The original BBP pi formula is
$$\pi = \sum_{n=0}^\infty \frac1{2^{4n}}\left(\frac{4}{8n+1}-\frac{2}{8n+4}-\frac{1}{8n+5}-\frac{1}{8n+6}\right)$$
Others are:
Order 4. Let $P_k = 4n+k$
$$\pi = \sum_{n=0}^\infty \frac{(-1)^n}{\color{blue}{2^{2n}}}\left(\frac{2}{P_1}+\frac{2}{P_2}+\frac{1}{P_3}\right)$$
Order 8. Let $Q_k = 8n+k$
$$\pi = \frac{1}{2}\sum_{n=0}^\infty \frac1{\color{blue}{2^{4n}}}\left(\frac{2^3}{Q_2}+\frac{4}{Q_3}+\frac{4}{Q_4}-\frac{1}{Q_7}\right)$$
Order 12. Let $R_k = 12n+k$
$$\pi = \frac{1}{2^3}\sum_{n=0}^\infty \frac{(-1)^n}{\color{blue}{2^{6n}}}\left(\frac{2^5}{R_2}+\frac{24}{R_3}+\frac{4}{R_6}+\frac{3}{R_9}+\frac{2}{R_{10}}\right)$$
Order 16. Let $S_k = 16n+k$
$$\pi = \frac{1}{2^5}\sum_{n=0}^\infty \frac{1}{\color{blue}{2^{8n}}}\left(\frac{2^7}{S_2}+\frac{2^6}{S_3}+\frac{2^6}{S_4}-\frac{2^4}{S_7}+\frac{2^3}{S_{10}}+\frac{4}{S_{11}}+\frac{4}{S_{12}}-\frac{1}{S_{15}}\right)$$
Order 20. Let $T_k = 20n+k$
$$\pi = \frac{1}{2^6}\sum_{n=0}^\infty \frac{(-1)^n}{\color{blue}{2^{10n}}}\left(\frac{2^9}{T_2}-\frac{160}{T_5}-\frac{2^7}{T_6}-\frac{2^3}{T_{10}}-\frac{2^3}{T_{14}}-\frac{5}{T_{15}}+\frac{2}{T_{18}}\right)$$
Order 24. (known)
$$\pi = \frac{1}{2^5}\sum_{n=0}^\infty \frac{1}{\color{blue}{2^{12n}}}\left(\sum_{k=1}^{23} \frac{a_k}{24n+k}\right)$$
Order 28. ($\color{red}{unknown}$)
$$\pi = \frac{1}{2^{(?)}}\sum_{n=0}^\infty \frac{(-1)^n}{\color{blue}{2^{14n}}}\left(\sum_{k=1}^{27} \frac{b_k}{28n+k}\right)$$
Order 32. (known)
$$\pi = \frac{1}{2^{12}}\sum_{n=0}^\infty \frac{1}{\color{blue}{2^{16n}}}\left(\sum_{k=1}^{31} \frac{c_k}{32n+k}\right)$$
where $a_k,c_k$ are integers and can be found here.
R: However, what (if any) is the sequence of integers $b_k$ such that order 28 is true?
P.S. I tried Mathematica's Integer relations feature but couldn't find any, though I may have used too little decimal precision.