The explicit formula is as follows: $$ S_a=\frac{1}{a^2}\left(\sum_{z^a=2^a}+2\sum_{p_a(z)=0}\right)e^{az} $$ where the polynomials $p_a$ are given by A244608. For example, $$ p_9(z)= 1 - 13604 z^9 - 13359 z^{18} + 247 z^{27} + z^{36} $$ whose roots looks like follows:

The first few solutions are \begin{align} S_1&=e^{2 x}\\ S_2&=\frac{e^{-2 x}}{4}+\frac{e^{2 x}}{4}\\ S_3&=\frac{2 e^{-x}}{9}+\frac{e^{2 x}}{9}+\frac{2}{9} e^{\frac{x}{2}-\frac{1}{2} i \sqrt{3} x}+\frac{2}{9} e^{\frac{x}{2}+\frac{1}{2} i \sqrt{3} x}+\frac{1}{9} e^{-x-i \sqrt{3} x}+\frac{1}{9} e^{-x+i \sqrt{3} x}\\ S_4&=\frac{e^{-2 x}}{16}+\frac{1}{8} e^{(-1-i) x}+\frac{1}{8} e^{(-1+i) x}+\frac{1}{16} e^{-2 i x}+\frac{1}{16} e^{2 i x}+\frac{1}{8} e^{(1-i) x}+\frac{1}{8} e^{(1+i) x}+\frac{e^{2 x}}{16} \end{align} as given by \begin{align} p_1(z)&=0\\ p_2(z)&=0\\ p_3(z)&=1+z^3\\ p_4(z)&=4+z^4\\ p_5(z)&=-1+11z^5+z^{10}\\ p_6(z)&=-27+26z^6+z^{12} \end{align} etc.

The explicit formula is as follows: $$ S_a=\frac{1}{a^2}\left(\sum_{z^a=2^a}+2\sum_{p_a(z)=0}\right)e^{az} $$ where the polynomials $p_a$ are given by A244608. For example, \begin{align} p_9(z)&= 1 - 13604 z^9 - 13359 z^{18} + 247 z^{27} + z^{36}\\ p_{10}(z)&=3125-383750 z^{10}-73749 z^{20}+502 z^{30}+z^{40} \end{align} whose roots in the complex plane look like follows:

The first few solutions are \begin{align} S_1&=e^{2 x}\\ S_2&=\frac{e^{-2 x}}{4}+\frac{e^{2 x}}{4}\\ S_3&=\frac{2 e^{-x}}{9}+\frac{e^{2 x}}{9}+\frac{2}{9} e^{\frac{x}{2}-\frac{1}{2} i \sqrt{3} x}+\frac{2}{9} e^{\frac{x}{2}+\frac{1}{2} i \sqrt{3} x}+\frac{1}{9} e^{-x-i \sqrt{3} x}+\frac{1}{9} e^{-x+i \sqrt{3} x}\\ S_4&=\frac{e^{-2 x}}{16}+\frac{1}{8} e^{(-1-i) x}+\frac{1}{8} e^{(-1+i) x}+\frac{1}{16} e^{-2 i x}+\frac{1}{16} e^{2 i x}+\frac{1}{8} e^{(1-i) x}+\frac{1}{8} e^{(1+i) x}+\frac{e^{2 x}}{16} \end{align} as given by \begin{align} p_1(z)&=0\\ p_2(z)&=0\\ p_3(z)&=1+z^3\\ p_4(z)&=4+z^4\\ p_5(z)&=-1+11z^5+z^{10}\\ p_6(z)&=-27+26z^6+z^{12} \end{align} etc. Quoting the OEIS entry, the coefficients are found as follows:

Let $\omega$ be a primitive $j$-th root of unity. Let $L(k)=\sum_{p=0}^{j-1} c(p)\omega^{kp}$ with $c(0)=2$ and $c(i)=C(j,i)$ if $i>0$. Then $p(j,X)=(X-L(1))(X-L(2))\dots(X-L([(n-1)/2]))$.

The explicit formula is as follows: $$ S_a=\frac{1}{a^2}\left(\sum_{z^a=2^a}+2\sum_{p_a(z)=0}\right)e^{az} $$ where the polynomials $p_a$ are given by A244608. For example, $$ p_9(z)= 1 - 13604 z^9 - 13359 z^{18} + 247 z^{27} + z^{36} $$ whose roots looks like follows:

The explicit formula is as follows: $$ S_a=\frac{1}{a^2}\left(\sum_{z^a=2^a}+2\sum_{p_a(z)=0}\right)e^{az} $$ where the polynomials $p_a$ are given by A244608. For example, $$ p_9(z)= 1 - 13604 z^9 - 13359 z^{18} + 247 z^{27} + z^{36} $$ whose roots looks like follows:

The first few solutions are \begin{align} S_1&=e^{2 x}\\ S_2&=\frac{e^{-2 x}}{4}+\frac{e^{2 x}}{4}\\ S_3&=\frac{2 e^{-x}}{9}+\frac{e^{2 x}}{9}+\frac{2}{9} e^{\frac{x}{2}-\frac{1}{2} i \sqrt{3} x}+\frac{2}{9} e^{\frac{x}{2}+\frac{1}{2} i \sqrt{3} x}+\frac{1}{9} e^{-x-i \sqrt{3} x}+\frac{1}{9} e^{-x+i \sqrt{3} x}\\ S_4&=\frac{e^{-2 x}}{16}+\frac{1}{8} e^{(-1-i) x}+\frac{1}{8} e^{(-1+i) x}+\frac{1}{16} e^{-2 i x}+\frac{1}{16} e^{2 i x}+\frac{1}{8} e^{(1-i) x}+\frac{1}{8} e^{(1+i) x}+\frac{e^{2 x}}{16} \end{align} as given by \begin{align} p_1(z)&=0\\ p_2(z)&=0\\ p_3(z)&=1+z^3\\ p_4(z)&=4+z^4\\ p_5(z)&=-1+11z^5+z^{10}\\ p_6(z)&=-27+26z^6+z^{12} \end{align} etc.