s)$ can be reformulated in terms of an auxiliary function related to $F(s)$. More precisely, define \begin{align*} q(s) &= \pi^{-s/2} \Gamma\left(\frac{s}{2}\right), \\ H(s) &= q(s) F(s) - q(1-s) F(2-s). \end{align*} Then $H(s)$ is a meromorphic function. With this notation, the functional equation for $\zeta(s)$ $$q(s)\zeta(s) = q(1-s)\zeta(1-s),$$ which can be rewritten as $$ q(s) F(s) + q(s) F(s+1) = q(1-s) F(1-s) + q(1-s) F(2-s), $$ or, rearranging terms, $$ q(s) F(s) - q(1-s) F(2-s) = q(1-s) F(1-s) - q(s) F(s+1), $$ which is simply the statement that $$ H(1-s) = H(s). $$

6. $F(s)$ has the Mellin transform representation $$ F(s) = \frac{1}{\Gamma(s-1)} \int_0^{\infty} \left(- e^x \log(1-e^{-x}) - 1\right) x^{s-2} dx \qquad (\operatorname{Re}(s)>2). $$ (Proof: expand the integrand in a series in powers of $e^{-x}$ and integrate termwise.)

Has anyone seen/studied this function before? It has the property that $$\zeta(s) = F(s)+F(s+1), \label{1} \tag{1} $$ where $\zeta(s)$ is the Riemann zeta function. This is trivial to check: $$ F(s)+F(s+1) = \sum_{n=1}^\infty \frac{n}{(n+1)n^s} + \sum_{n=1}^\infty \frac{1}{(n+1)n^s} = \sum_{n=1}^\infty \frac{n+1}{n+1} \cdot \frac{1}{n^s} = \zeta(s). $$ It also has several other properties which make it seem interesting. I'll start by giving some motivation and then describe the properties I noticed, and ask some related questions.

1. By rewriting \eqref{1} as $F(s) = \zeta(s) - F(s+1)$, one sees that this gives a way to analytically continue $F(s)$ to the region $\operatorname{Re}(s)>0$, then inductively to the region $\operatorname{Re}(s)>-1$, $\operatorname{Re}(s)>-2$, etc, using the formulas \begin{align*} F(s) &= \zeta(s) - F(s+1) \\ &= \zeta(s) - \zeta(s+1) + F(s+2) \\ &= \zeta(s) - \zeta(s+1) + \zeta(s+2) - F(s+3) = \ldots \\ &= \zeta(s) - \zeta(s+1) + \zeta(s+2) - \zeta(s+3) + \ldots + (-1)^k \zeta(s+k) + (-1)^{k+1} F(s+k+1) \end{align*} This shows that $F(s)$ can be analytically continued to a meromorphic function on $\mathbb{C}$, with poles at $s=1, 0, -1, -2, \ldots$. The pole at $s=n$ for integer $n\le 1$ is a simple pole with residue $(-1)^n$.

2. $F(s)$ has the special values \begin{align*} F(2) &= 1, \\ F(3) &= -1 + \zeta(2), \\ F(4) &= 1 - \zeta(2) + \zeta(3), \\ &\ \ \vdots \\ F(n) &= (-1)^n + \sum_{k=2}^{n-1} (-1)^{n+k+1} \zeta(k). \end{align*} (Proof: the evaluation of $F(2)$ is a trivial telescoping series, and the other ones follow from it by induction using \eqref{1}.)

3. $F(s)$ has no zeros in the region $\operatorname{Re}(s)>2$.

   Proof: if $\operatorname{Re}(s)>2$ then it is easily checked that the first term $\frac{1}{2}$ in the series defining $F(s)$ dominates the sum of the remaining terms.

