def _compute_p_max(m_max):
sqrt_m_max = np.sqrt(m_max)<EOL>p_low = int(np.floor(sqrt_m_max))<EOL>p_high = int(np.ceil(sqrt_m_max + <NUM_LIT:1>))<EOL>return max(p for p in range(p_low, p_high+<NUM_LIT:1>) if p*(p-<NUM_LIT:1>) <= m_max + <NUM_LIT:1>)<EOL>
Compute the largest positive integer p such that p*(p-1) <= m_max + 1.

Do this in a slightly dumb way, but safe and not too slow.

Parameters
----------
m_max : int
    A count related to bounds.