Say, I have a set of 6 +ve integers sorted in ascending order:

$A = \{2,4,4,4,5,7\}$

Now to make it easier to deal with (Minimum one starts with 1) I deducted one from all of them:

$\therefore B= \{1,3,3,3,4,6\}$

Now what I am trying to do is get rid of the repetition of numbers and I tried by keeping one number (4) and the rest same numbers in the incremental way (4,5) and the increased amount (+2) is added to the following numbers (4,6) after that.

$\therefore C = \{1,3,4,5,6,8\}$

But I wanted to keep the subset sum ascending ordering, which seems like doesn't now hold. For example:

For Set A: $(2), (4), (4), (4), (5), (2,4)... $

Sums are: $2,4,4,4,5,6 ...$ (Which has the desired ascending order)

For Set C: $(1), (3), (4), (1,3), (5), (6), (1,6)..... $

Sums are: $1,3,4,4,5,6,7...$ (Which has the desired ascending order)

Up to this point it is fine. But what if I want to generate them from the original set $A$, and the problem is the order doesn't not hold anymore!

For example:

For Set A: $(2), (4), (4), (4), (5), (2,4)... $

Now if we replace 2 with 1, 4 with (3,4,5), 5 with 8 and 7 with 8:

It becomes: $(1), (3), (4), (5), (1,3)...$

Now the sums are: $1,3,4,5,4...$ We see that subset sum ascending ordering doesn't hold any longer $(..,4,5,4..)$. (Which DOES NOT have the desired ascending order)

Now my question is: Is there any way I can get rid of the repetition numbers i.e. they will become unique and the subset sum ordering will remain intact???