一类递推数列中的平方数

The Primes in a Family of Recurrence Sequences

设{x n}是满足递推关系x0=1,x1=a>1,xn+2=2a x n+1-xn的数列.本文给出了:a=5,9,169以及9 801时所有可使xn是平方数的正整数n.

Let be {x n} a sequence which satisfy,x0=1,x1=a1,xn＋2=2a x n＋1-xn.In this paper,all positive integers that make the form to be squares were given when a=5,9,169 and 9 801.