Ramanujan–Sato series

Examples for levels 1–4 were given by Ramanujan in his 1917 paper. Given q=e^{2\pi i \tau} as in the rest of this article. Let, :\begin{align} j(\tau) &= \left(\frac{E_4(\tau)}{\eta^8(\tau)}\right)^3 = \frac{1}{q} + 744 + 196884q + 21493760q^2 +\cdots\\ j^*(\tau) &= 432\,\frac{\sqrt{j(\tau)}+ \sqrt{j(\tau)-1728}}{\sqrt{j(\tau)}- \sqrt{j(\tau)-1728}} = \frac{1}{q} - 120 + 10260q - 901120q^2 + \cdots \end{align} with the j-function j(τ), Eisenstein series E4, and Dedekind eta function η(τ). The first expansion is the McKay–Thompson series of class 1A () with a(0) = 744. Note that, as first noticed by J. McKay, the coefficient of the linear term of j(τ) almost equals 196883, which is the degree of the smallest nontrivial irreducible representation of the monster group, a relationship called monstrous moonshine. Similar phenomena will be observed in the other levels. Define :s_{1A}(k)=\binom{2k}{k}\binom{3k}{k}\binom{6k}{3k}=1, 120, 83160, 81681600,\ldots () :s_{1B}(k)=\sum_{j=0}^k\binom{2j}{j}\binom{3j}{j}\binom{6j}{3j}\binom{k+j}{k-j}(-432)^{k-j} =1, -312, 114264, -44196288,\ldots Then the two modular functions and sequences are related by :\sum_{k=0}^\infty s_{1A}(k)\,\frac{1}{(j(\tau))^{k+\frac12}}= \pm \sum_{k=0}^\infty s_{1B}(k)\,\frac{1}{(j^*(\tau))^{k+\frac12}} if the series converges and the sign chosen appropriately, though squaring both sides easily removes the ambiguity. Analogous relationships exist for the higher levels. Examples: :\frac{1}{\pi} = 12\,\boldsymbol{i}\,\sum_{k=0}^\infty s_{1A}(k)\,\frac{163\cdot3344418k+13591409}{\left(-640320^3\right)^{k+\frac12}},\quad j\left(\frac{1+\sqrt{-163}}{2}\right)=-640320^3=-262537412640768000 :\frac{1}{\pi} = 24\,\boldsymbol{i}\,\sum_{k=0}^\infty s_{1B}(k)\,\frac{-3669+320\sqrt{645}\,\left(k+\frac12\right)}{\left({-432}\,U_{645}^3\right)^{k+\frac12}},\quad j^*\left(\frac{1+\sqrt{-43}}{2}\right) = -432\,U_{645}^{3}=-432\left(\frac{127+5\sqrt{645}}{2}\right)^{3} where 645=43\times15, and U_n is a fundamental unit. The first belongs to a family of formulas that were rigorously proven by the Chudnovsky brothers in 1989 and later used to calculate 10 trillion digits of π in 2011. The second formula, and the ones for higher levels, were established by H.H. Chan and S. Cooper in 2012. ==Level 2==

Using Zagier's notation ==Level 3==

Define, :\begin{align} j_{3A}(\tau) &=\left(\left(\frac{\eta(\tau)}{\eta(3\tau)}\right)^{6}+3^3 \left(\frac{\eta(3\tau)}{\eta(\tau)}\right)^{6}\right)^2 = \frac{1}{q} + 42 + 783q + 8672q^2 +65367q^3+\cdots\\ j_{3B}(\tau) &= \left(\frac{\eta(\tau)}{\eta(3\tau)}\right)^{12} = \frac{1}{q} - 12 + 54q - 76q^2 - 243q^3 + 1188q^4 + \cdots\\ \end{align} where 782 is the smallest degree greater than 1 of the irreducible representations of the Fischer group Fi23 and, :s_{3A}(k)=\binom{2k}{k}\binom{2k}{k}\binom{3k}{k}=1, 12, 540, 33600, 2425500,\ldots () :s_{3B}(k)=\sum_{j=0}^k\binom{2j}{j}\binom{2j}{j}\binom{3j}{j}\binom{k+j}{k-j}(-27)^{k-j}=1, -15, 297, -6495, 149481,\ldots Examples: :\frac{1}{\pi} = 2\,\boldsymbol{i}\,\sum_{k=0}^\infty s_{3A}(k)\,\frac{267\cdot53k+827}{\left(-300^3\right)^{k+\frac12}},\quad j_{3A}\left(\frac{3+\sqrt{-267}}{6}\right) = -300^3 = -27000000 :\frac{1}{\pi} = \boldsymbol{i}\,\sum_{k=0}^\infty s_{3B}(k)\,\frac{12497-3000\sqrt{89}\, \left(k+\frac12\right)}{\left(-27\,U_{89}^{2}\right)^{k+\frac12}},\quad j_{3B}\left(\frac{3+\sqrt{-267}}{6}\right)=-27\,\left(500+53\sqrt{89}\right)^2=-27\,U_{89}^{2} ==Level 4==

Define, :\begin{align} j_{4A}(\tau)&=\left(\left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^{4}+4^2 \left(\frac{\eta(4\tau)}{\eta(\tau)}\right)^{4}\right)^2 = \left(\frac{\eta^2(2\tau)}{\eta(\tau)\,\eta(4\tau)} \right)^{24} =-\left(\frac{\eta\left(\frac{2\tau+3}{2}\right)}{\eta(2\tau+3)} \right)^{24} = \frac{1}{q} + 24+ 276q + 2048q^2 +11202q^3+\cdots\\ j_{4C}(\tau) &= \left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^{8} = \frac{1}{q} -8 + 2 q - 62q^3 + 216q^5 - 641q^7 + \ldots\\ \end{align} where the first is the 24th power of the Weber modular function \mathfrak{f}(2\tau). And, :s_{4A}(k)=\binom{2k}{k}^3=1, 8, 216, 8000, 343000,\ldots () :s_{4C}(k)=\sum_{j=0}^k\binom{2j}{j}^3\binom{k+j}{k-j}(-16)^{k-j}= (-1)^k \sum_{j=0}^k\binom{2j}{j}^2\binom{2k-2j}{k-j}^2 =1, -8, 88, -1088, 14296,\ldots () Examples: :\frac{1}{\pi} = 8\,\boldsymbol{i}\,\sum_{k=0}^\infty s_{4A}(k)\,\frac{6k+1}{\left(-2^9\right)^{k+\frac12}},\quad j_{4A}\left(\frac{1+\sqrt{-4}}{2}\right)=-2^9=-512 :\frac{1}{\pi} = 16\,\boldsymbol{i}\,\sum_{k=0}^\infty s_{4C}(k)\,\frac{1-2\sqrt{2}\, \left(k+\frac12\right)}{\left(-16\,U_{2}^{4}\right)^{k+\frac12}},\quad j_{4C}\left(\frac{1+\sqrt{-4}}{2}\right) = -16\,\left(1+\sqrt{2}\right)^4=-16\,U_{2}^{4} ==Level 5==

Define, :\begin{align} j_{5A}(\tau)&=\left(\frac{\eta(\tau)}{\eta(5\tau)}\right)^{6}+5^3 \left(\frac{\eta(5\tau)}{\eta(\tau)}\right)^{6}+22 =\frac{1}{q} + 16 + 134q + 760q^2 +3345q^3+\cdots\\ j_{5B}(\tau)&=\left(\frac{\eta(\tau)}{\eta(5\tau)}\right)^{6}= \frac{1}{q}- 6 + 9q + 10q^2 - 30q^3 + 6q^4 + \cdots \end{align} and, :s_{5A}(k)=\binom{2k}{k}\sum_{j=0}^k \binom{k}{j}^2\binom{k+j}{j} =1, 6, 114, 2940, 87570,\ldots :s_{5B}(k)=\sum_{j=0}^k(-1)^{j+k}\binom{k}{j}^3\binom{4k-5j}{3k}=1, -5, 35, -275, 2275, -19255,\ldots () where the first is the product of the central binomial coefficients and the Apéry numbers () Examples: :\frac{1}{\pi} = \frac{5}{9}\,\boldsymbol{i}\,\sum_{k=0}^\infty s_{5A}(k)\,\frac{682k+71}{(-15228)^{k+\frac12}},\quad j_{5A}\left(\frac{5+\sqrt{-5(47)}}{10}\right)=-15228=-(18\sqrt{47})^2 :\frac{1}{\pi} = \frac{6}{\sqrt{5}}\,\boldsymbol{i}\,\sum_{k=0}^\infty s_{5B}(k)\,\frac{25\sqrt{5}-141\left(k+\frac12\right)}{\left(-5\sqrt{5}\,U_{5}^{15}\right)^{k+\frac12}},\quad j_{5B}\left(\frac{5+\sqrt{-5(47)}}{10}\right)=-5\sqrt{5}\,\left(\frac{1+\sqrt{5}}{2}\right)^{15}=-5\sqrt{5}\,U_{5}^{15} ==Level 6==

