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As ever with a number puzzle, this was a slog through the calculations. But it held my interest better than most.
Neat idea and construction.

Thanks Jordan. To my mind, "increasing" does not include "equal to".

So, to confirm, the first few clues are actually correct formulae?

yes.

dylan - mathematically, this use of increase is correct - to distinguish you would say strictly increasing if equal wasnt allowed.

However given the majority of listener solvers probably dont have a maths degree ( i dont remember doing this till analysis classes at university), i agree its a bit misleading.

i also think the preamble is misleading in another way also (well it sent me down wrong road initially).

Interesting that this was deemed to be acceptable by the editors.

Does one assume that the individual clues are always given in the form A,B,C - ie the the first listed term is smaller than the second.
Puzzling over an early snag: 30dn would seem to give unique values for c, p and q, although looking at 12ac and 26 I can find no triples where the final term differs by the value I have for c; plenty which differ by p but not c. Have I misunderstood something?

All clues with an error that corresponds to a letter are used in the perimeter.

This tells you that there isn’t an error in 12a or 26

Candledave - indeed. I had figured that there are several clues that are exact triples - including 30dn, 12ac and 26 - which is why I am puzzled. Looking at a list of triples the differences between consecutive sums are 4, 8 or multiples of those. If the letters in 30dn are given in ascending order then c is not 4.

You are right c isn’t 4 but I’m afraid I don’t understand where you are stuck

Maybe this is too explicit but looking at 30dn, c, p and q (and indeed h) are all indices, therefore < 6, which gives a unique solution for 30dn where c=3. However from 12ac and 26 that suggests that there are two exact triples where the C term differs by the value of c, therefore 3. Looking through lists of triples that's not the case - they differ by 0, 4, 8 and larger multiples - and indeed cannot be the case because of the requirements of oddness and evenness for the numbers involved.
Am I being super-dumb?

I'm stuck in a different way, which I feel encouraged to share by the information already agreed. Unless I'm missing something obvious, 14 ends in an 8, but I don't think any of the possible solutions to 15d can begin with an 8. Can someone tell me where I'm going wrong?