CSL or not ?

Question

L={ (aⁿ)^m bⁿ | n,m>=1 }

Comments

If there is an upper limit on m then it is CFL. As L can be written as

L = L1 U L2 U L3 U L4 U...   finite times.

L1 = { $a^n$$b^n$ | n >=1}

L2 = { $a^{2n}$$b^n$ | n >=1}

L3 = { $a^{3n}$$b^n$ | n >=1}

L4 = { $a^{4n}$$b^n$ | n >=1}

....

Finite times.

We know that finite union of CFL is CFL.

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@ Hemant Parihar

It is CSL, as we can write C program to check whether the input is in this format or not.

But, if we are able to write a program , it means that it is decidable and hence REC , how can we be sure that it is CSL?

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@VS, Here we need to do the matching of a and b in some order. Which can be done using a TM whose tape size is linearly dependent on the input size. That's why CSL.