January 26, 2026
In this paper, we study various classes of partition functions such as those related to the parity of the number of parts, to differences of partition numbers, and to partitions with a repeated smallest part. We establish identities connecting these various classes of partitions. Moreover, our identities help us to extend the Euler’s partition theorem. An analogue of Legendre’s theorem of the partition-theoretic interpretation of Euler’s pentagonal number theorem is also derived. Both combinatorial and \(q\)-series proofs are given for our results.
Let \(A(n)\) denote the number of partitions of \(n\) into distinct parts, and let \(B(n)\) denote the number of partitions of \(n\) into odd parts. Then one of the most celebrated theorems in the theory of partitions is Euler’s theorem, which states that \[\begin{align} A(n)=B(n). \end{align}\]
Very recently, Andrews, Yee and the first author [1] introduced two further important classes of partitions to Euler’s theorem. More explicitly, they derived the following theorem.
Theorem 1. ([[1]][)]For \(n>0\), we have \[\begin{align} \label{main32theorem32eqn} A(n)=B(n)=C(n+1)=\frac{1}{2}D(n+1), \end{align}\qquad{(1)}\] where \(C(n)\) is the number of partitions of \(n\) with largest part even and parts not exceeding half of the largest part are distinct, and \(D(n)\) is the number of partitions of \(n\) into non-negative parts wherein the smallest part appear exactly twice and no other parts are repeated.
Analogues of this theorem in two different directions related to Glaisher’s partition theorem [2] are subsequently obtained by Andrews and Dhar [3] and by Lin and Zang [4].
Note that the partition function \(D(n)\) counts the number of partitions of \(n\) into non-negative parts wherein the smallest part appear exactly twice and no other parts are repeated. Now this naturally raises the question that what happens if the smallest part is allowed to appear an arbitrary number of times, say \(k\), while all other parts remain distinct? Let us denote the number of such partitions by \(D_k(n)\). This question serves as the starting point of the present paper, and it will soon become evident that it opens up several further directions to explore.
Before stating our results, we introduce the following partition functions based on the parity. Let \(B_{k}^e(n)\) (resp. \(B_{k}^o(n)\)) denote the number of partitions of \(n\) in which all parts are odd, except possibly for at most \(k-1\) distinct even parts lying in the interval \([2\ell+2,2\ell+2k-2]\), where \(2\ell-1\) is the largest odd part, the number of such distinct even parts being even (resp. odd).
Moreover, let \(C_{k}^e(n)\) (resp. \(C_{k}^o(n)\)) denote the number of partitions of \(n\) in which the partition contains an even part (say \(2\ell\)), all parts not exceeding \(\ell\) are distinct, and the only parts larger than \(2\ell\) are at most \(k-1\) distinct even parts lying in \([2\ell+2,2\ell+2k-2]\), the number of such distinct even parts being even (resp. odd).
Our first result is stated below.
Theorem 2. For any positive integers \(n\geq1\) and \(k\geq1\), we have \[\begin{align} B_{k}^e(n)=C_{k}^e(n+1),\nonumber \end{align}\] and \[\begin{align} B_{k}^o(n)=C_{k}^o(n+1).\nonumber \end{align}\]
Example 1. We take \(n=8\) and \(k=2\) to illustrate Theorem 2.
| \(B_{2}^e(8)\) | \(C_{2}^e(9)\) | \(B_{2}^o(8)\) | \(C_{2}^o(9)\) |
|---|---|---|---|
| \(7+1\) | \(8+1\) | \(4+1+1+1+1\) | \(4+2+2+1\) |
| \(5+3\) | \(6+3\) | ||
| \(5+1+1+1\) | \(6+2+1\) | ||
| \(3+3+1+1\) | \(4+4+1\) | ||
| \(1+1+1+1+1+1+1+1\) | \(4+3+2\) | ||
| \(3+1+1+1+1+1\) | \(2+2+2+2+1\) |
This table shows that \(B_2^e(8)=6=C_2^e(9)\) and \(B_2^o(8)=1=C_2^o(9)\).
Differences of partition numbers have been studied extensively in the literature, see, for example, [5]–[7], difference?, among others. In particular, differences of partition numbers according to parity are of fundamental importance (see [5], ac? and references therein). A classical example, due to Legendre [8] ([5]), arises from an interpretation of Euler’s pentagonal number theorem, namely, \[\begin{align} \label{legendre} P_e(d,n)-P_o(d,n)= \begin{cases} (-1)^m, & \mathrm{if}\;n=\frac{1}{2}m(3m\pm1), \\ 0, & \text{otherwise}, \end{cases} \end{align}\tag{1}\] where \(P_e(d,n)\) (resp. \(P_o(d,n)\)) denote the number of partitions of \(n\) into an even (resp. odd) number of distinct parts.
In this direction, our next result shows that differences of the partition numbers appearing in Theorem 2 are precisely ‘equal’ to \(D_k(n)\), the number of partitions into non-negative parts wherein smallest part is allowed to repeat exactly \(k\) times, while all other parts remain distinct, as defined earlier.
Theorem 3. Let \(k\) be any natural number. For all \(n\geq2^{k-1}k(2k-1)\), we have \[\begin{align} \label{BkCkDk32eqn} B_{k}^e(n)-B_{k}^o(n)=C_{k}^e(n+1)-C_{k}^o(n+1)=\frac{1}{2}D_{2k}(n+1). \end{align}\qquad{(2)}\]
Remark 1. Equation ?? immediately implies that \(B_{k}^e(n)>B_{k}^o(n)\) and \(C_{k}^e(n)>C_{k}^o(n)\) for \(n\geq2^{k-1}k(2k-1)\), a fact that does not seem to be easy to establish directly.
Our next result connects \(D_k(n)\) with another type of partitions.
