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David Roberts ♦
David Roberts ♦
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I can answer your first question. In arXiv:1208.4074arXiv:1208.4074 by Dabholkar, Murthy and Zagier you can find a formula that implies $H^{(2)}(\tau)= \frac{48 F_2^{(2)}(\tau)- 2 E_2(\tau)}{\eta(\tau)^3}$ where $E_2(\tau)$ is the quasi modular Eisenstein series and $F_2^{(2)}(\tau)= \sum_{r>s>0,r-s~ {\rm odd}} (-1)^r s q^{rs/2}$$F_2^{(2)}(\tau)= \sum_{r>s>0,r-s\ \mathrm{odd}} (-1)^r s\, q^{rs/2}$ which you can use to compute $H^{(2)}(\tau)$ to whatever order you desire. I have no idea concerning your questions 2,3, and 4.

I can answer your first question. In arXiv:1208.4074 by Dabholkar, Murthy and Zagier you can find a formula that implies $H^{(2)}(\tau)= \frac{48 F_2^{(2)}(\tau)- 2 E_2(\tau)}{\eta(\tau)^3}$ where $E_2(\tau)$ is the quasi modular Eisenstein series and $F_2^{(2)}(\tau)= \sum_{r>s>0,r-s~ {\rm odd}} (-1)^r s q^{rs/2}$ which you can use to compute $H^{(2)}(\tau)$ to whatever order you desire. I have no idea concerning your questions 2,3, and 4.

I can answer your first question. In arXiv:1208.4074 by Dabholkar, Murthy and Zagier you can find a formula that implies $H^{(2)}(\tau)= \frac{48 F_2^{(2)}(\tau)- 2 E_2(\tau)}{\eta(\tau)^3}$ where $E_2(\tau)$ is the quasi modular Eisenstein series and $F_2^{(2)}(\tau)= \sum_{r>s>0,r-s\ \mathrm{odd}} (-1)^r s\, q^{rs/2}$ which you can use to compute $H^{(2)}(\tau)$ to whatever order you desire. I have no idea concerning your questions 2,3, and 4.

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Jeff Harvey
Jeff Harvey
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I can answer your first question. In arXiv:1208.4074 by Dabholkar, Murthy and Zagier you can find a formula that implies $H^{(2)}(\tau)= \frac{48 F_2^{(2)}(\tau)- 2 E_2(\tau)}{\eta(\tau)^3}$ where $E_2(\tau)$ is the quasi modular Eisenstein series and $F_2^{(2)}(\tau)= \sum_{r>s>0,r-s~ {\rm odd}} (-1)^r s q^{rs/2}$ which you can use to compute $H^{(2)}(\tau)$ to whatever order you desire. I have no idea concerning your questions 2,3, and 4.