The 4α condensate state for 16 O is discussed with the THSR (Tohsaki-Horiuchi-Schuck-Röpke) wave function which has α-particle condensate character. Taking into account a proper treatment of resonances, it is found that the 4α THSR wave function yields a fourth 0 + state in the continuum above the 4α-breakup threshold in addition to the three 0 + states obtained in a previous analysis. It is shown that this fourth 0 + ((0 + 4 )THSR) state has an analogous structure to the Hoyle state, since it has a very dilute density and a large component of α + 12 C(0 + 2 ) configuration. Furthermore, single-α motions are extracted from the microscopic 16-nucleon wave function, and the condensate fraction and momentum distribution of α particles are quantitatively discussed. It is found that for the (0 + 4 )THSR state a large α-particle occupation probability concentrates on a single-α 0S orbit and the α-particle momentum distribution has a δ-function-like peak at zero momentum, both indicating that the state has a strong 4α condensate character. It is argued that the (0 + 4 )THSR state is the counterpart of the 0 + 6 state which was obtained as the 4α condensate state in the previous 4α OCM (Orthogonality Condition Model) calculation, and therefore is likely to correspond to the 0 + 6 state observed at 15.1 MeV.