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By Jan Mycielski, Pavel Pudlak, Alan S. Stern

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R . 5. — Pour r ≥ 1 on a : rA(t, u) = (r − 1)! 1A(t, u) . The D´emonstration. — Pour r = 1, il n’y a rien ` a prouver. Supposons r ≥ 2 et prenons pour µ le morphisme prolongeant l’application σ → θ∆Dσ = t|∆Dσ| au mono¨ıde S+ . Si l’on a σ ∈ Sn+r−1 et σw = g1 i1 g2 i2 . . gr−1 ir−1 gr 36 ´ ERATRICES ´ CHAPITRE IV : FONCTIONS GEN avec {i1 , i2 , . . , ir−1 } = [r − 1], il vient |∆Dgj | = |∆Dωgj | (j ∈ [r]), puisque ω est un morphisme injectif. D’autre part, puisque le mot g1 g2 , . . gr contient toutes les lettres du mot σw sup´erieures ou ´egales ` a r, on a r−1 r−1 |∆ ∆Dσ| = j |∆Dωgj |, d’o` u µ r σ = θ∆ ∆Dσ.

Pour r = 1, il n’y a rien ` a prouver. Supposons r ≥ 2 et prenons pour µ le morphisme prolongeant l’application σ → θ∆Dσ = t|∆Dσ| au mono¨ıde S+ . Si l’on a σ ∈ Sn+r−1 et σw = g1 i1 g2 i2 . . gr−1 ir−1 gr 36 ´ ERATRICES ´ CHAPITRE IV : FONCTIONS GEN avec {i1 , i2 , . . , ir−1 } = [r − 1], il vient |∆Dgj | = |∆Dωgj | (j ∈ [r]), puisque ω est un morphisme injectif. D’autre part, puisque le mot g1 g2 , . . gr contient toutes les lettres du mot σw sup´erieures ou ´egales ` a r, on a r−1 r−1 |∆ ∆Dσ| = j |∆Dωgj |, d’o` u µ r σ = θ∆ ∆Dσ.

Si l’on a σ ∈ Sn+r−1 et σw = g1 i1 g2 i2 . . gr−1 ir−1 gr 36 ´ ERATRICES ´ CHAPITRE IV : FONCTIONS GEN avec {i1 , i2 , . . , ir−1 } = [r − 1], il vient |∆Dgj | = |∆Dωgj | (j ∈ [r]), puisque ω est un morphisme injectif. D’autre part, puisque le mot g1 g2 , . . gr contient toutes les lettres du mot σw sup´erieures ou ´egales ` a r, on a r−1 r−1 |∆ ∆Dσ| = j |∆Dωgj |, d’o` u µ r σ = θ∆ ∆Dσ. 2, µ r {Sn+r−1 } = rAn+r−1 (t). 5 r´esulte de l’identit´e (11). 3. Autres interpr´ etations des polynˆ omes eul´ eriens Les techniques du chapitre pr´ec´edent pourraient ˆetre appliqu´ees ` a d’autres probl`emes d’´enum´eration.

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