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kouddy/SICP 1.10.part2

Created February 21, 2015 17:38
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SICP 1.10.part2
;;;(A 2 4)
;;; We know that (A 1 n) = 2^n, so steps to evaluate (A 1 n) are omitted here.
(A 1 (A 2 3))
(A 1 (A 1 (A 2 2)))
(A 1 (A 1 (A 1 (A 2 1))))
(A 1 (A 1 (A 1 2)))
...
(A 1 (A 1 4))
...
(A 1 16)
2^16 = 65536
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