Theorem 4. Let \(k\) and \(n\) be any natural numbers. Then, we have \[\begin{align} \label{AkDk32eqn} A_k(n)=\frac{1}{2}D_{k}(n+1), \end{align}\qquad{(3)}\] where \(A_k(n):=\frac{1}{2}\left(P_1(d,n)+P_{k-1}'(d,n)+P_2(d,n)+P_{k-1}^{''}(d,n)\right).\) Here, and throughout the paper, \(P_1(d,n)\) counts partitions of \(n\) into distinct parts whose smallest part exceeds \(1\) and \(P_2(d,n)\) denote the number of partitions of \(n\) into distinct parts for which the difference between the two smallest parts is at least \(2\). Moreover the function \(P_{k-1}'(d,n)\) enumerates partitions of \(n\) in which all parts are distinct except that the smallest part must be \(1\) with multiplicity \(k-1\); and finally \(P_{k-1}^{''}(d,n)\) denote the number of partitions of \(n\) into distinct parts except that the second smallest part appears with multiplicity \(k-1\), and the difference between the two smallest parts is \(1\).
Our next result unifies Theorem 3 and Theorem 4 to produce a generalization of Theorem 1 of Andrews, Kumar, and Yee [1].
Theorem 5. Let \(k\) be any natural number. For all \(n\geq2^{k-1}k(2k-1)\), we have \[\begin{align} \label{genral32result32eqn} A_{2k}(n)=B_{k}^e(n)-B_{k}^o(n)=C_{k}^e(n+1)-C_{k}^o(n+1)=\frac{1}{2}D_{2k}(n+1). \end{align}\qquad{(4)}\]
The special case \(k=1\) of Theorem 5 reduces to Theorem 1, as shown in the following corollary.
Corollary 1. The equation ?? holds true.
Our next result extends Theorem 1 by incorporating two additional partition functions.
Theorem 6. For \(n>0\), we have \[\begin{align} A(n)=B(n)=C(n+1)=\frac{1}{2}D(n+1)=E(n+2)=F(n+1),\nonumber \end{align}\] where \(E(n)\) denotes the number of partitions of \(n\) having all parts odd with unique largest part and \(F(n)\) denotes the number of partitions of \(n\) with unique largest even part and all other parts are odd.
Continuing our discussion of differences of partition numbers and parity of parts, we now present an elegant analogue of Legendre’s assertion 1 in the next result.
Theorem 7. Let \(D_k^e(n)\) (resp. \(D_k^o(n)\)) denote the number of partitions counted by \(D_k(n)\) with the additional condition that the number of parts greater than the smallest part is even (resp. odd). Then \[\begin{align} \label{legendre32analogue32eqn} D_k^e(n)-D_k^o(n)= \begin{cases} (-1)^m, & \mathrm{if}\;1\leq n=\frac{1}{2}m(3m\pm1)\leq k-1, \\ P_e^{(k)}(d,n)-P_o^{(k)}(d,n), & \mathrm{if}\;k-1<n\leq\frac{1}{2}k(k-1),\\ 0, & \text{otherwise}, \end{cases} \end{align}\qquad{(5)}\] where \(P_e^{(k)}(d,n)\) (resp. \(P_o^{(k)}(d,n)\)) denote the number of partitions of \(n\) into an even (resp. odd) number of distinct parts in which no part exceeds \(k-1\).
The theorem above gives the following important corollary.
Corollary 2. For \(n>\frac{k(k-1)}{2}\), among the partitions counted by \(D_k(n)\), those with an even number of parts exceeding the smallest part are equinumerous with those having an odd number of such parts.
Consequently, \(D_k(n)\) is always even for all \(n>\frac{k(k-1)}{2}\), where \(k\) is any positive integer.
Remark 1. Note that while Theorem 1 shows that \(D_2(n)\) (equivalently, \(D(n)\)) is even, Corollary 2 gives another proof of this fact. In fact, it reveals a stronger structural result, namely that for each \(k\), \(D_k(n)\) \((\)and hence \(D_2(n))\) decomposes naturally into two disjoint partition classes.
Example 2. Consider the case \(D_2(7)\) to demonstrate the above result. The admissible partitions of \(7\) counted by \(D_2(n)\) are \[0+0+7 ,\;0+0+6+1,\;0+0+5+2,\;0+0+4+3,\;0+0+4+2+1,\;1+1+5,\;1+1+2+3,\; 2+2+3.\] Among these, the four partitions \[0+0+6+1,\;0+0+5+2,\;0+0+4+3,\;1+1+2+3\] have even number of parts greater than the smallest part, whereas the remaining four partitions \[0+0+7,\;0+0+4+2+1,\;1+1+5,\;2+2+3\] have an odd number of parts greater than the smallest part. Hence, both classes are equinumerous.
Remark 1. In the view of Corollary 2 or Theorem 7, we can say that Theorem 1 above can be reformulated in a more refined way, namely, \[\begin{align} \label{main32theorem32re} A(n)=B(n)=C(n+1)=\frac{1}{2}D(n+1)=D_2^e(n+1)=D_2^o(n+1). \end{align}\qquad{(6)}\] More generally, our Theorem ?? can be recast in the following way: \[\begin{align} \label{genral32result32re} A_{2k}(n)=B_{k}^e(n)-B_{k}^o(n)=C_{k}^e(n+1)-C_{k}^o(n+1)=\frac{1}{2}D_{2k}(n+1)=D_{2k}^e(n+1)=D_{2k}^o(n+1). \end{align}\qquad{(7)}\]
Before stating our further results, we note that Andrews and Bachraoui [9] recently studied the function spt\(k_d(n)\), which counts the number of partitions of \(n\) in which the smallest part occurs exactly \(k\) times and the remaining parts do not repeat. Note that our function \(D_k(n)\) differs from spt\(k_d(n)\) only in that \(D_k(n)\) allows non-negative parts as well. We next derive some properties of \(D_k(n)\). It is clear that its generating function can be given by1 \[\begin{align} \label{generating32function32of32D95k} \sum_{n=0}^{\infty}D_k(n)q^n=\sum_{n=0}^\infty q^{nk}(-q^{n+1};q)_\infty, \end{align}\tag{2}\] here, and throughout the sequel, \[\begin{align} (A;q)_N=(1-A)(1-Aq)\cdots\left(1-Aq^{N-1}\right) \qquad (N\in\mathbb{N}), \end{align}\] and \[\begin{align} (A;q)_0=1, \end{align}\] and \[\begin{align} (A;q)_\infty=\lim_{N\to\infty}(A;q)_N. \end{align}\]
We here note that the study of integer partitions with a repeating smallest part has received considerable attention in recent years; see, for example, [3], [9]–[11], among others.
In the next theorem, we show that the generating function of \(D_k(n)\) in 2 can be written explicitly as follows.