4. Numerically using Mathematica I found some zeros of $F(s)$, at these points: \begin{align*} Z_1 &\approx 0.901294 + 14.11648 i, \\ Z_2 & \approx 0.85788022 + 21.0356764 i, \\ Z_3 & \approx 0.8389893 + 24.982853 i, \\ Z_4 & \approx 0.821207 + 30.4765 i, \\ Z_5 & \approx 0.812678 + 32.8777 i, \\ Z_6 & \approx 0.0987755 + 1.27788 i, \\ Z_7 & \approx -1.47031 + 1.65906 i. \end{align*}

5. The functional equation for $\zeta(s)$ can be reformulated in terms of an auxiliary function related to $F(s)$. More precisely, define \begin{align*} q(s) &= \pi^{-s/2} \Gamma\left(\frac{s}{2}\right), \\ H(s) &= q(s) F(s) - q(1-s) F(2-s). \end{align*} Then $H(s)$ is a meromorphic function. With this notation, the functional equation for $\zeta(s)$ $$q(s)\zeta(s) = q(1-s)\zeta(1-s), \label{2} \tag{2}$$ can be rewritten as $$ q(s) F(s) + q(s) F(s+1) = q(1-s) F(1-s) + q(1-s) F(2-s), $$ or, rearranging terms, $$ q(s) F(s) - q(1-s) F(2-s) = q(1-s) F(1-s) - q(s) F(s+1), $$ This is simply the statement that $$ H(1-s) = H(s). \label{3} \tag{3} $$ So we see that \eqref{2} and \eqref{3} are equivalent.

6. $F(s)$ has the Mellin transform representation $$ F(s) = \frac{1}{\Gamma(s-1)} \int_0^{\infty} \left(- e^x \log(1-e^{-x}) - 1\right) x^{s-2} dx \qquad (\operatorname{Re}(s)>2). $$ (Proof: expand the integrand in a series in powers of $e^{-x}$ and integrate termwise.)

1. By rewriting \eqref{1} as $F(s) = \zeta(s) - F(s+1)$, one sees that this gives a way to analytically continue $F(s)$ to the region $\operatorname{Re}(s)>0$, then inductively to the region $\operatorname{Re}(s)>-1$, $\operatorname{Re}(s)>-2$, etc, using the formulas \begin{align*} F(s) &= \zeta(s) - F(s+1) \\ &= \zeta(s) - \zeta(s+1) + F(s+2) \\ &= \zeta(s) - \zeta(s+1) + \zeta(s+2) - F(s+3) = \ldots \\ &= \zeta(s) - \zeta(s+1) + \zeta(s+2) - \zeta(s+3) + \ldots + (-1)^k \zeta(s+k) + (-1)^{k+1} F(s+k+1) \end{align*} This shows that $F(s)$ can be analytically continued to a meromorphic function on $\mathbb{C}$, with poles at $s=1, 0, -1, -2, \ldots$. The pole at $s=n$ for integer $n\le 1$ is a simple pole with residue $(-1)^n$.

2. $F(s)$ has the special values \begin{align*} F(2) &= 1, \\ F(3) &= -1 + \zeta(2), \\ F(4) &= 1 - \zeta(2) + \zeta(3), \\ &\ \ \vdots \\ F(n) &= (-1)^n + \sum_{k=2}^{n-1} (-1)^{n+k+1} \zeta(k). \end{align*} (Proof: the evaluation of $F(2)$ is a trivial telescoping series, and the other ones follow from it by induction using \eqref{1}.)

3. $F(s)$ has no zeros in the region $\operatorname{Re}(s)>2$.

   Proof: if $\operatorname{Re}(s)>2$ then it is easily checked that the first term $\frac{1}{2}$ in the series defining $F(s)$ dominates the sum of the remaining terms.

4. Numerically using Mathematica I found some zeros of $F(s)$, at these points: \begin{align*} Z_1 &\approx 0.901294 + 14.11648 i, \\ Z_2 & \approx 0.85788022 + 21.0356764 i, \\ Z_3 & \approx 0.8389893 + 24.982853 i, \\ Z_4 & \approx 0.821207 + 30.4765 i, \\ Z_5 & \approx 0.812678 + 32.8777 i, \\ Z_6 & \approx 0.0987755 + 1.27788 i, \\ Z_7 & \approx -1.47031 + 1.65906 i. \end{align*}

5. The functional equation for $\zeta(s)$ can be reformulated in terms of an auxiliary function related to $F(s)$. More precisely, define \begin{align*} q(s) &= \pi^{-s/2} \Gamma\left(\frac{s}{2}\right), \\ H(s) &= q(s) F(s) - q(1-s) F(2-s). \end{align*} Then $H(s)$ is a meromorphic function. With this notation, the functional equation for $\zeta(s)$ $$q(s)\zeta(s) = q(1-s)\zeta(1-s),$$ which can be rewritten as $$ q(s) F(s) + q(s) F(s+1) = q(1-s) F(1-s) + q(1-s) F(2-s), $$ becomes equivalent to the statement that $$ H(1-s) = H(s). $$

6. $F(s)$ has the Mellin transform representation $$ F(s) = \frac{1}{\Gamma(s-1)} \int_0^{\infty} \left(- e^x \log(1-e^{-x}) - 1\right) x^{s-2} dx \qquad (\operatorname{Re}(s)>2). $$ (Proof: expand the integrand in a series in powers of $e^{-x}$ and integrate termwise.)

