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The series which forms the basis of all the other series is:- 1,2,4,8,16,18. Some other combinations are:- 1,2,3,7,14,22 ; 1,2,4,7,15,20 ; 1,2,4,8,13,21. However, I obtained the basic combination by the following method:-

Step 1:- You definitely need "1". Step 2:- You need two and also 3. So, the next numbers is 2. Step 3:- Now you have 1,2. So, the next numbers is 4. Step 4:- Now you have 1,2,4. So, the next numbers is 8 Step 5:- Now you have 1,2,4,8. So, the next numbers is 16 Step 6:- Now you have 1,2,4,8,16. So, the next numbers is 18 So the series is 1,2,4,8,16,18.

My Question:- I don't know how to prove that there are more ( or exactly these many ) rigorously ( and why ). Any help would be appreciated.

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How exactly do we "obtain any number between 1 and 49?" Is it by adding together various elements of our set? If so, then your question is probably related to this concept: – Ryan Dec 17 '12 at 2:17

In step 3, you could also select $3$, as you have in one of your examples. Your approach shows that with $n$ numbers we can add up to $2^n-1$ and to do so we want to select the powers of $2$. When the upper limit is not one less than a power of $2$ you have more flexibility. As you have seen, you can lower the top one from $32$ to $18$ and still reach $49$. There are certainly other possibilities: One of them is $1,2,4,8,12,22$

Basically your constraints are: the six numbers must sum to at least $49,$ each one must be less than or equal to one more than the sum of the previous numbers. A little thought and an inductive proof should convince you these are the only constraints. I don't know how to count the number of possibilities easily.

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