Sum of Four Fourth Powers?

We define $(a_1,a_2,a_3,a_4)=\sum_{i=1}^{4}a_i^4$ . Also we make the assumption that $a_i < a_{i+1} \ \forall i$ , meaning we sort the components in ascending order.

not hard to see the following holds true

1- $(a_1,a_2,a_3,a_4) < (b_1,b_2,b_3,b_4) \iff a_i \leq b_i \ \forall i$

2- when $(a_1,a_2,a_3,a_4)$ is changed to $(b_1,b_2,b_3,b_4)$ by changing exactly two components $a_i,a_j \ , \ i< j$ such that $b_i=a_i-1,b_j=b_j+1$ , then $(a_1,a_2,a_3,a_4) < (b_1,b_2,b_3,b_4)$

then we need to observe that $(2,3,5,6)=2018$ . Also, $(a_1,a_2,a_3,7)$ is greater than $2018$ . therefore, the tuples of four $(a_1,a_2,a_3,a_4)$ , that we are looking for, have all components less than $7$ . the re are $\binom{6}{4}=15$ such tuples. Then, it seems intuitively correct to see which tuples are greater than $(2,3,5,6)$ . Using the two rules above

$(2,3,5,6) < (3,4,5,6) \ (1)$ $(2,3,5,6) < (2,4,5,6) \ (1)$ $(2,3,5,6) < (1,4,5,6) \ (2)$

same relations can be sorted, using the same rules, for tuples that are smaller than $(2,3,5,6)$ . So there are only three tuples greater than $(2,3,5,6)$ . Consequently, $(2,3,5,6)$ is the 12th element in the ordered set.