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+## Missing $ on a $((..)) expression? (or use ( ( for arrays).
+
+### Problematic code:
+
+```sh
+var=((foo+1))
+```
+
+### Correct code:
+
+```sh
+var=$((foo+1))
+```
+
+### Rationale:
+
+You appear to be missing the `$` on an assignment from an arithmetic expression `var=$((..))` .
+
+Without the `$`, this is an array expression which is either nested (ksh) or invalid (bash). 
+
+### Exceptions:
+
+If you are trying to define a multidimensional Ksh array, add spaces between the `( (` to clarify:
+
+    var=( (1 2 3) (4 5 6) )