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Home » \Rightarrow y = 3x - 3 + 4 = 3x + 1 - Aurero

\Rightarrow y = 3x - 3 + 4 = 3x + 1 - Aurero

📅 March 14, 2026 👤 scraface

Mar 14, 2026

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📌 $a_n = b_n + c_n + d_n$, but easier: define based on how the string ends.

📌 Use states:

📌 $S_n$: number of valid strings of length $n$ ending in 3

📌 $T_n$: ending in exactly one 2 (i.e., last is 2, previous ≠ 2)

📌 $U_n$: ending in exactly two consecutive 2s (i.e., last two are 22, but not 222)

📌 Recurrences:

📌 $S_n = S_{n-1} + T_{n-1} + U_{n-1}$ (append 3 to any valid string)

📌 $T_n = S_{n-1}$ (append 2 to a string ending in 3)

🔥 Popular Posts

直角三角形の直角を挟む辺の長さが8と15です。斜辺の長さは？
\[ c^2 = a^2 + b^2 \]
ここで、\(a = 8\)，\(b = 15\) です。
\[ c^2 = 8^2 + 15^2 = 64 + 225 = 289 \]
したがって、斜辺 \(c\) は：
\[ c = \sqrt{289} = 17 \]
3x - y = 5
まず、2番目の式から \(y\) を求めます：
\[ 3x - y = 5 \implies y = 3x - 5 \]
\(y = 3x - 5\) を1番目の式に代入します：

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