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 $1,24$. 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.
Best Answer
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.