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when do the patriots play the jets - RTA

📅 May 9, 2026 👤 scraface

May 09, 2026

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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 $
$ a_6 = 13 + 8 = 21 $
So 21 sequences with no two consecutive 3s.
Total sequences: $ 2^6 = 64 $
Thus, number with **at least one** "33" is $ 64 - 21 = 43 $
Therefore, the answer is $ \boxed{43} $.

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