Evenly spaced is the worst possible plan.

Two radio mics at fi and fj make third-order products at 2·fi − fj. Write an evenly spaced set as f0 + k·d and that product becomes f0 + (2k − j)·d — a point on the same grid the channels are on. Every product whose index lands in range is therefore sitting exactly on another channel.

The engine is free and needs no account. Eight transmitters, 400 kHz apart, in an ordinary UHF block: 124 of their 224 third-order products land exactly on another channel — 55.4%, and not one of them is bad luck. The step size does not help, because the grid moves with the spacing: at 250 kHz the count is identical. The fix is a theorem. Lay the same eight on a set whose pairwise differences are all distinct and zero products land, because 2·fi − fj = fk would need two pairs to share a difference. The whole cost is span: 10.2 MHz instead of 2.8, inside the same 24 MHz band. And a product follows the pair that makes it, so the channel worth moving is a culprit, not the victim whose receiver is buzzing. Every lane ships a worked example that costs nothing. Writing the report needs credits.