use integer

That's the piece I was overlooking. I was expecting that, as with purely 32-bit arithmetic, the number would be an IV unless it was either a fraction, or was too big to fit into an IV. But that's not the case - an integer becomes an NV as soon as 53 bits are exceeded (even though the IV can accommodate 64 bits). Of course, with 'use integer;', the integer retains it's IV status, even when it exceeds 64 bits.

Things start to make some sense ... though still appear a little buggy. (But maybe that's just because I need to readjust my vision :-)

Cheers,
Rob