Theorem 8. For any positive integer \(k\), we have \[\begin{align} \label{D95k32in32terms32of32polyn32eqn} \sum_{n=0}^{\infty}D_k(n)q^n&=\sum_{n=0}^\infty q^{nk}(-q^{n+1};q)_\infty\nonumber\\ &=2(-q;q)_\infty\sum_{j=0}^{k-1}(-1)^j(q^{k-j};q)_j+(-1)^k(q;q)_{k-1}. \end{align}\qquad{(8)}\]
The rightmost identity in the above result follows directly from its finite analogue \(q\)-series identity, stated in the next theorem.
Theorem 9. If \(k\ge 1\) and \(N\ge 0\) then \[\begin{align} \label{finite32analogue32eqn} \sum_{j=0}^{N}q^{kj}(-q^{j+1};q)_\infty =(-q;q)_\infty\sum_{j=0}^{k-1}(-1)^{j+k-1}\left(q^{j+1};q\right)_{k-j-1}\left(2-\frac{q^{(N+1)j}}{(-q;q)_N}\right). \end{align}\qquad{(9)}\]
The special case \(k=1\) of this identity is the following result of Andrews andrews32amm?: \[\begin{align} \label{andrews32amm} \sum_{j=0}^{N}\frac{q^{j}}{(-q;q)_j}=2-\frac{1}{(-q;q)_N}. \end{align}\tag{3}\]
The result in Theorem 8 produces the following identity.
Theorem 10. For \(n>k-1\), we have \[\begin{align} \label{D95k43D95k3232eqn} D_{k}(n)+D_{k-1}(n)=D_{k-1}(n-k+1)+2A(n). \end{align}\qquad{(10)}\]
We conclude this section by showing that Theorem 8 produces infinitely many results of the form presented below.
Corollary 3. For \(n\geq4\), we have \[\begin{align} \label{D3A32eqn} D_3(n)=2A(n-3)-2A(n-1)+2A(n). \end{align}\qquad{(11)}\]
This paper is organized as follows. Section 2 is devoted to combinatorial proofs. The \(q\)-series proofs of all the results stated in Section 1 are given in Section 3. We conclude the paper with some remarks and further questions in Section 4.
We first provide a bijection between the sets containing the partitions enumerated by \(D_k(n)\) and \(P_1(d,n-1)+P_2(d,n-1)+P_{k-1}'(d,n-1)+P_{k-1}''(d,n-1),\;n>1\).
Consider a partition \(\lambda=(\lambda_1,\lambda_2,\dots,\lambda_\ell)\) counted by \(D_{k}(n)\). Since \(D_{k}(n)\) can be viewed as counting partitions of \(n\) either into distinct parts or with the smallest part appearing exactly \(k\) times and and other parts remains distinct, we accordingly distinguish between these two cases, namely,
Case 1: all parts \(\lambda_i,\;1\leq i\leq\ell\), are distinct, or
Case 2: the smallest part is repeated \(k\) times and other parts are distinct.
Case 1: \(\lambda_1>\lambda_2>\dots>\lambda_\ell>0\). In this situation, we further distinguish two cases: \(\lambda_\ell>1\) and \(\lambda_\ell=1\).
Subcase 1: First, suppose that \(\lambda_\ell>1\). Subtracting \(1\) from the smallest part \(\lambda_\ell\) produces a partition of \(n-1\) into \(\ell\) distinct parts in which the difference between the two smallest parts is at least \(2\). Hence, the resulting partition is counted by \(P_2(d,n-1)\).
Subcase 2: Next consider the case \(\lambda_\ell=1\). We subtract 1 from the smallest part \(\lambda_\ell\). In this process, we will get a partition of \(n-1\) into \(\ell-1\) distinct parts where the smallest part is greater than \(1\). Thus, the resulting partition is enumerated by \(P_1(d,n-1)\).
Case 2: \(\lambda_1>\lambda_2>\dots>\lambda_{(\ell-(k-1))}=\lambda_{(\ell-(k-2))}=\dots=\lambda_\ell>0\). Similar to the previous case, we again distinguish two cases: \(\lambda_\ell>1\) and \(\lambda_\ell=1\).
Subcase 1: First, suppose that \(\lambda_{(\ell-(k-1))}=\lambda_{(\ell-(k-2))}=\dots=\lambda_{\ell-1}=\lambda_\ell>1\). Upon subtracting 1 from \(\lambda_\ell\), we get a partition counted by \(P_{k-1}''(d,n-1)\) as all the parts are distinct except that the second smallest part appears \(k-1\) times and the difference between the two smallest parts is 1.
Subcase 2: Now consider the case \(\lambda_1>\lambda_2>\dots>\lambda_{(\ell-(k-1))}=\lambda_{(\ell-(k-2))}=\dots=\lambda_{\ell-1}=\lambda_\ell=1\). If we subtract \(1\) from \(\lambda_\ell\), then it leads to a partition counted by \(P_{k-1}'(d,n-1)\) as all the parts are distinct except that the smallest part is \(1\) with multiplicity \(k-1\).
This completes the proof.
Example 3. *To illustrate the above bijection, we consider the case \(n=8\) and \(k=4\). The table below shows how partitions counted by \(D_4(8)\) are mapped to other classes of partitions under this bijection. The partition appearing next to each partition of \(D_4(8)\) is its image.*
| \(D_4(8)\) | \(P_1(d,7)\) | \(P_{k-1}'(d,7)\) | \(P_2(d,7)\) | \(P_{k-1}''(d,7)\) |
|---|---|---|---|---|
| 8 | 7 | |||
| 6+2 | 6+1 | |||
| 7+1 | 7 | |||
| 5+3 | 5+2 | |||
| 5+2+1 | 5+2 | |||
| 4+3+1 | 4+3 | |||
| 4+1+1+1+1 | 4+1+1+1 | |||
| 2+2+2+2 | 2+2+2+1 |
Note that when \(k=1\), \(B_{1}^e(n)=B(n)\), \(C_{1}^e(n+1)=C(n+1)\) and \(B_{1}^o(n)=0\), \(C_{1}^o(n)=0\), where \(B(n)\) and \(C(n)\) are as defined in Theorem 1. A bijection between the sets enumerated by \(B(n)\) and \(C(n+1)\) was provided in [1] through Glaisher’s bijection [2]. We will be referring to this bijection multiple times in our proof.
