关于FE字的一个问题

On a problem of FE words

讨论了 FE字的一个猜想limk→∞A(k)A(k-1 ) =limk→∞B(k)B(k-1 ) =32 (1 )成立的条件 ,得到以下结论 :1 )若 V1 (W)～ V2 (W) (|W|→∞ ) ,对 W是 FE字成立 ,则 (1 )成立 ,这里 Vi(W)表示 FE字 W中字符 i(i=1 ,2 )的个数 .2 )记 f(n) =min{V2 (W) |W∈H∞ ,|W|=n}g(n) =max{V2 (W) |W∈ H∞ ,|W|=n}若 f(n)～ g(n) (n→∞ )则 V1 (W)～V2 (W) (|W|→∞ )

In this paper we have proved following results: 1)If V 1(W)～V 2(W) (|W|→∞), for all FE words. Then lim k→∞A(k)A(k-1)= lim k→∞B(k)B(k-1)=32 Where V i(W) denote the number of occurrences of i in FE word W. 2)let f(n)=min｛V 2(W)|W∈H ∞,|W|=n｝g(n)=max｛V 2(W)|W∈H ∞,|W|=n｝ if f(n)～g(n) (n→∞), then V 1(W) ～V 2(W)(|W|→∞).