1. By rewriting \eqref{1} as $F(s) = \zeta(s) - F(s+1)$, one sees that this gives a way to analytically continue $F(s)$ to the region $\operatorname{Re}(s)>0$, then inductively to the region $\operatorname{Re}(s)>-1$, $\operatorname{Re}(s)>-2$, etc, using the formulas \begin{align*} F(s) &= \zeta(s) - F(s+1) \\ &= \zeta(s) - \zeta(s+1) + F(s+2) \\ &= \zeta(s) - \zeta(s+1) + \zeta(s+2) - F(s+3) = \ldots \\ &= \zeta(s) - \zeta(s+1) + \zeta(s+2) - \zeta(s+3) + \ldots + (-1)^k \zeta(s+k) + (-1)^{k+1} F(s+k+1) \end{align*} This shows that $F(s)$ can be analytically continued to a meromorphic function on $\mathbb{C}$, with poles at $s=1, 0, -1, -2, \ldots$. The pole at $s=n$ for integer $n\le 1$ is a simple pole with residue $(-1)^n$.

2. $F(s)$ has the special values \begin{align*} F(2) &= 1, \\ F(3) &= -1 + \zeta(2), \\ F(4) &= 1 - \zeta(2) + \zeta(3), \\ &\ \ \vdots \\ F(n) &= (-1)^n + \sum_{k=2}^{n-1} (-1)^{n+k+1} \zeta(k). \end{align*} (Proof: the evaluation of $F(2)$ is a trivial telescoping series, and the other ones follow from it by induction using \eqref{1}.)

3. $F(s)$ has no zeros in the region $\operatorname{Re}(s)>2$.

   Proof: if $\operatorname{Re}(s)>2$ then it is easily checked that the first term $\frac{1}{2}$ in the series defining $F(s)$ dominates the sum of the remaining terms.

4. Numerically using Mathematica I found some zeros of $F(s)$, at these points: \begin{align*} Z_1 &\approx 0.901294 + 14.11648 i, \\ Z_2 & \approx 0.85788022 + 21.0356764 i, \\ Z_3 & \approx 0.8389893 + 24.982853 i, \\ Z_4 & \approx 0.821207 + 30.4765 i, \\ Z_5 & \approx 0.812678 + 32.8777 i, \\ Z_6 & \approx 0.0987755 + 1.27788 i, \\ Z_7 & \approx -1.47031 + 1.65906 i. \end{align*}

5. The functional equation for $\zeta(s)$ can be reformulated in terms of an auxiliary function related to $F(s)$. More precisely, define \begin{align*} q(s) &= \pi^{-s/2} \Gamma\left(\frac{s}{2}\right), \\ H(s) &= q(s) F(s) - q(1-s) F(2-s). \end{align*} Then $H(s)$ is a meromorphic function. With this notation, the functional equation for $\zeta(s)$ $$q(s)\zeta(s) = q(1-s)\zeta(1-s),$$ which can be rewritten as $$ q(s) F(s) + q(s) F(s+1) = q(1-s) F(1-s) + q(1-s) F(2-s), $$ or, rearranging terms, $$ q(s) F(s) - q(1-s) F(2-s) = q(1-s) F(1-s) - q(s) F(s+1), $$ which is simply the statement that $$ H(1-s) = H(s). $$

6. $F(s)$ has the Mellin transform representation $$ F(s) = \frac{1}{\Gamma(s-1)} \int_0^{\infty} \left(- e^x \log(1-e^{-x}) - 1\right) x^{s-2} dx \qquad (\operatorname{Re}(s)>2). $$ (Proof: expand the integrand in a series in powers of $e^{-x}$ and integrate termwise.)