The proof is carried out by induction on \(k\). The base case \(k=1\) follows from the preceding discussion. We first describe the bijective map for the case \(k=2\) before treating the general case.
The case \(k=2\): Observe that \(B_{2}^e(n)=B(n)\) and \(C_{2}^e(n+1)=C(n+1)\). Since Theorem 1 asserts that \(B(n)=C(n+1)\), the bijection constructed in [1] provides a bijection between the sets of partitions counted by \(B_{2}^e(n)\) and \(C_{2}^e(n+1)\).
We next show a bijection between the sets of partitions enumerated by \(B_{2}^o(n)\) and \(C_{2}^o(n+1)\). To that end, let \(\lambda\) be a partition counted by \(C_{2}^o(n)\). Then it is of the form: \[(2\ell+2)+(2\ell)\;+\;\mathrm{the parts not exceeding \ell are distinct, while parts in (\ell,2\ell] are unrestricted}.\] Now, clearly the partition, \[(2\ell)+\;\mathrm{the parts not exceeding \ell are distinct, while parts in (\ell,2\ell] are unrestricted}.\] is counted by \(C_{2}^e(n-2\ell-2)\). Using the bijection between \(B_{2}^e(n)\) and \(C_{2}^e(n+1)\), this partition can be mapped to a partition \(\mu\) counted by \(B_{2}^e(n-2\ell-3)\). Appending the part \(2\ell+2\) to \(\mu\) then yields a partition enumerated by \(B_{2}^o(n-1)\).
Conversely, consider a partition counted by \(B_{2}^o(n)\). Such a partition is of the form \[(2m)+(2m-3)+ \mathrm{ odd parts}.\] Next observe that the partition, \[(2m-3)+\mathrm{ odd parts}.\] is counted by \(B_{2}^e(n-2m)\). By applying the bijection between \(B_{2}^e(n)\) and \(C_{2}^e(n+1)\), this partition corresponds to a partition \(\mu\) counted by \(C_{2}^e(n-2m+1)\) . Finally, adjoining the part \(2m\) to \(\mu\) produces a partition counted by \(C_{2}^o(n+1)\).
This provides the required bijection for the case \(k=2\).
The general case \(k\): Let \(\lambda\) be a partition counted by \(C_{k}^e(n+1)\), then \(\lambda\) is of the form: \[\begin{align} \underbrace{\mathrm{parts belonging to the set}\; S=\{2\ell+2k-2,2\ell+2k-4,\dots,2\ell+2\}}_{\text{number of parts are even}}\\ +\;2\ell\;+\mathrm{the parts not exceeding}\;\ell\;\mathrm{are distinct, while parts in}\; (\ell,2\ell]\;\mathrm{are unrestricted}. \end{align}\] First of all, note that if no part from the set \(S\) appears in \(\lambda\), then such a partition is precisely counted by \(C(n+1)\). In this case, the known aforementioned bijection maps it to a partition counted by \(B(n)\), and hence they are, in particular, counted by \(B_k^e(n)\).
Now suppose that at least one part from \(S\) appears in \(\lambda\), and let \(m\in S\) be the largest such part. Writing \[m+ \mathrm{remaining parts.}\] we see that the remaining parts contain an odd number of distinct parts from the set \(S\). Consequently, the remaining parts form a partition counted by \(C_{k-1}^o(n-m+1)\). By the induction hypothesis, this partition can be mapped to a partition \(\mu\) counted by \(B_{k-1}^o(n-m)\). Finally, adjoining the part \(m\) to \(\mu\) yields a partition counted by \(B_k^e(n)\).
Conversely, let \(\lambda\) be a partition counted by \(B_{k}^e(n)\), then \(\lambda\) is of the form, \[\begin{align} &\underbrace{\text{parts belonging to the set}\;S=\{2l+2k-2,2l+2k-4,\dots,2l+2\}}_{\text{number of parts are even}}\\ &\qquad\qquad\quad+\;(2l-1)\;+\text{remaining parts are odd}. \end{align}\] Similar to the previous case, if no part from the set \(S\) appears in \(\lambda\), then such a partition is precisely counted by \(B(n)\). In this case, the known aforementioned bijection maps it to a partition counted by \(C(n+1)\), and hence they are, in particular, counted by \(C_k^e(n)\).
Now suppose that at least one part from \(S\) appears in \(\lambda\), and let \(m\in S\) be the largest such part. Writing \[m+ \mathrm{remaining parts}.\] Clearly, these remaining parts represent a partition counted by \(B_{k-1}^o(n-m)\). Now by the induction hypothesis, this partition can be mapped to a partition \(\mu\) counted by \(C_{k-1}^o(n-m+1)\). Finally, adjoining the part \(m\) to \(\mu\) yields a partition counted by \(C_k^e(n+1)\).
In same way, one can give a bijective proof for \[B_{k}^o(n)=C_{k}^o(n+1).\] This completes the proof.
Example 4. For bijection between \(B_{k}^e(n)\) and \(C_{k}^e(n+1)\) for \(k=3\). Consider a partition \((8,6,4,4)\) is a partition counted by \(C_{3}^e(22)\). Clearly, in this case \(2\ell=4\) and the subpartition partition \((6,4,4)\) is counted by \(C_{2}^o(14)\). This subpartition can be mapped to the partition \((6,3,1,1,1,1)\) which is counted by \(B_{2}^o(13)\), by converting the pair \((4,4)\) into \((3,1,1,1,1)\) as follows:
subtract 1 from the largest part \(4\).
apply Glaisher’s bijections to the remaining part \(4\) by writing \(4=2^2.1\) (See [1], [2] for more details).
Finally, adjoining the removed part \(8\) to this new partition, we obtain \((8,6,3,1,1,1,1)\) which is counted by \(B_{3}^e(21)\).
