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$a_1=1,\forall n\ge 2(a_n=\sqrt{n+a_{n-1}})$.I want to find the smallest natural number n such that $a_n>10$.I find that the sequence is increasing,and I guess:For natural number k,there are only 2k items $a_{j+1},a_{j+2},\cdots,a_{j+2k}$ such that $k\le a_i<k+1,i=j+1,j+2,\cdots,j+2k$，but I have no idea how to prove it.I tried Mathematical induction but I failed ,Please help!I found the answer is 91 using python,but I can't prove it yet.I think the problem differs from Asymptotic behavior of the sequence $a_n=\sqrt{n+a_{n-1}}$ i focus on the specific properties while that problem cares about asymptotic behavior.

EC9
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    Check this: https://math.stackexchange.com/q/3304064/42969, or this: https://math.stackexchange.com/q/1237171/42969. – Martin R Mar 14 '21 at 06:49
  • I think the inequalities in that question is not accurate enough for my question,as I only need to estimate finite items in the sequence. – EC9 Mar 14 '21 at 08:59

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I find another way to solve the problem. Suppose $n\ge 91$,$$a_n = \sqrt{91+a_{n-1}}=\sqrt{91+\sqrt{90+a_{n-2}}}>\sqrt{91+\sqrt{90}}>10.$$Suppose $n \le 90$, $a_{90}=\sqrt{90+a_{98}}>a_{98}\Rightarrow a_{98}<10$.hence,$a_{90}=\sqrt{90+a_{98}}<\sqrt{90+10}=10$. Though I know how to solve the problem, i wonder how to prove my guess that there are only 2k items in the sequence such that integer part of them are k.

EC9
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