以拉曼努扬精神扩展陈与库珀的一个特性

Florian Münkel, Lerna Pehlivan, Kenneth S. Williams
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引用次数: 0

摘要

Chan 和 Cooper 证明,如果整数({c(n) (n=0,1,2,(点)})由 $$\begin{aligned} 给出,那么 c(n)q^n = (prod_{n=1}^\infty)。\sum _{n=0}^\infty c(n)q^n = \prod _{n=1}^\infty \frac{1}{left( 1-q^n\right) ^2\left( 1-q^{3n}\right) ^2}, \end{aligned}$$那么$$begin{aligned}。\sum _{n=0}^infty c(2n+1)q^n = 2 \prod _{n=1}^infty \frac{left( 1-q^{2n}\right) ^4\left( 1-q^{6n}\right) ^4}{left( 1-q^^n\right) ^6\left( 1-q^{3n}\right) ^6}。\end{aligned}$$ 我们还证明了许多类似的结果,并将它们应用于系数全等性质的确定。
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Extensions of an identity of Chan and Cooper in the spirit of Ramanujan

Chan and Cooper proved that if the integers \({c(n) (n=0,1,2,\ldots )}\) are given by

$$\begin{aligned} \sum _{n=0}^\infty c(n)q^n = \prod _{n=1}^\infty \frac{1}{\left( 1-q^n\right) ^2\left( 1-q^{3n}\right) ^2}, \end{aligned}$$

then

$$\begin{aligned} \sum _{n=0}^\infty c(2n+1)q^n = 2 \prod _{n=1}^\infty \frac{\left( 1-q^{2n}\right) ^4\left( 1-q^{6n}\right) ^4}{\left( 1-q^n\right) ^6\left( 1-q^{3n}\right) ^6}. \end{aligned}$$

We prove many other results of this type and apply them to the determination of congruence properties of the coefficients.

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