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Jun 18, 2014 at 4:40 vote accept XL _At_Here_There
May 14, 2014 at 9:27 history edited XL _At_Here_There CC BY-SA 3.0
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May 14, 2014 at 9:06 history edited XL _At_Here_There CC BY-SA 3.0
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May 14, 2014 at 8:59 history edited XL _At_Here_There CC BY-SA 3.0
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May 14, 2014 at 8:25 answer added Dima Pasechnik timeline score: 1
May 14, 2014 at 8:20 comment added Dima Pasechnik (continuing) let $F$ be the GF for a recursively enumerable language with non-recursively enumerable complement, and $A$ the GF for the language of all words. Then $A-F$ is the GF for the complement of $F$, i.e. for a non-Chomsky language.
May 14, 2014 at 8:12 history edited XL _At_Here_There CC BY-SA 3.0
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May 14, 2014 at 8:08 comment added Dima Pasechnik If it is the way I described in my comment then there certainly are GFs not corresponding to any language in Chomsky hierarchy, as the latter must be recursively enumerable, a property known not to be closed under taking the complement.
May 14, 2014 at 8:06 comment added XL _At_Here_There @DimaPasechnik,yes,that is what I mean
May 14, 2014 at 8:05 history edited XL _At_Here_There CC BY-SA 3.0
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May 14, 2014 at 8:05 comment added Dima Pasechnik I am not sure if it is well-known how does a language $L$ correspond to a generating function $F(t)=\sum_i a_i t^i$. Is $a_i$ equal to the number of length $i$ words in $L$?
May 14, 2014 at 7:54 history edited XL _At_Here_There CC BY-SA 3.0
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May 14, 2014 at 7:21 history asked XL _At_Here_There CC BY-SA 3.0