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*To*: <obm-l@xxxxxxxxxxxxxx>*Subject*: Problema do Rousseau*From*: "Luis Lopes" <llopes@xxxxxxxxxxxxx>*Date*: Wed, 24 Jan 2001 20:43:49 -0200*References*: <200012151211.KAA10599@ls02.esquadro.com.br> <3A509587.2A1AAB06@lbm.com.br>*Reply-To*: obm-l@xxxxxxxxxxxxxx*Sender*: owner-obm-l@xxxxxxxxxxxxxx

Sauda,c~oes, O prof. Rousseau me prop^os o seguinte problema: The following problem appears in the most recent Pi Mu Epsilon Journal, and so far I haven't been able to solve it. The Smarnandache function S is defined as follows: S(n) is the smallest integer m such that n divides m!. Prove that there is always a prime between S(n) and S(n+1) (including the end values). The first few values are S(1) = 0, S(2)=2, S(3)=3, S(4)=4, S(5)=5, S(6)=3, S(7)=7, S(8)=4, S(9)=6, S(10)=5. In general S(n) = max S(p^k) where the maximum is taken over all factors p_1^{k_1}, p_2^{k_2}, \ldots, in the canonical prime factorization of n. Of course, if either S(n) or S(n+1) is prime, there is nothing to prove. Thus the first "interesting" case above is S(8)=4, S(9)=6, where p=5 is the prime between S(8) and S(9). [ ]'s Lu'is

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