DLMF:12.14#Ex3 (Q4247)

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DLMF:12.14#Ex3
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    α n + 2 = a α n + 1 - 1 2 ( n + 1 ) ( 2 n + 1 ) α n , subscript 𝛼 𝑛 2 𝑎 subscript 𝛼 𝑛 1 1 2 𝑛 1 2 𝑛 1 subscript 𝛼 𝑛 {\displaystyle{\displaystyle\alpha_{n+2}=a\alpha_{n+1}-\tfrac{1}{2}(n+1)(2n+1)% \alpha_{n},}}
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