There are several children, the candy of the first child is 2 more than the candy of the second child, the candy of the second child is 2 more than the candy of the third child, ..., that is, the former child is always more than the latter Each child has 2 more candies. They formed a circle in order to play games, starting from the first child to give 2 pieces of candy to the second child, and the second child gave 4 pieces of candy to the third child. When the game progresses to a certain child who receives the candy from the previous child but cannot give the candy according to the regulations, there are two adjacent children The ratio of the number of candies is 13:1, how many children are there at most?
有若干名小朋友，第一名小朋友的糖果比第二名小朋友的糖果多2块，第二名小朋友的糖果比第三名小朋友的糖果多2块，…，即前一名小朋友总比后一名小朋友多2块糖果．他们按次序围成圆圈做游戏，从第一名小朋友开始给第二名小朋友2块糖果，第二名小朋友给第三名小朋友4块糖果，…，即每一名小朋友总是将前面传来的糖果再加上自己的2块传给下一名小朋友，当游戏进行到某一名小朋友收到上一名小朋友传来的糖果但无法按规定给出糖果时，有两名相邻小朋友的糖果数的比是13：1，问最多有多少名小朋友？
<result>25</result>