Hi, Noel wrote: > I fed '26 3 ^p' into 'dc' to see just how big it was - and got > "17576", a 16-bit word array of which would fit into a PDP-11 64KB > address space. I'm a fan of dc(1), but also of units(1) and since both seem underused here's an alternative method. $ units -1v '26^3 16 bit' 64KiB 26^3 16 bit = 0.53637695 * 64KiB -- Cheers, Ralph.