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$ b_n = c_{n-1} $ (can only append 3 if previous digit is not 3)
$ c_n = a_{n-1} $ (append 7 to any valid sequence)
So $ a_n = b_n + c_n = c_{n-1} + a_{n-1} $, but better to write:
From recurrence:
$ a_n = a_{n-1} + a_{n-2} $ — standard Fibonacci.
Base cases:
$ n = 1 $: sequences: "3", "7" → both valid → $ a_1 = 2 $
$ n = 2 $: "37", "73", "77", "33" → exclude "33" → 3 valid → $ a_2 = 3 $
$ a_4 = 5 + 3 = 8 $
$ a_5 = 8 + 5 = 13 $