Lyndon words over $GF(5)$ with given trace and subtrace

Here we consider the number $L(n;t,s)$ of length $n$ Lyndon words $a_1 a_2 \ldots a_n$ over the alphabet consisting of the elements of the field $GF(5)$ that have trace $t$ and subtrace $s$. The trace of a Lyndon word is the sum of its digits over the field, i.e., $t = a_1 + a_2 + \cdots + a_n$. The subtrace is the sum of the products of all $n(n-1)/2$ pairs of digits taken over the field, i.e., $s=\sum_{1\leq i\lt j\leq n} a_i a_j$. Note that $GF(5)=\mathbb{Z}_5$.

(trace,subtrace)
$n$ (0,0) (0,1)
(0,4)
(0,2)
(0,3)
(1,0)
(2,0)
(3,0)
(4,0)
(1,1)
(2,4)
(3,4)
(4,1)
(1,2)
(2,3)
(3,3)
(4,2)
(1,3)
(2,2)
(3,2)
(4,3)
(1,4)
(2,1)
(3,1)
(4,4)
1 100 1 000 0
2 010 1 001 0
3 022 2 202 2
4 657 5 657 7
5 242030 25 252525 25
6 10499107 104 99107107 99
7 432450450 450 450450450 432
8 195019651935 1935 196519501965 1935
9 873686668666 8666 873686668666 8666
10 392983898538990 39050 390503905039050 39050
11 177784177500177500 177784 177500177500177500 177500
12 813748814006813490 814006 813490813490814006 813748
13 375504837562503756250 3756250 375625037550483756250 3756250
14 174384001743727517439507 17437275 174384001743727517439507 17439507
15 813801928137499081385410 81380200 813802008138020081380200 81380200

Examples

The two 5-ary Lyndon words of trace 2, subtrace 4 and length 3 are $\{124, 142\}$. The five 5-ary Lyndon words of trace 1, subtrace 2 and length 4 are $\{0222, 1113, 1244, 1424, 1442\}$. The seven 5-ary Lyndon words of trace 3, subtrace 1 and length 4 are $\{0044, 0233, 0323, 0332, 1124, 1142, 1214\}$.

Enumeration (OEIS)