Modular functions In 2002, Takeshi Sato one of which was, :T_{6A}-T_{6B}-T_{6C}-T_{6D}+2T_{6E} = 0 or using the above eta quotients jn, :j_{6A}-j_{6B}-j_{6C}-j_{6D}+2j_{6E} = 22 A similar relation exists for level 10. α Sequences For the modular function '''j6A', one can associate it with three different sequences. (A similar situation happens for the level 10 function 'j10A'''.) Let, :\alpha_1(k)=\binom{2k}{k}\sum_{j=0}^k \binom{k}{j}^3 =1, 4, 60, 1120, 24220,\ldots (, labeled as s6 in Cooper's paper) :\alpha_2(k)=\binom{2k}{k}\sum_{j=0}^k \binom{k}{j}\sum_{m=0}^j\binom{j}{m}^3=\binom{2k}{k}\sum_{j=0}^k \binom{k}{j}^2\binom{2j}{j} =1, 6, 90, 1860, 44730,\ldots () :\alpha_3(k)=\binom{2k}{k}\sum_{j=0}^k \binom{k}{j}(-8)^{k-j}\sum_{m=0}^j\binom{j}{m}^3 =1, -12, 252, -6240, 167580, -4726512,\ldots The three sequences involve the product of the central binomial coefficients c(k)=\tbinom{2k}{k} with: first, the Franel numbers \textstyle\sum_{j=0}^k \tbinom{k}{j}^3; second, , and third, (-1)^k . Note that the second sequence, α2(k) is also the number of 2n-step polygons on a cubic lattice. Their complements, :\alpha'_2(k)=\binom{2k}{k}\sum_{j=0}^k \binom{k}{j}(-1)^{k-j}\sum_{m=0}^j\binom{j}{m}^3 =1, 2, 42, 620, 12250,\ldots :\alpha'_3(k)=\binom{2k}{k}\sum_{j=0}^k \binom{k}{j}(8)^{k-j}\sum_{m=0}^j\binom{j}{m}^3 =1, 20, 636, 23840, 991900,\ldots There are also associated sequences, namely the Apéry numbers, :s_{6B}(k)=\sum_{j=0}^k \binom{k}{j}^2\binom{k+j}{j}^2 =1, 5, 73, 1445, 33001,\ldots () the Domb numbers (unsigned) or the number of 2n-step polygons on a diamond lattice, :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 () and the Almkvist-Zudilin numbers, :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 () where :\frac{(3j)!}{j!^3}=\binom{2j}{j}\binom{3j}{j} Identities The modular functions can be related as, : P = \sum_{k=0}^\infty \alpha_1(k)\,\frac{1}{\left(j_{6A}(\tau)\right)^{k+\frac12}} = \sum_{k=0}^\infty \alpha_2(k)\,\frac{1}{\left(j_{6A}(\tau)+4\right)^{k+\frac12}} = \sum_{k=0}^\infty \alpha_3(k)\,\frac{1}{\left(j_{6A}(\tau)-32\right)^{k+\frac12}} : Q = \sum_{k=0}^\infty s_{6B}(k)\,\frac{1}{\left(j_{6B}(\tau)\right)^{k+\frac12}}= \sum_{k=0}^\infty s_{6C}(k)\,\frac{1}{\left(j_{6C}(\tau)\right)^{k+\frac12}}= \sum_{k=0}^\infty s_{6D}(k)\,\frac{1}{\left(j_{6D}(\tau)\right)^{k+\frac12}} if the series converges and the sign chosen appropriately. It can also be observed that, :P = Q = \sum_{k=0}^\infty \alpha'_2(k)\,\frac{1}{\left(j_{6A}(\tau)-4\right)^{k+\frac12}} = \sum_{k=0}^\infty \alpha'_3(k)\,\frac{1}{\left(j_{6A}(\tau)+32\right)^{k+\frac12}} which implies, :\sum_{k=0}^\infty \alpha_2(k)\,\frac{1}{\left(j_{6A}(\tau)+4\right)^{k+\frac12}} = \sum_{k=0}^\infty \alpha'_2(k)\,\frac{1}{\left(j_{6A}(\tau)-4\right)^{k+\frac12}} and similarly using α3 and α'3. Examples One can use a value for '''j6A''' in three ways. For example, starting