“…Satisfy the reduction formula(Sharma, 2016;Acu and Agrawal, 2019;Acu and Tachev, 2021;Acu et al, 2023;Adell and Cárdenas-Morales, 2022 ) 𝑆(𝑣, 𝑛, 𝑥, 𝑦) = 𝑥𝑆(𝜈 − 1, 𝑛, 𝑥, 𝑦) + 𝑛𝛽𝑆(𝑣, 𝑛 − 1, 𝑥 + 𝛽, 𝑦) By repeated use of the reduction formula, we can show that 𝑆(1, 𝑛, 𝑥, 𝑦) = ∑ 𝑛 𝑘=0 ( 𝑛 𝑣 ) 𝑣! 𝛽 𝑘 (𝑥 + 𝑦 + 𝑛𝛽) 𝑛−𝑣 as 𝑥𝑆(0, 𝑛, 𝑥, 𝑦) = (𝑥 + 𝑦 + 𝑛𝛽) 𝑛…”