Conversely, the partition \((8,6,3,1,1,1,1)\) which is counted by \(B_{3}^e(21)\) can be mapped back to \((8,6,4,4)\) counted by \(C_{3}^e(22)\). To that end, we convert the subpartition \((3,1,1,1,1)\) into \((4,4)\) using the bijection map between \(B(n)\) and \(C(n+1)\) as follows [1]:
adding 1 to the largest part \(3\), obtaining \(4\),
here \(k(1)=2\) (Note that \(k(a)\) is the smallest integer such that \(2^k>\frac{N}{a}\) and \(\mu=2N-1\) is the largest odd part present in the partition and odd part \(a\le N\) repeat with multiplicity \(f\)). In our case, the part \(a=1\) occurs with multiplicity \(f=4\) and the binary representation of 4 is \(2^2\). Hence, the binary expansion yields (for further details of the method see [[1]][)] \[(1,1,1,1)\longrightarrow4.\]
Example 5. Take \(k=3\) and \(n=7\). It is clear that \(B_3^o(7)=1\), with the unique admissible partition being \(4+1+1+1\), and that \(C_3^o(8)=1\), whose only allowed partition is \(4+2+2\). Hence, there is no alternative choice, these two partitions must correspond to each other under the bijection. This can also be verified directly using the bijection described above.
For the even case, the bijection is described in the following table. The partition written in front of each partition denotes its image under the bijection.
| \(C_{3}^e(8)\) | \(B_{3}^e(7)\) |
|---|---|
| \(8\) | \(7\) |
| \(6+2\) | \(5+1+1\) |
| \(4+4\) | \(3+1+1+1+1\) |
| \(2+2+2+2\) | \(1+1+1+1+1+1+1\) |
Let \(\lambda\) be a partition counted by \(D_{k}(n)\) or counted by \(D_{k-1}(n)\).
Case 1: if \(\lambda\) is counted by \(D_{k}(n)\) then it is of the form \(\lambda_1>\lambda_2>\dots>\lambda_{\ell-k+1}=\lambda_{\ell-k+2}=\dots=\lambda_{\ell+1}=\lambda_\ell\).
If \(\lambda_\ell=1\), we subtract \(1\) from each of the first \(k-1\) smallest parts to obtain a partition counted by \(D_{k-1}(n-k+1)\), in which the smallest part is \(0\), all parts are distinct, and the second smallest part is equal to \(1\).
If \(\lambda_\ell>1\), then subtracting \(1\) from the first \(k-1\) smallest parts yields a partition counted by \(D_{k-1}(n-k+1)\), in which the two smallest parts are consecutive and the smallest part is at least \(1\).
If \(\lambda_\ell=0\) then clearly it is a partition counted by \(A(n)\).
Case 2: if \(\lambda\) is counted by \(D_{k-1}(n)\), then it is of the form \(\lambda_1>\lambda_2>\dots>\lambda_{\ell-k+2}=\dots=\lambda_{\ell+1}=\lambda_\ell\) (smallest part repeats \(k-1\) times).
If \(\lambda_\ell=1\), we subtract \(1\) from each of the first \(k-1\) smallest parts to obtain a partition counted by \(D_{k-1}(n-k+1)\), where the smallest part is \(0\) and all parts are distinct with second smallest part\(>1\).
If \(\lambda_\ell>1\), then subtracting \(1\) each of the first \(k-1\) smallest parts to obtain a partition counted by \(D_{k-1}(n-k+1)\), whose two smallest parts are not consecutive.
If \(\lambda_\ell=0\) then clearly it is a partition counted by \(A(n)\).
By union of cases \(1\) and case \(2\), we get all partitions counted by \(D_{k-1}(n-k+1)\) and \(2A(n)\). This completes the proof of the theorem.
For \(0\leq m\leq k-1\), let \(C_{k,m}(n)\) denote the number of partitions of \(n\) which have an even part (say \(2\ell\)) and parts which are not exceeding than \(\ell\) are distinct, and only \(m\) parts are larger than \(2\ell\) which are distinct and even lying in the interval \([2\ell+2,2\ell+2k-2]\). Moreover, let \(B_{k,m}(n)\) denote the number of partitions of \(n\) having odd parts, and only except \(m\) parts (even and distinct) which lies in the interval \([2\ell+2,2\ell+2k-2]\) (\(2\ell-1\) is the largest odd part). Consider the generating function, (where \(x\) counts the number of parts after \(2\ell\)) \[\begin{align} \sum_{n=1}^{\infty}\frac{(-q;q)_n(-xq^{2n+2};q^2)_{k-1}}{(q^{n+1};q)_n}q^{2n}=\sum_{n=1}^{\infty}\left(\sum_{m=0}^{k-1}C_{k,m}(n)x^m\right)q^n. \end{align}\] Now the left-hand side of the above equation can be rewritten as \[\begin{align} \sum_{n=1}^{\infty}\frac{(-q;q)_n(-xq^{2n+2};q^2)_{k-1}}{(q^{n+1};q)_n}q^{2n}&=\sum_{n=1}^{\infty}\frac{(-xq^{2n+2};q^2)_{k-1}}{(q;q^2)_n}q^{2n}\\ &=q\sum_{n=1}^{\infty}\frac{(-xq^{2n+2};q^2)_{k-1}}{(q;q^2)_n}q^{2n-1}\\ &=q\sum_{n=1}^{\infty}\left(\sum_{m=0}^{k-1}B_{k,m}(n)x^m\right)q^n\\ &=\sum_{n=1}^{\infty}\left(\sum_{m=0}^{k-1}B_{k,m}(n)x^m\right)q^{n+1}. \end{align}\] On comparing the coefficients of \(x^{2j}q^n\) both sides, we deduce that \[\begin{align} B_{k,2j}(n)=C_{k,2j}(n+1). \end{align}\] Now summing over over \(j\), we obtain \[B_{k}^e(n)=C_{k}^e(n+1).\] In the same way, we can prove that \[B_{k}^o(n)=C_{k}^o(n+1).\]
0◻
The first equality is simple consequence of Theorem 2.
