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📅 May 15, 2026 👤 scraface

May 15, 2026

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Compute up to $a_8$:
a_3 = a_2 + a_1 = 3 + 2 = 5 \\
a_4 = a_3 + a_2 = 5 + 3 = 8 \\
a_5 = a_4 + a_3 = 8 + 5 = 13 \\
a_6 = a_5 + a_4 = 13 + 8 = 21 \\
a_7 = a_6 + a_5 = 21 + 13 = 34 \\
a_8 = a_7 + a_6 = 34 + 21 = 55
So, number of 8-digit numbers **without** consecutive 3s is 55.
Thus, the number **with** at least one pair of consecutive 3s is:
- 55 = \boxed{201}

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