(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?
Best Answer
For $s_{10C}$, Maple finds this $7$-term recurrence:
{(15625*n^3 + 46875*n^2 + 46875*n + 15625)*u(n) + (11250*n^3 + 61875*n^2 + 115625*n + 73125)*u(n + 1) + (4575*n^3 + 36600*n^2 + 99150*n + 90950)*u(n + 2) + (1116*n^3 + 11718*n^2 + 41434*n + 49322)*u(n + 3) + (183*n^3 + 2379*n^2 + 10371*n + 15157)*u(n + 4) + (18*n^3 + 279*n^2 + 1445*n + 2501)*u(n + 5) + (n^3 + 18*n^2 + 108*n + 216)*u(n + 6), u(0) = 1, u(1) = -1, u(2) = 1, u(3) = -1, u(4) = 1, u(5) = 23}\begin{align} 0 = &\left( 15625\,{n}^{3}+46875\,{n}^{2}+46875\,n+15625 \right) u \left( n \right) \\ &+ \left( 11250\,{n}^{3}+61875\,{n}^{2}+115625\,n+ 73125 \right) u \left( n+1 \right) \\ &+ \left( 4575\,{n}^{3}+36600\,{n}^{ 2}+99150\,n+90950 \right) u \left( n+2 \right) \\ &+ \left( 1116\,{n}^{3}+ 11718\,{n}^{2}+41434\,n+49322 \right) u \left( n+3 \right) \\ &+ \left( 183\,{n}^{3}+2379\,{n}^{2}+10371\,n+15157 \right) u \left( n+4 \right) \\ &+ \left( 18\,{n}^{3}+279\,{n}^{2}+1445\,n+2501 \right) u \left( n+5 \right) \\ &+ \left( {n}^{3}+18\,{n}^{2}+108\,n+216 \right) u \left( n+6 \right) , \\ &u \left( 0 \right) =1,u \left( 1 \right) =-1,u \left( 2 \right) =1,u \left( 3 \right) =-1,u \left( 4 \right) =1,u \left( 5 \right) =23 \end{align}
or \begin{align} 0 = & \,5^6\, \left( k-2 \right) ^{3}u \left( k-3 \right) \\ &+ 5^4 \left( 18\,k^3 - 63\,k^2 + 77\,k -33 \right) u \left( k-2 \right) \\ &+ 5^2 \left( 183\,k^3 - 183\,k^2 + 123\,k - 25 \right) u \left( k-1 \right) \\ &+2\, \left( 2\,k+1 \right) \left( 279\,{k}^{2}+279\,k+175 \right) u \left( k \right) \\ &+ \left( 183\,{k}^{3}+732\,{k}^{2}+1038\,k +514 \right) u \left( k+1 \right) \\ &+ \left( 18\,{k}^{3}+117\,{k}^{2}+ 257\,k+191 \right) u \left( k+2 \right) \\ &+ \left( k+3 \right) ^{3}u \left( k+3 \right) \end{align}