Hi Vyacheslav,
If you tally ‘nu-fission’ by itself, this should actually give you a score that is equivalent to k-eff (or k-inf for an infinite medium problem). All tallies in OpenMC are “per source particle”. Since k-effective is production/losses, and by definition all source particles will be lost through leakage or absorption, it follows that production/losses = nu-fission/1. “Production” from reactions like (n,2n) is treated during a single source history, so nu-fission will already account for that.
Best regards,
Paul