We next prove that \[\begin{align} \frac{1}{2}D_{2k}(n)=C_k^e(n)-C_k^o(n),\quad n>2^{k-1}k(2k-1),\nonumber \end{align}\] To prove this, observe that \[\begin{align} \label{before32euler} &\frac{1}{2}\sum_{m=0}^{\infty}q^{2mk}(-q^{m+1};q)_\infty+\frac{1}{2}(q;q)_{2k-1}\nonumber\\ &=\frac{1}{2}(-q;q)_\infty\sum_{m=0}^{\infty}\frac{q^{2mk}}{(-q;q)_m}+\frac{1}{2}(q;q)_\infty\sum_{m=0}^{\infty}\frac{q^{2mk}}{(q;q)_m}, \end{align}\tag{4}\] where we used the Euler’s identity [5] \[\begin{align} \label{euler} \frac{1}{(t;q)_\infty}=\sum_{m=0}^\infty\frac{t^m}{(q;q)_m}, \end{align}\tag{5}\] with replacing \(t\) by \(q^{2k}\). Simplifying the right-hand side of 4 , we arrive at \[\begin{align} \label{after32euler} &\frac{1}{2}\sum_{m=0}^{\infty}q^{2mk}(-q^{m+1};q)_\infty+\frac{1}{2}(q;q)_{2k-1}\nonumber\\ &=\frac{1}{2}(q^2;q^2)_\infty\;\sum_{m=0}^{\infty}\frac{q^{2mk}}{(q^2;q^2)_m}\left(\frac{1}{(q^{m+1};q)_\infty}+\frac{1}{(-q^{m+1};q)_\infty}\right)\nonumber\\ &=\frac{1}{2}(q^2;q^2)_\infty\;\sum_{m=0}^{\infty}\frac{q^{2mk}}{(q^2;q^2)_m}\left(\sum_{n=0}^{\infty}\frac{q^{n(m+1)}}{(q;q)_n}+\sum_{n=0}^{\infty}\frac{(-1)^nq^{n(m+1)}}{(q;q)_n}\right), \end{align}\tag{6}\] where the final step is obtained by applying 5 twice, first with \(t=q^{m+1}\) and then with \(t=-q^{m+1}\). Equation 6 can be rewritten as \[\begin{align} \frac{1}{2}\sum_{m=0}^{\infty}q^{2mk}(-q^{m+1};q)_\infty+\frac{1}{2}(q;q)_{2k-1} &=(q^2;q^2)_\infty\sum_{m,n=0}^{\infty}\frac{1}{2}\left((1+(-1)^n)\right)\frac{q^{2km+n+nm}}{(q;q)_n(q^2;q^2)_m}\\ &=(q^2;q^2)_\infty\sum_{m,n=0}^{\infty}\frac{1}{2}\frac{q^{2km+2n+2nm}}{(q;q)_{2n}(q^2;q^2)_m}\\ &=(q^2;q^2)_\infty\sum_{n=0}^{\infty}\frac{q^{2n}}{(q;q)_{2n}}\sum_{m=0}^{\infty}\frac{q^{2nm+2km}}{(q^2;q^2)_{m}}. \end{align}\] Employing 5 again for the inner sum on the right-hand of the above equation, we deduce that \[\begin{align} \frac{1}{2}\sum_{m=0}^{\infty}q^{2mk}(-q^{m+1};q)_\infty+\frac{1}{2}(q;q)_{2k-1}&=(q^2;q^2)_\infty\sum_{n=0}^{\infty}\frac{q^{2n}}{(q;q)_{2n}(q^{2n+2k};q^2)_\infty}\\ &=\sum_{n=0}^{\infty}\frac{(-q;q)_n(q^{2n+2};q^2)_{k-1}}{(q^{n+1};q)_n}q^{2n}\\ &=\sum_{n=1}^{\infty}\frac{(-q;q)_n(q^{2n+2};q^2)_{k-1}}{(q^{n+1};q)_n}q^{2n}+(q^2;q^2)_{k-1}. \end{align}\] Now, this clearly implies that \[\begin{align} \frac{1}{2}\sum_{n=0}^{\infty} D_{2k}(n)q^n+\frac{1}{2}(q;q)_{2k-1}=\sum_{n=0}^\infty \left(C_k^e(n)-C_k^o(n)\right)q^n+(q^2;q^2)_{k-1}, \end{align}\] Finally, comparing coefficients of \(q^n\) for \(n>2^{k-1}k(2k-1)\), we completes the proof of the second equality of ?? .
0◻
Note that \[\begin{align} \label{enum} &\sum_{n=1}^{\infty}\left(P_1(d,n)+P_{k-1}'(d,n)+P_2(d,n)+P_{k-1}^{''}(d,n)\right)q^{n+1} \nonumber\\ &=q\sum_{n=1}^{\infty}P_1(d,n)q^{n}+q\sum_{n=1}^{\infty}P_{k-1} '(d,n)q^{n}+q\sum_{n=1}^{\infty}P_2(d,n)q^{n}+q\sum_{n=1}^{\infty}P_{k-1}^{''}(d,n)q^{n} \nonumber\\ &=q\sum_{n=1}^{\infty}q^{n+1}(-q^{n+2};q)_\infty+q^{k}(-q^{2};q)_\infty+q\sum_{n=1}^{\infty}q^n(-q^{n+2};q)_\infty+q\sum_{n=1}^{\infty}q^{n+(n+1)(k-1)}(-q^{n+2};q)_\infty \nonumber\\ &=\sum_{n=1}^{\infty}(1+q)q^{n+1}(-q^{n+2};q)_\infty+q^{k}(-q^{2};q)_\infty+\sum_{n=2}^{\infty}q^{nk}(-q^{n+1};q)_\infty \nonumber\\ &=\sum_{n=1}^{\infty}(1+q)q^{n+1}(-q^{n+2};q)_\infty+\sum_{n=1}^{\infty}q^{nk}(-q^{n+1};q)_\infty. \end{align}\tag{7}\] Now observing that the first term in the ultimate line is the generating function for the number of partitions of \(n>1\) into distinct parts, whereas the second term enumerates the partitions of \(n\) in which the smallest part is repeated \(k\) times and all other parts are distinct. Together, these two classes account for the partitions enumerated by \(D_k(n)\).
Now upon comparing the coefficients on both sides of 7 , we complete the proof of our claim in ?? . 0◻
Let \(k=1\) in ?? . Using the facts that \(B_1^e(n)=B(n),\;B_1^o(n)=0=C_1^o(n)\) and \(C_1^e(n)=C(n)\) and \(A_2(n)=A(n)\), \(D_2(n)=D(n)\), we arrive at ?? .
