*(For brevity, the level-6 functions have been migrated to [another post](https://mathoverflow.net/q/448777/12905).)*


**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](https://www.researchgate.net/publication/266859001_Level_10_analogues_of_Ramanujan%27s_series_for_1p)*", 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*](https://en.wikipedia.org/wiki/Ramanujan%E2%80%93Sato_series#Level_10) 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**


1. What is the recurrence relation for $s\_{10C}$?
2. 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?



On level $10$ of the McKay-Thompson series of the Monster
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}


