贵刊文[1]利用函数f(x)的“不动点”巧妙地求出了形如a_n=(a·a_(n-1)+b)/(c·a_(n-1)+d)(a,b,c,d均不为零且(d-a)~+4dc≥0),及a_n=(a·a_(n-1)~2+b)/(2a·a_(n-1)+c) (a,b,c均不为零且c~2+4ad>0)的数列通项公式,读后深受启发.经过研究,本人发现利用函数f(x)的“不动点”还可求出如下一种形式的数列通项公式: Your article [1] uses the “fixed point” of the function f (x) to find out the shape of a_n = (a · a_ (n-1) + b) / (c · a_ ) + d) (a, b, c and d are not zero and (da) to + 4dc ≧ 0) and a_n = (a · a_n- 1 to 2 + b) n-1) + c) (a, b, c are non-zero and c ~ 2 + 4ad> 0), which is inspired by the research. “Fixed point ” can also be calculated as follows a sequence of formulas: