a(1)=1，a(n+1)=a(n)+1／a(n)（n=1,2,…），求 b(n)=[a(n)] 和 ∑(k=1,n)b(k)
设\(a_1=1{,}\ a_{n+1}=a_n+\frac{1}{a_n}\ \left( n=1{,}\ 2{,}\ \ldots\right)\)，\(b_n\)是\(a_n\)的整数部分。是否可以找到\(b_n\)的通项公式和\(\sum_{k=1}^nb_k\)？ a(n)分子=1, 2, 5, 29, 941, 969581, 1014556267661, 1099331737522548368039021,
1280590510388959061548230114212510564911731118541,
172699903806694372485750863858638650428153927937609103408648......
a(n)分母=1, 1, 2, 10, 290, 272890, 264588959090, 268440386798659418988490,
295105036840595214385430531020664149472669868290,
377908709746050392481071609609580527436122569261424131112048......
详见OEIS——A073833
分子。Numerator]
分母。Denominator]
b(n)。Table]/
Denominator]], {n, 28, 28}]
b(n)={1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 8, ......}
期待高手出手！谢谢！
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