Correction to: Infinitely Many Congruences for k-Regular Partitions with Designated Summands

1 Correction to: Bull Braz Math Soc, New Series https://doi.org/10.1007/s00574-019-00156-x

In our original paper da Silva and Sellers (2019), Eq. (42) was not correctly quoted from one of the references, which led to some minor errors that do not affect the final results.

In da Silva and Sellers (2019) we proved infinitely many congruences for the number of k-regular partitions with designated summands, denoted by $PD_k(n)$. In order to do so, we made use of some known 2- and 3-dissections of certain quotients of eta functions. One of the 3-dissections was wrongly quoted from Chan (2010), namely Eq. (42). Below we indicate the small changes that are necessary to correct the minor errors caused by having wrongly quoted Eq. (6) from Chan (2010).

• 1. Eq. (42) should be replaced by

  $$\begin{aligned} \displaystyle \frac{1}{f_{1}f_{2}}&= \displaystyle \frac{f_{9}^{9}}{f_{3}^6f_{6}^2f_{18}^{3}} + q\displaystyle \frac{f_{9}^{6}}{f_{3}^5f_{6}^3} + 3q^2\displaystyle \frac{f_{9}^{3}f_{18}^{3}}{f_{3}^4f_{6}^4} -2q^3\displaystyle \frac{f_{18}^{6}}{f_{3}^3f_{6}^5} +4q^{4}\displaystyle \frac{f_{18}^{9}}{f_{3}^2f_{6}^6f_{9}^{3}}. \end{aligned}$$

• 2. The two identities after (46) should be replaced by

  $$\begin{aligned} \sum _{n=0}^{\infty } PD_{6}(n)q^n&= \frac{f_{6}^2f_{12}f_{18}}{f_{3}f_{36}} \left( \displaystyle \frac{f_{9}^{9}}{f_{3}^6f_{6}^2f_{18}^{3}} + q\displaystyle \frac{f_{9}^{6}}{f_{3}^5f_{6}^3} + 3q^2\displaystyle \frac{f_{9}^{3}f_{18}^{3}}{f_{3}^4f_{6}^4} \right. \\&\quad \left. -2q^3\displaystyle \frac{f_{18}^{6}}{f_{3}^3f_{6}^5} +4q^{4}\displaystyle \frac{f_{18}^{9}}{f_{3}^2f_{6}^6f_{9}^{3}} \right) , \end{aligned}$$

  and

  $$\begin{aligned} \sum _{n=0}^{\infty } PD_{6}(3n+1)q^{3n+1} = q\frac{f_9^6f_{12}f_{18}}{f_{3}^6f_{6}f_{36}} + 4q^4\frac{f_{12}f_{18}^{10}}{f_{3}^3f_{6}^4f_9^3f_{36}}. \end{aligned}$$

• 3. The proof of (13) should be replaced by: In order to prove (13), we replace (42) in (49) to get

  $$\begin{aligned} \sum _{n=0}^{\infty } PD_{12}(3n+1)q^{3n+1} = \frac{f_{6}f_{12}f_{24}f_{36}}{f_{3}f_{72}} \left( q\frac{f_{9}^6}{f_{3}^5f_{6}^3} + 4q^{4}{\frac{f_{18}^9}{f_{3}^2f_{6}^6f_{9}^3}} \right) , \end{aligned}$$

  from which we obtain

  $$\begin{aligned} \sum _{n=0}^{\infty } PD_{12}(3n+1)q^{n}&= \frac{f_{2}f_{4}f_{8}f_{12}}{f_{1}f_{24}} \left( \frac{f_{3}^6}{f_{1}^5f_{2}^3} + 4q{\frac{f_{6}^9}{f_{1}^2f_{2}^6f_{3}^3}} \right) \\&= \frac{f_{3}^6f_{4}f_{8}f_{12}}{f_{1}^6f_2^2f_{24}} + 4q{ \frac{f_{4}f_{6}^9f_8f_{12}}{f_1^3f_{2}^5f_3^3f_{24}}} \\&\equiv \frac{f_1^2}{f_3^2} \frac{f_{4}f_{6}^4f_{8}f_{12}}{f_{2}^6f_{24}} + 4qf_1f_3 {\frac{f_{6}^7f_{8}f_{12}}{f_{2}^5f_{24}}} \pmod {8}, \end{aligned}$$

  using the fact that, by (43),

  $$\begin{aligned} \frac{f_{4}f_{6}^9f_{8}f_{12}}{f_{1}^4f_2^5f_3^4f_{24}} \equiv \frac{f_{6}^7f_{8}f_{12}}{f_{2}^5f_{24}} \pmod {2}. \end{aligned}$$

  Now, using (32) and (34) it follows that

  $$\begin{aligned} \sum _{n=0}^{\infty } PD_{12}(6n+4)q^{n}&\equiv -2\frac{f_4^2f_{6}^2}{f_{1}^4} + 4{\frac{f_3^6f_{4}^3f_6^5}{f_1^4f_{2}^2f_{12}^3}} \nonumber \\&\equiv -2\frac{f_4^2f_{6}^2}{f_{1}^4} + 4\frac{f_{4}^3f_6^8}{f_1^4f_2^2f_{12}^3} \pmod {8}, \end{aligned}$$

  (1)

  since, by (43),

  $$\begin{aligned} \frac{f_3^6f_{4}^3f_6^5}{f_1^4f_{2}^2f_{12}^3} \equiv \frac{f_{4}^3f_6^8}{f_1^4f_2^2f_{12}^3} \pmod {2}. \end{aligned}$$

  From (25), using (23) and (24), we have

  $$\begin{aligned} \frac{1}{f_{1}^{4}} = \frac{f_{4}^{14}}{f_{2}^{14}f_{8}^{4}} + 4q\frac{f_{4}^{2}f_{8}^{4}}{f_{2}^{10}}. \end{aligned}$$

  (2)

  Hence the odd part of $\frac{1}{f_1^4}$ is divisible by 4. Thus taking the odd part on both sides of (52), we obtain (13) when $k=1$.

The original article was updated.