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_xgw | 7 years ago
Not necessarily. You can make recursive generators which never "terminate". For example, here's a codegolfed version of a recursive generator which represents a n + 1 sequence:
p=function*a(x){yield x;yield*a(x+1)}(0)
loige|7 years ago
I should probably make it more explicit
PS: I like your n+1 example.