I'm looking for an optimal (or much better than I have now) increasing sequence
$t_1, t_2, .., t_{13}: t_i \in N, 0 \le t_i<t_{i+1}$
where $3t_{12} + 4t_{13}$ is minimized, subject to the below constraint.
I hope I'm expressing it right, but what I want is a sequence of integer numbers, such that in any way you add exactly 7 of those numbers together (the same number up to 4 times), that sum will be different from any other 7 of those numbers added together. I think that constraint would be
$\forall A=a_1,a_2,..,a_{13}, B=b_1,b_2,..,b_{13} ,$
$a_i, b_i \in N; 0\le a_i\le 4; 0 \le b_i \le 4,$
$\sum A=\sum B = 7 :$
$\sum a_i t_i = \sum b_i t_i => A=B$
The best solution I've found so far can be expressed by the sequence
$T(1) = 0$
$T(2) = 1$
$T(n) = 4 T(n-1) + 3T(n-2) + 1$
which yields an optimization value of 113,088,216.
If someone can show a lower bound for the optimal solution that is not significantly less than this value, I consider this solved as well.
The usage of this sequence is quite interesting and is explained in here: http://www.codeproject.com/Articles/1068755/Quick-card-poker-ranking-in-Java
The following is a 'greedy' list of $t_i$, where $t_n$ is the lowest value that fits with all the lower $t_i$ for four $n$-card suits. I think a lower $t_{13}$ is possible if I allow leeway in earlier $t$-values, but my routine is slow.
0, 1, 5, 24, 106, 472, 2058, 8768, 29048, 70347, 233028, 583164, 1472911
That is without the $\%C$ mentioned in the comments, so it saves on that calculation. $4t_{13}+3t_{12}=7641136$, which needs an array about one percent larger than the one in comments.
Or you can keep the $\%C$ step, and take all the $\sum t_i$ numbers modulo 4731770.