with, :\Delta=j_{6A}\left(\sqrt{\frac{-17}{6}}\right)=198^2-4=\left(140\sqrt{2}\right)^2=39200 and noting that 3\cdot17=51 then, :\begin{align} \frac{1}{\pi} &= \frac{24\sqrt{3}}{35}\,\sum_{k=0}^\infty \alpha_1(k)\,\frac{51\cdot11k+53}{(\Delta)^{k+\frac12}}\\ \frac{1}{\pi} &= \frac{4\sqrt{3}}{99}\,\sum_{k=0}^\infty \alpha_2(k)\,\frac{17\cdot560k+899}{(\Delta+4)^{k+\frac12}}\\ \frac{1}{\pi} &= \frac{\sqrt{3}}{2}\,\sum_{k=0}^\infty \alpha_3(k)\,\frac{770k+73}{(\Delta-32)^{k+\frac12}}\\ \end{align} as well as, :\begin{align} \frac{1}{\pi} &= \frac{12\sqrt{3}}{9799}\,\sum_{k=0}^\infty \alpha'_2(k)\,\frac{11\cdot51\cdot560k+29693}{(\Delta-4)^{k+\frac12}}\\ \frac{1}{\pi} &= \frac{6\sqrt{3}}{613}\,\sum_{k=0}^\infty \alpha'_3(k)\,\frac{51\cdot770k+3697}{(\Delta+32)^{k+\frac12}}\\ \end{align} though the formulas using the complements apparently do not yet have a rigorous proof. For the other modular functions, :\frac{1}{\pi} = 8\sqrt{15}\,\sum_{k=0}^\infty s_{6B}(k)\,\left(\frac12-\frac{3\sqrt{5}}{20}+k\right)\left(\frac{1}{\phi^{12}}\right)^{k+\frac12}, \quad j_{6B}\left(\sqrt{\frac{-5}{6}}\right)=\left(\frac{1+\sqrt{5}}{2}\right)^{12}=\phi^{12} :\frac{1}{\pi} = \frac12\,\sum_{k=0}^\infty s_{6C}(k)\,\frac{3k+1}{32^k}, \quad j_{6C}\left(\sqrt{\frac{-1}{3}}\right)=32 :\frac{1}{\pi} = 2\sqrt{3}\,\sum_{k=0}^\infty s_{6D}(k)\,\frac{4k+1}{81^{k+\frac12}}, \quad j_{6D}\left(\sqrt{\frac{-1}{2}}\right)=81 ==Level 7==

Define :s_{7A}(k)=\sum_{j=0}^k \binom{k}{j}^2\binom{2j}{k}\binom{k+j}{j} =1, 4, 48, 760, 13840,\ldots () and, :\begin{align} j_{7A}(\tau) &=\left(\left(\frac{\eta(\tau)}{\eta(7\tau)}\right)^{2}+7 \left(\frac{\eta(7\tau)}{\eta(\tau)}\right)^{2}\right)^2=\frac{1}{q} +10 + 51q + 204q^2 +681q^3+\cdots\\ j_{7B}(\tau)&=\left(\frac{\eta(\tau)}{\eta(7\tau)}\right)^{4}= \frac{1}{q}- 4 + 2q + 8q^2 - 5q^3 - 4q^4 - 10q^5 + \cdots \end{align} Example: :\frac{1}{\pi} = \frac{\sqrt{7}}{22^3}\,\sum_{k=0}^\infty s_{7A}(k)\, \frac{11895k+1286}{\left(-22^3\right)^{k}}, \quad j_{7A}\left(\frac{7+\sqrt{-427}}{14}\right) = -22^3+1 = -\left(39\sqrt{7}\right)^2=-10647 No pi formula has yet been found using j7B. ==Level 8==

Modular functions Levels 2, 4, 8 are related since they are just powers of the same prime. Define, :\begin{align} j_{4B}(\tau) &=\sqrt{j_{2A}(2\tau)} = \left( \sqrt{j_{4D}(\tau)} + \frac{8}{\sqrt{j_{4D}(\tau)}} \right)^2 -16 = \left( \sqrt{j_{8A}(\tau)} - \frac{4}{\sqrt{j_{8A}(\tau)}} \right)^2 = \left( \sqrt{j_{8A'}(\tau)} + \frac{4}{\sqrt{j_{8A'}(\tau)}} \right)^2\\ &=\left(\frac{\eta(2\tau)}{\eta(4\tau)}\right)^{12}+2^6 \left(\frac{\eta(4\tau)}{\eta(2\tau)}\right)^{12} = \frac{1}{q} + 52q + 834q^3 + 4760q^5 + 24703q^7+\cdots\\ j_{4D}(\tau)&= \left(\frac{\eta(2\tau)}{\eta(4\tau)}\right)^{12} =\frac{1}{q} - 12q + 66q^3 - 232q^5 + 639q^7 - 1596q^9 + \cdots\\ j_{8A}(\tau)&=\left(\frac{\eta(2\tau)\,\eta(4\tau)}{\eta(\tau)\,\eta(8\tau)}\right)^{8}=\frac{1}{q} + 8 + 36q + 128q^2 + 386q^3 +1024q^4+\cdots\\ j_{8A'}(\tau)&=\left(\frac{\eta(\tau)\,\eta^2(4\tau)}{\eta^2(2\tau)\,\eta(8\tau)}\right)^{8}=\frac{1}{q} - 8 + 36q - 128q^2 + 386q^3 -1024q^4+\cdots\\ j_{8B}(\tau)&=\left(\frac{\eta^2(4\tau)}{\eta(2\tau)\,\eta(8\tau)}\right)^{12}=\sqrt{j_{4A}(2\tau)}=\frac{1}{q} + 12q + 66q^3 + 232q^5 + 639q^7+\cdots\\ j_{8E}(\tau)&=\left(\frac{\eta^3(4\tau)}{\eta(2\tau)\,\eta^2(8\tau)}\right)^{4} =\frac{1}{q} + 4q + 2q^3 - 8q^5 - q^7 + 20q^9 - 2q^{11} - 40q^{13} +\cdots \end{align} Just like for level 6, five of these functions have a linear relationship, :j_{4B}-j_{4D}-j_{8A}-j_{8A'}+2j_{8E} = 0 But this is not one of the nine Conway-Norton-Atkin linear dependencies since j_{8A'} is not a moonshine function. However, it is related to one as, :j_{8A'}(\tau) = -j_{8A}\Big(\tau+\tfrac12\Big) Sequences :s_{4B}(k)=\binom{2k}{k}\sum_{j=0}^k 4^{k-2j}\binom{k}{2j}\binom{2j}{j}^2 =\binom{2k}{k}\sum_{j=0}^k \binom{k}{j}\binom{2k-2j}{k-j}\binom{2j}{j}=1, 8, 120, 2240, 47320,\ldots :s_{4D}(k)=\binom{2k}{k}^3=1, 8, 216, 8000, 343000,\ldots :s_{8A}(k)=\sum_{j=0}^k \binom{k}{j}^2\binom{2j}{k}^2 =1, 4, 40, 544, 8536,\ldots () :s_{8B}(k)=\sum_{j=0}^k \binom{2j}{j}^3\binom{2k-4j}{k-2j} =1, 2, 14, 36, 334,\ldots where the first is the product of the central binomial coefficient and a sequence related to an arithmetic-geometric mean (). Identities The modular functions can be related as, : \pm\sum_{k=0}^\infty s_{4B}(k)\,\frac{1}{\left(j_{4B}(\tau)+16\right)^{k+\frac12}} = \sum_{k=0}^\infty s_{4D}(k)\,\frac{1}{\left(j_{4D}(\tau)\right)^{2k+\frac12}} = \sum_{k=0}^\infty s_{8A}(k)\,\frac{1}{\left(j_{8A}(\tau)\right)^{k+\frac12}} = \sum_{k=0}^\infty (-1)^k s_{8A}(k)\,\frac{1}{\left(j_{8A'}(\tau)\right)^{k+\frac12}} if the series converges and signs chosen appropriately. Note also the different exponent of \left(j_{4D}(\tau)\right)^{2k+\frac12} from the others. Examples Recall that j_{2A}\left(\tfrac\sqrt{-58}{2}\right)=396^4, while j_{4B}\left(\tfrac\sqrt{-58}{4}\right)=396^2. Hence, :\frac{1}{\pi} = \frac{2\sqrt{2}}{13}\,\sum_{k=0}^\infty s_{4B}(k)\,\frac{70\cdot99\,k+579}{\left(396^2+16\right)^{k+\frac12}},\qquad j_{4B}\left(\frac\sqrt{-58}{4}\right)=396^2 :\frac{1}{\pi} = 2\sqrt{2}\,\sum_{k=0}^\infty s_{8A}(k)\,\frac{-222+70\sqrt{58}\,\left(k+\frac12\right)}{\left(4\left(99+13\sqrt{58}\right)^{2}\right)^{k+\frac12}},\qquad j_{8A}\left(\frac\sqrt{-58}{4}\right)=4\left(99+13\sqrt{58}\right)^{2}=4U_{58}^2 :\frac{1}{\pi} = 2\,\sum_{k=0}^\infty (-1)^k s_{8A}(k)\,\frac{-222\sqrt{2}+13\times58\,\left(k+\frac12\right)}{\left(4\left(1+\sqrt{2}\right)^{12}\right)^{k+\frac12}},\qquad j_{8A'}\left(\frac\sqrt{-58}{4}\right)=4\left(1+\sqrt{2}\right)^{12}=4U_{2}^{12}, For another level 8 example, :\frac{1}{\pi} = \frac1{16}\sqrt{\frac35}\,\sum_{k=0}^\infty s_{8B}(k)\,\frac{210k+43}{(64)^{k+\frac12}},\qquad j_{8B}\left(\frac\sqrt{-7}{4}\right)=2^6=64 ==Level 9==

