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Added two more lines MMA code
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Fred Hucht
Fred Hucht
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This is an incomplete answer, the last step is missing (yet).

We can differentiate the OPs equation to get \begin{align}\tag{1}\label{eq:1} (1-z^2) H'(z)+2 z H(z^2)-(z+1) H(z)=0. \end{align}\begin{align}\tag{1}\label{eq:1} (1-z^2) H'(z)-(z+1) H(z)+2 z H(z^2)=0. \end{align} The series expansion of $H(z)$ around $z=0$, \begin{align}\tag{2}\label{eq:2} H(z)=\sum_{n=0}^\infty h_n z^n, \end{align} fulfills the recurrence relation (assuming $H(0)=1$) \begin{align} h_0&=h_1=1\\ h_n&=\frac 1 n\left[ (n-1)h_{n-2}+h_{n-1} + ((-1)^n+1)h_{n/2-1}\right], \tag{3}\label{eq:3} \end{align} such that \begin{align}\tag{4}\label{eq:4} H(z)=1+z+\frac{2 z^3}{3}-\frac{z^4}{3}+\frac{7 z^5}{15}-\frac{z^6}{5}+\frac{13 z^7}{35}-\frac{31 z^8}{105}+\frac{281 z^9}{945}+O\left(z^{10}\right) \end{align} Some Mathematica code:

Clear[hn]; hn[0]=1; hn[1]=1;
hn[n_Integer]:=hn[n] = ((n-1) hn[n-2] + hn[n-1] - If[EvenQ[n], 2 hn[n/2-1], 0])/n
Sum[hn[n]Hs[z_]=Sum[hn[n] z^n, {n, 0, 10}]
(1-z^2) Hs'[z] + 2 z Hs[z^2] - (z+1) Hs[z] + O[z]^10

So we have to show that \begin{align}\tag{5}\label{eq:5} H(1)=\sum_{n=0}^\infty h_n = \frac 1 {1-\ln 2}. \end{align} Numerically, this is the case, the partial sums of \eqref{eq:5} converge with $O(1/n)$.

This is an incomplete answer, the last step is missing (yet).

We can differentiate the OPs equation to get \begin{align}\tag{1}\label{eq:1} (1-z^2) H'(z)+2 z H(z^2)-(z+1) H(z)=0. \end{align} The series expansion of $H(z)$ around $z=0$, \begin{align}\tag{2}\label{eq:2} H(z)=\sum_{n=0}^\infty h_n z^n, \end{align} fulfills the recurrence relation (assuming $H(0)=1$) \begin{align} h_0&=h_1=1\\ h_n&=\frac 1 n\left[ (n-1)h_{n-2}+h_{n-1} + ((-1)^n+1)h_{n/2-1}\right], \tag{3}\label{eq:3} \end{align} such that \begin{align}\tag{4}\label{eq:4} H(z)=1+z+\frac{2 z^3}{3}-\frac{z^4}{3}+\frac{7 z^5}{15}-\frac{z^6}{5}+\frac{13 z^7}{35}-\frac{31 z^8}{105}+\frac{281 z^9}{945}+O\left(z^{10}\right) \end{align} Some Mathematica code:

Clear[hn]; hn[0]=1; hn[1]=1;
hn[n_Integer]:=hn[n] = ((n-1) hn[n-2] + hn[n-1] - If[EvenQ[n], 2 hn[n/2-1], 0])/n
Sum[hn[n] z^n, {n, 0, 10}]

So we have to show that \begin{align}\tag{5}\label{eq:5} H(1)=\sum_{n=0}^\infty h_n = \frac 1 {1-\ln 2}. \end{align} Numerically, this is the case, the partial sums of \eqref{eq:5} converge with $O(1/n)$.

This is an incomplete answer, the last step is missing (yet).

We can differentiate the OPs equation to get \begin{align}\tag{1}\label{eq:1} (1-z^2) H'(z)-(z+1) H(z)+2 z H(z^2)=0. \end{align} The series expansion of $H(z)$ around $z=0$, \begin{align}\tag{2}\label{eq:2} H(z)=\sum_{n=0}^\infty h_n z^n, \end{align} fulfills the recurrence relation (assuming $H(0)=1$) \begin{align} h_0&=h_1=1\\ h_n&=\frac 1 n\left[ (n-1)h_{n-2}+h_{n-1} + ((-1)^n+1)h_{n/2-1}\right], \tag{3}\label{eq:3} \end{align} such that \begin{align}\tag{4}\label{eq:4} H(z)=1+z+\frac{2 z^3}{3}-\frac{z^4}{3}+\frac{7 z^5}{15}-\frac{z^6}{5}+\frac{13 z^7}{35}-\frac{31 z^8}{105}+\frac{281 z^9}{945}+O\left(z^{10}\right) \end{align} Some Mathematica code:

Clear[hn]; hn[0]=1; hn[1]=1;
hn[n_Integer]:=hn[n] = ((n-1) hn[n-2] + hn[n-1] - If[EvenQ[n], 2 hn[n/2-1], 0])/n
Hs[z_]=Sum[hn[n] z^n, {n, 0, 10}]
(1-z^2) Hs'[z] + 2 z Hs[z^2] - (z+1) Hs[z] + O[z]^10

So we have to show that \begin{align}\tag{5}\label{eq:5} H(1)=\sum_{n=0}^\infty h_n = \frac 1 {1-\ln 2}. \end{align} Numerically, this is the case, the partial sums of \eqref{eq:5} converge with $O(1/n)$.

Source Link
Fred Hucht
Fred Hucht
  • 3.7k
  • 1
  • 12
  • 30

This is an incomplete answer, the last step is missing (yet).

We can differentiate the OPs equation to get \begin{align}\tag{1}\label{eq:1} (1-z^2) H'(z)+2 z H(z^2)-(z+1) H(z)=0. \end{align} The series expansion of $H(z)$ around $z=0$, \begin{align}\tag{2}\label{eq:2} H(z)=\sum_{n=0}^\infty h_n z^n, \end{align} fulfills the recurrence relation (assuming $H(0)=1$) \begin{align} h_0&=h_1=1\\ h_n&=\frac 1 n\left[ (n-1)h_{n-2}+h_{n-1} + ((-1)^n+1)h_{n/2-1}\right], \tag{3}\label{eq:3} \end{align} such that \begin{align}\tag{4}\label{eq:4} H(z)=1+z+\frac{2 z^3}{3}-\frac{z^4}{3}+\frac{7 z^5}{15}-\frac{z^6}{5}+\frac{13 z^7}{35}-\frac{31 z^8}{105}+\frac{281 z^9}{945}+O\left(z^{10}\right) \end{align} Some Mathematica code:

Clear[hn]; hn[0]=1; hn[1]=1;
hn[n_Integer]:=hn[n] = ((n-1) hn[n-2] + hn[n-1] - If[EvenQ[n], 2 hn[n/2-1], 0])/n
Sum[hn[n] z^n, {n, 0, 10}]

So we have to show that \begin{align}\tag{5}\label{eq:5} H(1)=\sum_{n=0}^\infty h_n = \frac 1 {1-\ln 2}. \end{align} Numerically, this is the case, the partial sums of \eqref{eq:5} converge with $O(1/n)$.