0◻
As it is already known, from Theorem 1, that \(A(n)=B(n)=C(n+1)=\frac{1}{2}D(n+1)\), it is enough to show that \(C(n)=F(n)\) and \(C(n)=E(n+1)\). To that end, note that \[\begin{align} \sum_{n=1}^{\infty}C(n)q^n&=\sum_{n=1}^{\infty}\frac{(-q;q)_n}{(q^{n+1};q)_n}q^{2n}\\ &=\sum_{n=1}^{\infty}\frac{(q^2;q^2)_n}{(q;q)_{2n}}q^{2n}\\ &=\sum_{n=1}^{\infty}\frac{q^{2n}}{(q;q^2)_n}\\ &=\sum_{n=1}^{\infty}F(n)q^n. \end{align}\]
Next consider, \[\begin{align} \sum_{n=0}^{\infty}E(n)q^n &=\sum_{n=0}^{\infty}\frac{q^{2n+1}}{(q;q^2)_n}\\ &=q\sum_{n=0}^{\infty}\frac{q^{2n}}{(q;q^2)_n}\\ &=q+q\sum_{n=1}^{\infty}\frac{q^{2n}}{(q;q^2)_n}\\ &=q+\sum_{n=1}^{\infty}C(n)q^{n+1}. \end{align}\]
Bijective proof of \(B(n)=F(n+1)\) and \(B(n)=E(n+2)\): We obtain a simple bijection between \(B(n)\) and \(F(n+1)\) by adding \(1\) to the largest part of a partition counted by \(B(n)\) and, conversely, subtracting \(1\) from the largest part of a partition counted by \(F(n+1)\). Similarly, a bijection between \(B(n)\) and \(E(n+2)\) is obtained by adding \(2\) to, and subtracting \(2\) from, the largest part of the corresponding partitions. 0◻
It is clear that \[\begin{align} \label{k-1} \sum_{n=1}^\infty\left(D_k^e(n)-D_k^o(n)\right)q^n&=\sum_{n=0}^\infty q^{nk}\left(q^{n+1};q\right)_\infty\nonumber\\ &=(q;q)_{k-1}, \end{align}\tag{8}\] where we employed the \(q\)-binomial theorem [5] \[\begin{align} \sum_{n=0}^\infty\frac{(a;q)_n}{(q;q)_n}z^n=\frac{(az;q)_\infty}{(z;q)_\infty}. \end{align}\] Let \(n\leq k-1\). Observe that the coefficient of \(q^n\) on the right-hand side of 8 is \(P_e(d,n)-P_o(d,n)\). Hence, on comparing coefficients of \(q^n\) on both sides of 8 , we have \[\begin{align} \label{equal} D_k^e(n)-D_k^o(n)=P_e(d,n)-P_o(d,n), \;n\leq k-1. \end{align}\tag{9}\] Now, for \(1\leq n\leq k-1\) of the form \(n=\frac{1}{2}m(3m\pm1)\), applying 1 in 9 , we prove the first part.
Now let \(k-1< n\leq \frac{1}{2}k(k-1)\). Observe that coefficient of \(q^n\) on the right-hand side of 8 is \(P_e^{(k)}(d,n)-P_o^{(k)}(d,n)\). Hence, comparing the coefficients of 8 , we arrive at the second part.
The last part is easy as the coefficients of \(q^n\) on the right-hand side of 8 are zero for every \(n>\frac{1}{2}k(k-1)\). This completes the proof. 0◻
Let \(n>\frac{1}{2}k(k-1)\). The result in Corollary is quite straightforward from the third part of ?? . 0◻
Note that ?? can be rephrased as \[\begin{align} \sum_{n=0}^{N}\frac{q^{kn}}{(-q;q)_n}&=\sum_{i=0}^{k-1}(-1)^{k-1+i}(q^{i+1};q)_{k-1-i}\left(2-\frac{q^{(N+1)i}}{(-q;q)_N}\right). \end{align}\]
We will prove this by induction on \(k\) for any non-negative integer \(N\). Note that our result in ?? holds for \(k=1\) due to the result of Andrews given in 3 . Now suppose ?? is true for \(k-1\). Next, consider \[\begin{align} \sum_{n=0}^{N}\frac{q^{kn}}{(-q;q)_n}&= \sum_{n=0}^{N}\frac{q^{(k-1)n}(-1+(1+q^n))}{(-q;q)_n}\\ &=-\sum_{n=0}^{N}\frac{q^{(k-1)n}}{(-q;q)_n}+ \sum_{n=0}^{N}\frac{q^{(k-1)n}(1+q^n)}{(-q;q)_n}\\ &=-\sum_{n=0}^{N}\frac{q^{(k-1)n}}{(-q;q)_n}+\sum_{n=1}^{N}\frac{q^{(k-1)n}(1+q^n)}{(-q;q)_n}+2\\ &=-\sum_{n=0}^{N}\frac{q^{(k-1)n}}{(-q;q)_n}+\sum_{n=1}^{N}\frac{q^{(k-1)n}}{(-q;q)_{n-1}}+2. \end{align}\] Replacing \(n\) by \(n+1\) in the second sum on the right-hand side and then adding and subtracting the corresponding term with \(n=N+1\), we obtain \[\begin{align} \sum_{n=0}^{N}\frac{q^{kn}}{(-q;q)_n}&=-\sum_{n=0}^{N}\frac{q^{(k-1)n}}{(-q;q)_n}+q^{k-1}\sum_{n=0}^{N}\frac{q^{(k-1)n}}{(-q;q)_n}-\frac{q^{(k-1)(N+1)}}{(-q;q)_N}+2\\ &=-(1-q^{k-1})\sum_{n=0}^{N}\frac{q^{(k-1)n}}{(-q;q)_n}-\frac{q^{(k-1)(N+1)}}{(-q;q)_N}+2. \end{align}\] Now, applying ?? for \(k-1\), which is valid by the induction hypothesis, we obtain \[\begin{align} \sum_{n=0}^{N}\frac{q^{kn}}{(-q;q)_n}&=-(1-q^{k-1})\left(\sum_{i=0}^{k-2}(-1)^{k-2+i}(q^{i+1};q)_{k-2-i}\left(2-\frac{q^{(N+1)i}}{(-q;q)_N}\right)\right)-\frac{q^{(k-1)(N+1)}}{(-q;q)_N}+2\\ &=\sum_{i=0}^{k-2}(-1)^{k-1+i}(q^{i+1};q)_{k-1-i}\left(2-\frac{q^{(N+1)i}}{(-q;q)_N}\right)+\left(2-\frac{q^{(k-1)(N+1)}}{(-q;q)_N}\right)\\ &=\sum_{i=0}^{k-1}(-1)^{k-1+i}(q^{i+1};q)_{k-1-i}\left(2-\frac{q^{(N+1)i}}{(-q;q)_N}\right). \end{align}\] This proves [finite32analogue] for general \(k\), hence we complete the proof. 0◻
