For n=1, the only expression is i itself (i.e., sqrt(-1), the imaginary unit), so a(1) = 0.
For n=2, the only expression is i^i, which evaluates to the real number e^(-Pi/2), so a(2) = 1.
For n=3, the only expressions are i^(i^i) and (i^i)^i, neither of which evaluates to a real number, so a(3) = 0.
For n=4, four of the A000108(4-1) = 5 parenthesized expressions yield real values, so a(4) = 4:
i^((i^i)^i) = ((i^i)^i)^i; and
(i^i)^(i^i) = (i^(i^i))^i.
For n=5, of the A000108(5-1) = 14 parenthesized expressions, none yield a real number, so a(5) = 5.
For n=6, 24 parenthesized expressions yield real values, so a(6) = 24:
i^((i^i)^((i^i)^i)) = i^((i^((i^i)^i))^i) = i^((((i^i)^i)^i)^i) = ((i^i)^i)^((i^i)^i) = ((i^i)^((i^i)^i))^i = ((i^((i^i)^i))^i)^i = ((((i^i)^i)^i)^i)^i;
(i^i)^(i^((i^i)^i)) = (i^i)^(((i^i)^i)^i);
(i^i)^((i^i)^(i^i)) = (i^i)^((i^(i^i))^i) = (i^((i^i)^(i^i)))^i = (i^((i^(i^i))^i))^i;
(i^(i^i))^((i^i)^i) = (i^((i^i)^i))^(i^i) = (((i^i)^i)^i)^(i^i) = (((i^i)^i)^(i^i))^i = (((i^i)^(i^i))^i)^i = (((i^(i^i))^i)^i)^i;
((i^i)^(i^i))^(i^i) = ((i^(i^i))^i)^(i^i) = ((i^(i^i))^(i^i))^i; and
(i^(i^((i^i)^i)))^i = (i^(((i^i)^i)^i))^i.
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