(For brevity, the level-6 functions have been migrated to another post.)
I. Level-10 functions
Given the Dedekind eta function $\eta(\tau)$. To recall, for level-6,
$$j_{6A} = \left(\sqrt{j_{6B}} + \frac{\color{blue}{-1}}{\sqrt{j_{6B}}}\right)^2 =\left(\sqrt{j_{6C}} + \frac{\color{blue}8}{\sqrt{j_{6C}}}\right)^2 = \left(\sqrt{j_{6D}} + \frac{\color{blue}9}{\sqrt{j_{6D}}}\right)^2-4$$
For level-10,
$$j_{10A} = \left(\sqrt{j_{10D}} + \frac{\color{blue}{-1}}{\sqrt{j_{10D}}}\right)^2 = \left(\sqrt{j_{10B}} + \frac{\color{blue}4}{\sqrt{j_{10B}}}\right)^2 = \left(\sqrt{j_{10C}} + \frac{\color{blue}5}{\sqrt{j_{10C}}}\right)^2-4$$
where,
\begin{align} j_{10B}(\tau) &= \left(\frac{\eta(\tau)\,\eta(5\tau)}{\eta(2\tau)\,\eta(10\tau)}\right)^{4}\qquad \\ j_{10C}(\tau) &= \left(\frac{\eta(\tau)\,\eta(2\tau)}{\eta(5\tau)\,\eta(10\tau)}\right)^{2} \\ j_{10D}(\tau) &= \left(\frac{\eta(2\tau)\,\eta(5\tau)}{\eta(\tau)\,\eta(10\tau)}\right)^{6} \\ j_{10E}(\tau) &= \left(\frac{\eta(2\tau)\,\eta^5(5\tau)}{\eta(\tau)\,\eta^5(10\tau)}\right) \end{align}
Conway and Norton found these moonshine functions obey (or a version thereof),
$$j_{10A}+2j_{10E} = j_{10B}+j_{10C}+j_{10D}+6$$
II. Sequences
Just like for level-6, we can use the relations above to get four sequences. In Cooper's paper, "Level 10 Analogues of Ramanujan's series for 1/Pi", he discussed $s_{10}=s_{10A}$ and two related sequences found by Zudilin (p.10), but not the other three below,
\begin{align} s_{10A}(k) &=\sum_{m=0}^k \binom{k}{m}^4\\ s_{10D}(j) &=\sum_{k=0}^j (-u)^{j-k}\binom{j+k}{j-k}\,s_{10A}(k)\\ s_{10B}(j) &=\sum_{k=0}^j (-v)^{j-k}\binom{j+k}{j-k}\,s_{10A}(k)\\ s_{10C}(n) &=\sum_{j=0}^n\sum_{k=0}^j (-w)^{n-j}\binom{n+j}{n-j}\binom{j}{k}\binom{2j}{j}\binom{2k}{k}^{-1}s_{10A}(k) \end{align}
where $u = \color{blue}{-1}$, $v = \color{blue}4$, $w = \color{blue}5$. Using the variable $h$ for uniformity, the first few terms are,
\begin{align} s_{10A}(h) &=1, 2, 18, 164, 1810, 21252,\ldots\\ s_{10D}(h) &=1, 3, 25, 267, 3249, 42795, 594145,\ldots\\ s_{10B}(h) &=1, -2, 10, -68, 514, -4100, 33940,\ldots\\ s_{10C}(h) &=1, -1, 1, -1, 1, 23, -263, 1343, -2303,\ldots \end{align}
such that all $s_{10}(0) = 1.$ The sequences $(s_{10A}, s_{10D}, s_{10B}, s_{10C})$ have an $m$-term recurrence relation with $m=3,5,5,7$ (with the last one courtesy of G. Edgar's answer below).
III. Pi formulas
A. These four sequences can be used to generate new Ramanujan-Sato formulas for $1/\pi$ of level 10. For example, let $\tau = \sqrt{-19/10}$, then,
\begin{align} j_{10A}(\tau) &= 76^2\\ j_{10D}(\tau) &= (2+\sqrt5)^6\\ j_{10B}(\tau) &= 4(3+\sqrt{10})^4\\ j_{10C}(\tau) &= 5(1+\sqrt{2})^8\qquad \end{align}
to get (the first one is known),
\begin{align} \frac1{\pi} &= \frac{5}{\sqrt{95}}\,\sum_{n=0}^\infty s_{10A}(n)\,\frac{\;408n+47}{(76^2)^{n+1/2}}\\[4pt] \frac1{\pi} &= \frac{2\sqrt{95}}{17\sqrt{5}}\sum_{n=0}^\infty s_{10D}(n)\,\frac{408n+47-\psi_1}{\big((2+\sqrt5)^6\big)^{n+1/2}}\\[4pt] \frac1{\pi} &= \frac{\sqrt{95}}{6\sqrt{10}}\sum_{n=0}^\infty s_{10B}(n)\,\frac{408n+47+\psi_2\;}{\big(4(3+\sqrt{10})^4\big)^{n+1/2}}\\[4pt] \frac1{\pi} &= \frac{1}{\sqrt{95}}\;\sum_{n=0}^\infty s_{10C}(n)\,\frac{An+B+\psi_3}{\;\big(5(1+\sqrt{2})^8\big)^{n+1/2}}\\[4pt] \end{align}
where $\psi_1 = \frac{157}{38(2+\sqrt5)^3},$ and $\psi_2 = \frac{157}{19(3+\sqrt{10})^2}.$ (The fourth to be added later.)
B. Furthermore, if within the radius of convergence, it seems that,
$$\sum_{h=0}^\infty s_{10A}(h)\,\frac{1}{\;\big(j_{10A}\big)^{h+1/2}} = \sum_{h=0}^\infty s_{10B}(h)\,\frac{1}{\;\big(j_{10B}\big)^{h+1/2}} = \\ \sum_{h=0}^\infty s_{10C}(h)\,\frac{1}{\;\big(j_{10C}\big)^{h+1/2}} = \sum_{h=0}^\infty s_{10D}(h)\,\frac{1}{\;\big(j_{10D}\big)^{h+1/2}}\;$$
IV. Questions
- What is the recurrence relation for $s_{10C}$?
- Using the four given sequences of level $10$, is the last relation really true? And do their closed-forms have simpler versions, just like for level-6?
