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代进 t= log n /(loglog n)³之后，第一项 t(2+loglogn) 是 O(log n /(loglogn)²)
第二项log 2*log n / log t = log 2*log n /(loglog n - 3logloglog n)
= log2*log n /loglog n + log2*logn*3logloglogn / (loglog n)(loglog n - 3logloglog n)
后一部分是 O(log n*logloglog n / (loglogn)²)
加在一起是 log2*log n /(loglog n) + O(log n*logloglog n /(loglog n)²)，也可以写成图二的形式，方便比较
这个结论相当于 log τ(n)*loglog n / log n 的上极限≤log 2，可以证明结果正好是log 2
取t= (log n/loglog n)^(1/2) 的话，后一项应该太大了，得到的结果会变成2log 2 |
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