Letter to the Editor | Open Access
Comment on “Some Congruence Properties of a Restricted Bipartition Function ”
This commentary is intended to correct the errors that happened in . Recall that a partition of a positive integer is a nonincreasing sequence of positive integers whose sum is . A bipartition of a positive integer is a pair of partitions such that the sum of all the parts is . Let be the number of bipartitions of . This function has some beautiful arithmetic properties (see, e.g., ). Recently, there is more and more research on the bipartitions with certain restrictions on each partition, for example, [3–7]. In this short review, we consider the number of bipartitions with the restriction that each part of is divisible by . Let denote the number of bipartitions of such that part of is divisible by . The generating function for is given by  where
We will mainly focus on the case when , where is any positive integer. In , the discussion on congruence identities for was based on an incorrect lemma. We will point out all the incorrect points in that paper in the next section.
2. Comments on the Results of Saikia and Boruah
In , Saikia and Boruah considered the congruence identities for by employing the following lemma.
Lemma 13 of  where is the Rogers-Ramanujan continued fraction and
However, the lemma of this version is wrong, The correct version should be as follows.
Lemma 1. Let be defined as above:
Since Lemma of  is not correct, the discussion in Section is not correct.
First, we can say that the conclusion shown in Theorem of , which says is correct. However, in the course of the proof, there is one mistake. The equation which is equation () of , is not correct. It should be
However, the second conclusion and the third conclusion of Theorem of  are not correct. The proof of Theorem (ii, iii) of  relies on the claim that However, as we have pointed out above, we do not have the factor in . This is the reason why the second conclusion and the third conclusion of Theorem of  cannot hold.
Actually, we do have some 3-dissection formulas to find how and modulo-3 look like. Let us show them now.
Thanks to , we have If we let , and in Entry 31 of [12, p. 48], we can obtain where According to these two formulas, we get that When extracting terms involving , dividing by and replacing by , we obtain When extracting terms involving , dividing by and replacing by , we obtain Up to now, we have obtained the forms of and modulo-3. However, we cannot get any congruence properties of or by the forms of and modulo-3 above.
Conflicts of Interest
The author declares no conflicts of interest.
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Copyright © 2017 Qing Zou. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.