Determinining the floor

[SANZEED]

find floor
Let $a_{0}=$ 1996 and $a_{n+1}=a_{n}/[a_{n}^{2}+1]$ for $n=1,2,3,...,$.prove that $[a_{n}]=1996-n$ for $n=1,2,3,...,999$.Here $[x]$ denotes the greatest positive integer less than or equal to $x$.

[Phlembac Adib Hasan]

Hei, again problem.You said $a_{n+1}=\frac {a_n} {a_n^2+1}$.if it is not valid for $n=0$, there is no possible way to connect $a_0$ and $a_i$.So I can take any value of $a_1$ and so on.If the recurrence relation is valid for $n=0$, then $a_1=\frac {a_0} {[a_0^2+1]} =\frac {1996} {1996^2+1} \neq 1996-1=1995 $

[SANZEED]

Sorry.The recurrence holds for $n=0$ too.And yes,$[a_{1}]=1995$.And i put [] before and after $a_{n}^{2}+1$ not to indicate floor, but to indicate bracket.

[Phlembac Adib Hasan]

There are still problems.Just check this for some $n$s.

[SANZEED]

I don't understand.....It's a problem from the national math olympiad of Russia,1995.I myself solved it and checked the official solution. I would be greatful if you send me a private message involving your check.

[Phlembac Adib Hasan]

I don't understand.....It's a problem from the national math olympiad of Russia,1995.I myself solved it and checked the official solution. I would be greatful if you send me a private message involving your check.

The fraction has to be inversed.Then it is true and I did it by induction.