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sudokulover. If 10A ends with 6 It means that I have both 10A and 15d as the same?????

Your post at 13.23 answers this dilemma jackdaw.

Jackdaw, that's why, in my previous reply to your post, I advised caution in accepting sudoku lover's choice.

Thank you all! Must pay more attention to the rubric in future. Now for 12A.

I have just come to this, and I just cannot understand the clues - being stupid no doubt. If the "clue gives the sum of two positive integers" does this refer to the number to the left of the brackets? It cannot, because the clues beginning "2" could only both be 1+1. So what am I misunderstanding?

Melbourne once you have the positive integers 1 and 1, you use these to generate the primes. So the primes in this case could be 5 or 7 (using the formula 6n + 1). You then enter the lower prime followed by the difference between the two of them. Here we could have 50, 52 or 70. To get you started, 4Dn is 52 and 19Dn is 70. Good luck!

You are correct in that 2 must be 1+1, so u and v are both 1, but from that you can produce 3 possible entries. The preamble needs a fair bit of thought to get to grips with, but once you do follow what is involved the puzzle is definitely at the easier end of the range of Listener numericals.

Hi Melbourne to add to unclued's reply.
The clues give the sum of u+v.
Primes are therefore (6u +/- 1) and (6v +/- 1). Adding the possible combinations (6u - 1) + (6v -1) = 6(u + v) -2,(6u + 1) + (6v + 1) = 6(u+v) + 2, (6u - 1) + (6v + 1) &
(6u + 1) + (6v - 1) both give 6 (u + v).
In practice, if you multiply each clue by 6 then the 2 associated primes either equal this exactly or +/- 2.
So in 1a 26 x 6 = 156; so the 2 associated primes add to 154, 156 or 158. Hope this helps.

Thanks everyone - I had a slight panic coming to it so late. And it would never have occurred to me to treat them as Ginge said.
(Finished this week's very quickly, so plenty of time to spare)