Minimum possible sum of weights of the heaviest and the $2^{nd}$ heaviest kid

Question

Question:

The average weight of any $2$ kids in a group of $6$ kids is a distinct natural number. The minimum weight of any kid is $11$ kg. Find the minimum possible sum of the weights of the heaviest and the $2^{nd}$ heaviest kid in the group.

Solution:

By hit and try we get the values as $11,13,15,19,25,35$ kgs.
As after taking $11,13,15$ we can't take $17$, so next take $19$ and so on.

Is there any other (better, that is, more logical) way to do this question (without hit and try)?
Because the hit and try approach is too time taking.

Answer 1

Partial Answer:

Start with $11,13,15$ obviously and let them belong to the set $S$ (arranged in ascending order). Then all we need to be concerned with is $3$ numbers: $S_1, S_{n-1}, S_n$ (where last 2 terms represent the last 2 elements from set $S$).

Then the next term to be included in the set $S$ is given by: $$\bf S_{n+1}=S_n+S_{n-1}-S_1+2$$

So, the next term after $15$ is: $=15+13-11+2=19$
Similarly the next term is given by: $=19+15-11+2=25$
And finally the last term is given by: $=25+19-11+2=35$

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The logic:
The term $S_{n+1}$ should be such: $$\frac{S_n+S_{n-1}}{2}+1=\frac{S_{n+1}+S_1}{2}$$

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If anybody's interested in knowing the $1^{st}$ $25$ such numbers, they are as follows:
(Click to enlarge the image.)

Answer 2

A relevant OEIS sequence is http://oeis.org/A080430.
"a(n) = least odd number such that all pairwise sums a(i) + a(j), i < j <= n, are distinct."

That sequence starts $1,3,5,9,15,25$, ... We can add $10$ to each value there to get values for the given problem here. Adding a constant to every term does not affect the uniqueness of the pairwise sums.

Beyond the $6$th term, that sequence does not give us the correct answer. By using only the next lowest possible number at each step we are only considering the lowest possible weight for the heaviest kid, but we want the lowest possible sum of weights for the two heaviest kids

$\{11, 13, 15, 19, 27, 37, 47\}$ is the best solution for $7$ kids.
$\{11, 13, 15, 19, 27, 39, 49, 59\}$ is the best solution for $8$ kids.

These were found using a brute force program.