For all $ n in mathbb N $, $ Delta_n = (0,2,3, …, n) $,$ D_n = X + X ^ 2 + …. X ^ $ Y $ I_n = mathbb N cap (0, n) $

For all

$ f in mathbb N ^ {I_n} $ strictly growing, we say that $ P = Sigma_ {i in I_n} a_i.X ^ {s_i} en mathbb Z (X) $ is **1-equivalent** to $ P = Sigma_ {i in I_n} a_i.X ^ {$} Y **2 equivalents** to $ P = Sigma_ {i in I_n} to _ { sigma (i)}. X ^ s_i + D_ {s_n} $ For any $ sigma $ permutation of $ I_ {s_n} $

$ Z subset mathbb Z (X) $ is the kind of $ 0 $ according to him $ 3-equivalence $ Witch is defined as the equivalence ratio generated by these two superior equivalence relationships.

We say that $ P = Sigma_ {i in I_n} a_i.X ^ {i} en mathbb Z (X) $,have **braquetting property** yes for every positive integer $ k leq n $

we have

$ Sigma_ {i in I_k} a_i geq 0 $ Y $ Sigma_ {i in I_n} a_i = 0 $

We say he has the **property score** If any $ Q $ that is to say $ 3 equivalent $ to $ P $ have the **braquetting property**

One can show that the elements of $ Z $ have the **property score**, and the question is the reciprocal:

Is any $ P in mathbb Z (X) $ which has the property of punctuation, in Z?

Note that $ (a_0, a_1 + 1, a_2 + 2, …, a_n + n) $ it is the list of exits of a complete digraph (tournament) if and only if $ a_0 + a_1X + a_2X ^ 2 + …. + a_nX ^ n $ have the **property score**