After the McKay–Thompson series of levels $1$, $2$, $3$, $4$ of the Monster were mentioned in this MO post, level $6$ has very interesting relations as well. (Level 10 is in this post.)
I. Level-6 functions
Given the Dedekind eta function $\eta(\tau)$, define,
$$j_{6A} = \left(\sqrt{j_{6B}} + \frac{\color{blue}{-1}}{\sqrt{j_{6B}}}\right)^2 =\left(\sqrt{j_{6C}} + \frac{\color{blue}8}{\sqrt{j_{6C}}}\right)^2 = \left(\sqrt{j_{6D}} + \frac{\color{blue}9}{\sqrt{j_{6D}}}\right)^2-4$$
where,
\begin{align} j_{6B} &= \left(\frac{\eta(2\tau)\,\eta(3\tau)}{\eta(\tau)\,\eta(6\tau)}\right)^{12}\\ j_{6C} &= \left(\frac{\eta(\tau)\,\eta(3\tau)}{\eta(2\tau)\,\eta(6\tau)}\right)^{6}\\ j_{6D} &= \left(\frac{\eta(\tau)\,\eta(2\tau)}{\eta(3\tau)\,\eta(6\tau)}\right)^{4}\\ j_{6E} &= \left(\frac{\eta(2\tau)\,\eta^3(3\tau)}{\eta(\tau)\,\eta^3(6\tau)}\right)^{3}. \end{align}
Then Conway and Norton found (or a version thereof) that,
$$j_{6A}+2j_{6E} = j_{6B}+j_{6C}+j_{6D}+14.$$
They found similar relations for levels $6$, $10$, $12$, $18$, $30$.
II. Sequences
I found we can use the blue numbers of the square and near-square relations above to get four known sequences. Starting with Cooper's sequence $s_6=s_{6A}$, one derives the Apéry numbers, Domb numbers, and Almkvist–Zudilin numbers,
\begin{align} s_{6A}(k) &=\binom{2k}{k}\sum_{m=0}^k \binom{k}{m}^3 =1, 4, 60, 1120, 24220,\ldots\\ s_{6B}(j) &=\sum_{k=0}^j (\color{blue}{-u})^{j-k}\binom{j+k}{j-k}\,s_{6A}(k) =1, 5, 73, 1445, 33001,\ldots\\ s_{6C}(j) &=\sum_{k=0}^j (\color{blue}{-v})^{j-k}\binom{j+k}{j-k}\,s_{6A}(k) =1, -4, 28, -256, 2716,\ldots\\ s_{6D}(n) &=\sum_{j=0}^n\sum_{k=0}^j (\color{blue}{-w})^{n-j}\binom{n+j}{n-j}\binom{j}{k}\binom{2j}{j}\binom{2k}{k}^{-1}s_{6A}(k) \\ &=1, -3, 9, -3, -279, 2997,\ldots \end{align}
where $u = \color{blue}{-1}$, $v = \color{blue}8$, and $w = \color{blue}9$, respectively.
III. Alternative forms
However, the three sequences derived from $s_{6A}$ have simpler alternative closed-forms,
\begin{align} s_{6A}(k) &=\binom{2k}{k}\sum_{j=0}^k \binom{k}{j}^3 =1, 4, 60, 1120, 24220,\ldots\\ s_{6B}(k) &=\sum_{j=0}^k \binom{k}{j}^2\binom{k+j}{j}^2 =1, 5, 73, 1445, 33001,\ldots\\ s_{6C}(k) &= (-1)^k \sum_{j=0}^k \binom{k}{j}^2 \binom{2(k-j)}{k-j} \binom{2j}{j} =1, -4, 28, -256, 2716,\ldots\\ s_{6D}(k) &=\sum_{j=0}^k (-1)^{k-j}\,3^{k-3j}\,\frac{(3j)!}{j!^3} \binom{k}{3j} \binom{k+j}{j} =1, -3, 9, -3, -279, 2997,\ldots. \end{align}
All have a $3$-term recurrence relations. Two significant roles of these sequences are, if within the radius of convergence, then,
$$\sum_{k=0}^\infty s_{6A}(k)\,\frac{1}{\;\big(j_{6A}\big)^{k+1/2}} = \sum_{k=0}^\infty s_{6B}(k)\,\frac{1}{\;\big(j_{6B}\big)^{k+1/2}} = \\ \sum_{k=0}^\infty s_{6C}(k)\,\frac{1}{\;\big(j_{6C}\big)^{k+1/2}} = \sum_{k=0}^\infty s_{6D}(k)\,\frac{1}{\;\big(j_{6D}\big)^{k+1/2}}$$
as well as yielding formulas for $1/\pi$. (See also Ramanujan–Sato series.)
IV. Questions
- How do we transform the closed-forms of Section 2 into the alternative ones in Section 3?
- In general, is there a method to simplify binomial sums, especially to reduce the triple summation $s_{6D}$ into into simpler version in Section 3? (And the level-$10$ sequences do not yet have alternative closed-forms.)