Define, :\begin{align} j_{3C}(\tau) &= \left(j(3\tau)\right)^\frac13 =-6+\left(\frac{\eta^2(3\tau)}{\eta(\tau)\,\eta(9\tau)}\right)^6 -27 \left(\frac{\eta(\tau)\,\eta(9\tau)}{\eta^2(3\tau)}\right)^6=\frac{1}{q} + 248q^2 + 4124q^5 +34752q^8+\cdots\\ j_{9A}(\tau) &= \left(\frac{\eta^2(3\tau)}{\eta(\tau)\,\eta(9\tau)}\right)^6 = \frac{1}{q} + 6 + 27q + 86q^2 + 243q^3 + 594q^4+\cdots\\ \end{align} The expansion of the first is the McKay–Thompson series of class 3C (and related to the cube root of the j-function), while the second is that of class 9A. Let, :s_{3C}(k)=\binom{2k}{k}\sum_{j=0}^k (-3)^{k-3j}\binom{k}{j}\binom{k-j}{j}\binom{k-2j}{j} =\binom{2k}{k}\sum_{j=0}^k(-3)^{k-3j}\binom{k}{3j}\binom{2j}{j}\binom{3j}{j} = 1, -6, 54, -420, 630,\ldots :s_{9A}(k)=\sum_{j=0}^k\binom{k}{j}^2\sum_{m=0}^j\binom{k}{m}\binom{j}{m}\binom{j+m}{k} =1, 3, 27, 309, 4059,\ldots where the first is the product of the central binomial coefficients and (though with different signs). Examples: :\frac{1}{\pi} = \frac{-\boldsymbol{i}}{9}\sum_{k=0}^\infty s_{3C}(k)\,\frac{602k+85}{\left(-960-12\right)^{k+\frac12}},\quad j_{3C}\left(\frac{3+\sqrt{-43}}{6}\right)=-960 :\frac{1}{\pi} = 6\,\boldsymbol{i}\,\sum_{k=0}^\infty s_{9A}(k)\,\frac{4-\sqrt{129}\,\left(k+\frac12\right)}{\left( -3\sqrt{3U_{129}}\right)^{k+\frac12}},\quad j_{9A}\left(\frac{3+\sqrt{-43}}{6}\right)=-3\sqrt{3}\left(53\sqrt{3}+14\sqrt{43}\right) = -3\sqrt{3U_{129}} ==Level 10==

Modular functions Define, :\begin{align}j_{10A}(\tau) &=\left(\sqrt{j_{10D}(\tau)} - \frac{1}{\sqrt{j_{10D}(\tau)}}\right)^2 = \left(\sqrt{j_{6B}(\tau)} + \frac{4}{\sqrt{j_{10B}(\tau)}}\right)^2 = \left(\sqrt{j_{10C}(\tau)} + \frac{5}{\sqrt{j_{10C}(\tau)}}\right)^2-4 =\frac{1}{q} + 4 + 22q + 56q^2 +\cdots \end{align} :\begin{align}j_{10B}(\tau) &= \left(\frac{\eta(\tau)\eta(5\tau)}{\eta(2\tau)\eta(10\tau)}\right)^{4}=\frac{1}{q} - 4 + 6q - 8q^2 + 17q^3 - 32q^4 +\cdots \end{align} :\begin{align}j_{10C}(\tau) &= \left(\frac{\eta(\tau)\eta(2\tau)}{\eta(5\tau)\eta(10\tau)}\right)^{2}=\frac{1}{q} - 2 - 3q + 6q^2 + 2q^3 + 2q^4+\cdots\end{align} :\begin{align}j_{10D}(\tau) &= \left(\frac{\eta(2\tau)\eta(5\tau)}{\eta(\tau)\eta(10\tau)}\right)^{6}=\frac{1}{q} + 6 + 21q + 62q^2 + 162q^3 +\cdots \end{align} :\begin{align}j_{10E}(\tau) &= \left(\frac{\eta(2\tau)\eta^5(5\tau)}{\eta(\tau)\eta^5(10\tau)}\right)=\frac{1}{q} + 1 + q + 2q^2 + 2q^3 - 2q^4 +\cdots\end{align} Just like j_{6A}, the function j_{10A} is a square or a near-square of the others. Furthermore, there are also linear relations between these, :T_{10A}-T_{10B}-T_{10C}-T_{10D}+2T_{10E} = 0 or using the above eta quotients jn, :j_{10A}-j_{10B}-j_{10C}-j_{10D}+2j_{10E} = 6 β sequences Let, :\beta_{1}(k)=\sum_{j=0}^k \binom{k}{j}^4 =1, 2, 18, 164, 1810,\ldots (, labeled as s10 in Cooper's paper) :\beta_{2}(k)=\binom{2k}{k}\sum_{j=0}^k \binom{2j}{j}^{-1}\binom{k}{j}\sum_{m=0}^j \binom{j}{m}^4 =1, 4, 36, 424, 5716,\ldots :\beta_{3}(k)=\binom{2k}{k}\sum_{j=0}^k \binom{2j}{j}^{-1}\binom{k}{j} (-4)^{k-j}\sum_{m=0}^j \binom{j}{m}^4 =1, -6, 66, -876, 12786,\ldots their complements, :\beta_{2}'(k)=\binom{2k}{k}\sum_{j=0}^k \binom{2j}{j}^{-1}\binom{k}{j} (-1)^{k-j}\sum_{m=0}^j \binom{j}{m}^4 =1, 0, 12, 24, 564, 2784,\ldots :\beta_{3}'(k)=\binom{2k}{k}\sum_{j=0}^k \binom{2j}{j}^{-1}\binom{k}{j} (4)^{k-j}\sum_{m=0}^j \binom{j}{m}^4 =1, 10, 162, 3124, 66994,\ldots and, :s_{10B}(k)=1, -2, 10, -68, 514, -4100, 33940,\ldots :s_{10C}(k)=1, -1, 1, -1, 1, 23, -263, 1343, -2303,\ldots :s_{10D}(k)=1, 3, 25, 267, 3249, 42795, 594145,\ldots though closed forms are not yet known for the last three sequences. Identities The modular functions can be related as, :U = \sum_{k=0}^\infty \beta_1(k)\,\frac{1}{\left(j_{10A}(\tau)\right)^{k+\frac12}} = \sum_{k=0}^\infty \beta_2(k)\,\frac{1}{\left(j_{10A}(\tau)+4\right)^{k+\frac12}} = \sum_{k=0}^\infty \beta_3(k)\,\frac{1}{\left(j_{10A}(\tau)-16\right)^{k+\frac12}} :V = \sum_{k=0}^\infty s_{10B}(k)\,\frac{1}{\left(j_{10B}(\tau)\right)^{k+\frac12}} = \sum_{k=0}^\infty s_{10C}(k)\,\frac{1}{\left(j_{10C}(\tau)\right)^{k+\frac12}} = \sum_{k=0}^\infty s_{10D}(k)\,\frac{1}{\left(j_{10D}(\tau)\right)^{k+\frac12}} if the series converges. In fact, it can also be observed that, :U = V =\sum_{k=0}^\infty \beta_2'(k)\,\frac{1}{\left(j_{10A}(\tau)-4\right)^{k+\frac12}} = \sum_{k=0}^\infty \beta_3'(k)\,\frac{1}{\left(j_{10A}(\tau)+16\right)^{k+\frac12}} Since the exponent has a fractional part, the sign of the square root must be chosen appropriately though it is less an issue when jn is positive. Examples Just like level 6, the level 10 function '''j10A''' can be used in three ways. Starting with, :j_{10A}\left(\sqrt{\frac{-19}{10}}\right) = 76^2 = 5776 and noting that 5\cdot19=95 then, :\begin{align} \frac{1}{\pi} &= \frac{5}{\sqrt{95}}\,\sum_{k=0}^\infty \beta_1(k)\,\frac{408k+47}{\left(76^2\right)^{k+\frac12}}\\ \frac{1}{\pi} &= \frac{1}{17\sqrt{95}}\,\sum_{k=0}^\infty \beta_2(k)\,\frac{19\cdot 1824k+3983}{\left(76^2+4\right)^{k+\frac12}}\\ \frac{1}{\pi} &= \frac{1}{6\sqrt{95}}\,\,\sum_{k=0}^\infty \beta_3(k)\,\,\frac{19\cdot 646k+1427}{\left(76^2-16\right)^{k+\frac12}}\\ \end{align} as well as, :\begin{align} \frac{1}{\pi} &= \frac{5}{481\sqrt{95}}\,\sum_{k=0}^\infty \beta_2'(k)\,\frac{19\cdot 10336k+22675}{\left(76^2-4\right)^{k+\frac12}}\\ \frac{1}{\pi} &= \frac{5}{181\sqrt{95}}\,\sum_{k=0}^\infty \beta_3'(k)\,\frac{19\cdot 3876k+8405}{\left(76^2+16\right)^{k+\frac12}} \end{align} though the ones using the complements do not yet have a rigorous proof. A conjectured formula using one of the last three sequences is, :\frac{1}{\pi} = \frac{\boldsymbol{i}}{\sqrt{5}}\,\sum_{k=0}^\infty s_{10C}(k)\frac{10k+3}{\left(-5^2\right)^{k+\frac12}},\quad j_{10C}\left(\frac{1+\,\boldsymbol{i}}{2}\right) = -5^2 which implies there might be examples for all sequences of level 10. ==Level 11==

