International Journal of Mathematics and Mathematical Sciences

Research Article

Division Problem of a Regular Form: The Case 𝑥𝟐𝑢=𝜆𝑥𝑣

Table 1

Δ 𝑛 Δ 2 𝑛 = ( − 1 ) 𝑛 + 1 𝜆 2 Γ ( 𝜏 + 1 / 2 ) Γ ( 𝑛 + 𝜏 + 1 / 2 ) Γ 2 ( 𝑛 + 1 ) ( ( 𝜔 − 1 ) Λ 𝑛 + 1 ) 2 , 𝑛 ≥ 0 ,
Δ 2 𝑛 + 1 = ( − 1 ) 𝑛 𝜆 Γ 2 Γ ( 𝜏 + 1 / 2 ) 2 ( 𝑛 + 𝜏 + 3 / 2 ) Γ ( 𝑛 + 1 ) , 𝑛 ≥ 0 .
𝑎 𝑛 𝑎 2 𝑛 = ( 𝑛 + 𝜏 + 1 / 2 ) 2 ℎ 𝜆 Γ ( 𝜏 + 1 / 2 ) 𝑛 ( ( 𝜔 − 1 ) Λ 𝑛 + 1 ) 2 , 𝑛 ≥ 0 , 𝑎 2 𝑛 + 1 = − 𝜆 Γ ( 𝜏 + 1 / 2 ) 𝑛 + 1 ℎ 𝑛 + 1 ( ( 𝜔 − 1 ) Λ 𝑛 + 1 ) 2 , 𝑛 ≥ 0 .
𝑏 𝑛 𝑏 2 𝑛 = 𝑛 + 𝜏 + 1 / 2 , 𝑛 ≥ 0 , 𝑏 2 𝑛 + 1 = ( 𝑛 + 1 ) ( 𝜔 − 1 ) Λ 𝑛 + 1 + 1 ( 𝜔 − 1 ) Λ 𝑛 + 1 , 𝑛 ≥ 0 .
𝑐 𝑛 𝑐 0 = 0 , 𝑐 1 = − 𝜔 𝜆 , 𝑐 2 𝑛 + 2 = 1 𝜆 𝑛 + 𝜏 + 1 / 2 ℎ Γ ( 𝜏 + 1 / 2 ) 𝑛 ( ( 𝜔 − 1 ) Λ 𝑛 + 1 ) 2 , 𝑛 ≥ 0 ,
𝑐 2 𝑛 + 3 = − 𝜆 ( 𝑛 + 1 ) Γ ( 𝜏 + 1 / 2 ) ( 𝑛 + 𝜏 + 1 / 2 ) ℎ 𝑛 ( ( 𝜔 − 1 ) Λ 𝑛 + 1 + 1 ) ( ( 𝜔 − 1 ) Λ 𝑛 + 1 ) , 𝑛 ≥ 0 .
𝛾 1 = − 𝜆 2 𝜔 2 , 𝛾 2 = − ( 𝜏 + 1 / 2 ) 2 𝜆 2 𝜔 4 ,
𝛾 𝑛 + 1 𝛾 2 𝑛 + 3 𝜆 = − 2 Γ 2 ( 𝜏 + 1 / 2 ) ℎ 2 𝑛 + 1 ( ( 𝜔 − 1 ) Λ 𝑛 + 1 + 1 ) 2 ( ( 𝜔 − 1 ) Λ 𝑛 + 1 ) 2 , 𝑛 ≥ 0 ,
𝛾 2 𝑛 + 4 1 = − 𝜆 2 Γ 2 ( 𝜏 + 1 / 2 ) ( 𝑛 + 𝜏 + 3 / 2 ) 2 ℎ 2 𝑛 + 1 ( ( 𝜔 − 1 ) Λ 𝑛 + 1 + 1 ) 4 , 𝑛 ≥ 0 .
𝛽 0 = 𝜔 𝜆 , 𝛽 1 = − 𝜔 𝜆 − 𝜏 + 1 / 2 𝜆 𝜔 2 ,
𝛽 𝑛 𝛽 2 𝑛 + 3 = − 𝜆 ( 𝑛 + 1 ) Γ ( 𝜏 + 1 / 2 ) ( 𝑛 + 𝜏 + 1 / 2 ) ℎ 𝑛 ( ( 𝜔 − 1 ) Λ 𝑛 + 1 + 1 ) ( ( 𝜔 − 1 ) Λ 𝑛 + 1 ) − 𝑛 + 𝜏 + 3 / 2 ℎ 𝜆 Γ ( 𝜏 + 1 / 2 ) 𝑛 + 1 ( ( 𝜔 − 1 ) Λ 𝑛 + 1 + 1 ) 2 , 𝑛 ≥ 0 ,
𝛽 2 𝑛 + 2 = 1 𝜆 1 Γ ( 𝜏 + 1 / 2 ) ( 𝑛 + 𝜏 + 1 / 2 ) ℎ 𝑛 ( ( 𝜔 − 1 ) Λ 𝑛 + 1 ) 2 + 𝜆 Γ ( 𝜏 + 1 / 2 ) ℎ 𝑛 + 1 ( ( 𝜔 − 1 ) Λ 𝑛 + 1 + 1 ) ( ( 𝜔 − 1 ) Λ 𝑛 + 1 ) , 𝑛 ≥ 0 .