MathDB
a_{n+1} =\frac{a_n^2}{a_n +1} , prove [a_n] = 1994- n

Source: Slovenia TST 1998 p6

February 15, 2020
Sequencerecurrence relationalgebra

Problem Statement

Let a0=1998a_0 = 1998 and an+1=an2an+1a_{n+1} =\frac{a_n^2}{a_n +1} for each nonnegative integer nn. Prove that [an]=1994n[a_n] = 1994- n for 0n10000 \le n \le 1000