Note that ?? can be rewritten as \[\begin{align} &\sum_{n=0}^{N}q^{kn}(-q^{n+1};q)_\infty\\ &=2(-q;q)_\infty\sum_{j=0}^{k-1}(-1)^{k-1+j}(q^{j+1};q)_{k-1-j}-\frac{(-q;q)_\infty}{(-q;q)_N}\sum_{j=0}^{k-1}(-1)^{k-1+j}(q^{j+1};q)_{k-1-j}q^{(N+1)j}\\ &=2(-q;q)_\infty\sum_{j=0}^{k-1}(-1)^{k-1+j}(q^{j+1};q)_{k-1-j}-(-1)^{k-1}(q;q)_{k-1}\frac{(-q;q)_\infty}{(-q;q)_N}\\ &\qquad-(-q;q)_\infty\sum_{j=1}^{k-1}(-1)^{k-1+j}(q^{j+1};q)_{k-1-j}\frac{q^{(N+1)j}}{(-q;q)_N}. \end{align}\] Taking \(N\to\infty\) in the above equation and noting that the last finite sum vanishes, since \(\lim_{N\to\infty}q^{(N+1)j}=0\) for \(j\geq1\), we obtain \[\begin{align} &\sum_{n=0}^{\infty}q^{kn}(-q^{n+1};q)_\infty\\ &=2(-q;q)_\infty\sum_{j=0}^{k-1}(-1)^{k-1+j}(q^{j+1};q)_{k-1-j}+(-1)^{k}(q;q)_{k-1}\\ &=2(-q;q)_\infty\sum_{j=0}^{k-1}(-1)^{k-1+j}(q^{j+1};q)_{k-1-j}+(-1)^{k}(q;q)_{k-1}. \end{align}\] Finally, the result follows upon replacing \(j\) by \(k-1-j\) in the finite sum on the right-hand side of the above expression. 0◻
Observe that ?? can be rewritten as \[\begin{align} \sum_{n\ge0}^{}{D_k(n)q^n} &=2(-q;q)_{\infty}+2(-q;q)_\infty(1-q^{k-1})\sum_{j=1}^{k-1}(-1)^j(q^{k-j};q)_{j-1}+(-1)^k(q;q)_{k-1}\\ &=2(-q;q)_{\infty}-2(-q;q)_\infty(1-q^{k-1})\sum_{j=0}^{k-2}(-1)^j(q^{k-j-1};q)_{j}-(-1)^{k-1}(1-q^{k-1})(q;q)_{k-2}\\ &=2(-q;q)_{\infty}-\left\{2(-q;q)_\infty\sum_{j=0}^{k-2}(-1)^j(q^{k-j-1};q)_{j}+(-1)^{k-1}(q;q)_{k-2}\right\}\\ &\qquad+q^{k-1}\left\{2(-q;q)_\infty \sum_{j=0}^{k-2}(-1)^j(q^{k-j-1};q)_{j}+(-1)^{k-1}(q;q)_{k-2}\right\}\\ &=2\sum_{n\ge0}^{}A(n))q^n-\sum_{n\ge0}^{}D_{k-1}(n)q^n+\sum_{n\ge0}^{}D_{k-1}(n))q^{n+k-1}, \end{align}\] where the last step follows upon employing ?? twice. Now, comparing coefficients of \(q^n\) on both sides completes the proof. 0◻
Substituting \(k=3\) in ?? , we obtain \[\begin{align} \sum_{n=0}^{\infty}D_3(n)q^n&=((q^{2}-1)2q+2)(-q;q)_\infty-(1-q)(1-q^2)\\ &=(2q^3-2q+2)(-q;q)_\infty-1+q^2+q-q^3\\ &=2\sum_{n=3}^{\infty}A(n-3)q^n-2\sum_{n=1}^{\infty}A(n-1)q^n+2\sum_{n=0}^{\infty}A(n)q^n-1+q^2+q-q^3. \end{align}\] Equation ?? now follows upon comparing coefficients for \(n\geq4\). 0◻
In this work we have introduced and studied several classes of partitions, including those involving the parity of parts, differences of partition numbers, and partitions with a repeated smallest part. These types of partition classes have been extensively investigated in the literature. We have provided analytic proofs of all our results, together with bijective proofs for many of them.
We propose the following questions for further investigation.
We provided bijection for every equality in ?? except for those involving \(B_k^e(n)-B_k^o(n)\) or \(C_k^e(n+1)-C_k^o(n+1)\) and \(\frac{1}{2}D_{2k}(n+1)\) or \(A_{2k}(n)\). It would therefore be interesting to construct explicit bijections between these classes of partitions.
By decomposing \(D(n)\) into two classes \(D_2^e(n)\) and \(D_2^o(n)\), we reformulated Theorem 1 as given in ?? . It would be great to find bijections between the sets counted by \(D_2^e(n+1)\) (or \(D_2^o(n+1)\)) and those counted by \(A(n), B(n)\) or \(C(n)\). The same question applies for general \(k\) in ?? .
Theorem 7 gives the following identity \[\begin{align} D_k^e(n)-D_k^o(n)= P_e^{(k)}(d,n)-P_o^{(k)}(d,n), \quad \mathrm{if}\;k-1<n\leq\frac{1}{2}k(k-1),\;k>2. \end{align}\] Can one find a direct bijective proof of this equality?
Authors would like to thank Atul Dixit and Aritram Dhar for their valuable comments on an earlier version of the paper. The first author is partially supported by Grant ANRF/ECRG/ 2024/003222/PMS of the Anusandhan National Research Foundation, Govt. of India, and the FIG grants of IIT Roorkee. The second author is supported by UGC fellowship for her PhD. Both the authors sincerely thank these institutions for their support.
Here we use the convention that \(D_k(0)=1\).↩︎