Define the McKay–Thompson series of class 11A, :j_{11A}(\tau)= (1+3F)^3+\left(\frac{1}{\sqrt{F}}+3\sqrt{F}\right)^2=\frac{1}{q} + 6 + 17q + 46q^2 + 116q^3 +\cdots or sequence () and where, :F = \frac{\eta(3\tau)\,\eta(33\tau)}{\eta(\tau)\,\eta(11\tau)} and, :s_{11A}(k) = 1, 4, 28, 268, 3004, 36784, 476476,\ldots () No closed form in terms of binomial coefficients is yet known for the sequence but it obeys the recurrence relation, :(k + 1)^3 s_{k + 1} = 2(2k + 1)\left(5k^2 + 5k + 2\right)s_k - 8k\left(7k^2 + 1\right)s_{k - 1} + 22k(k - 1)(2k - 1)s_{k - 2} with initial conditions s(0) = 1, s(1) = 4. Example: :\frac{1}{\pi}=\frac{\boldsymbol{i}}{22}\sum_{k=0}^\infty s_{11A}(k)\,\frac{221k+67}{(-44)^{k+\frac12}},\quad j_{11A}\left(\frac{1+\sqrt\frac{-17}{11}}{2}\right)=-44 ==Higher levels==

As pointed out by Cooper, there are analogous sequences for certain higher levels. ==Similar series==

R. Steiner found examples using Catalan numbers C_k , :\frac{1}{\pi} = \sum_{k=0}^\infty \left(2C_{k-n}\right)^2 \frac{(4z)k+\left(4^{2n-3}-(4n-3)z\right)}{16^k}\qquad z \in \Z,\quad n\ge2,\quad n \in \N and for this a modular form with a second periodic for k exists: :k=\frac{(-20-12\boldsymbol{i})+16n}{16},\qquad k=\frac{(-20+12\boldsymbol{i})+16n}{16} Other similar series are :\frac{1}{\pi} = \sum_{k=0}^\infty \left(2C_{k-2}\right)^2 \frac{3k+\frac14}{16^k} :\frac{1}{\pi} = \sum_{k=0}^\infty \left(2C_{k-1}\right)^2 \frac{(4z+1)k-z}{16^k} \qquad z \in \Z :\frac{1}{\pi} = \sum_{k=0}^\infty \left(2C_{k-1}\right)^2 \frac{-1k+\frac12}{16^k} :\frac{1}{\pi} = \sum_{k=0}^\infty \left(2C_{k-1}\right)^2 \frac{0k+\frac14}{16^k} :\frac{1}{\pi} = \sum_{k=0}^\infty \left(2C_{k-1}\right)^2 \frac{{\frac{k}{5}+\frac15}}{16^k} :\frac{1}{\pi} = \sum_{k=0}^\infty \left(2C_{k-1}\right)^2 \frac{{\frac{k}{3}+\frac16}}{16^k} :\frac{1}{\pi} = \sum_{k=0}^\infty \left(2C_{k-1}\right)^2 \frac{{\frac{k}{2}+\frac18}}{16^k} :\frac{1}{\pi} = \sum_{k=0}^\infty \left(2C_{k-1}\right)^2 \frac{2k-\frac14}{16^k} :\frac{1}{\pi} = \sum_{k=0}^\infty \left(2C_{k-1}\right)^2 \frac{3k-\frac12}{16^k} :\frac{1}{\pi} = \sum_{k=0}^\infty \left(2C_k\right)^2 \frac{{\frac{k}{16}+\frac{1}{16}}}{16^k} with the last (comments in ) found by using a linear combination of higher parts of Wallis-Lambert series for \tfrac{4}{\pi} and Euler series for the circumference of an ellipse. Using the definition of Catalan numbers with the gamma function the first and last for example give the identities :\frac14 = \sum_{k=0}^\infty {\left (\frac{\Gamma(\frac12+k)}{\Gamma(2+k)}\right)}^2 \left(4zk-(4n-3)z+4^{2n-3}\right)\qquad z \in \Z,\quad n\ge2,\quad n \in \N ... :4 = \sum_{k=0}^\infty {\left (\frac{\Gamma(\frac12+k)}{\Gamma(2+k)}\right)}^2 (k+1). The last is also equivalent to, :\frac{1}{\pi} = \frac14 \sum_{k=0}^\infty \frac{\binom{2k}{k}^2}{k+1}\, \frac{1}{16^k} and is related to the fact that, : \lim_{k \rightarrow \infty} \frac{16^k}{k \binom{2k}{k}^2} = \pi which is a consequence of Stirling's approximation. ==See also==