On the Subpartitions of the Ordinary Partitions, Ii

In this note, we provide a new proof for the number of partitions of n having subpartitions of length with gap d. Moreover, by generalizing the definition of a subpartition, we show what is counted by q-expansion ∞ n=1 1 1 − q n ∞ n=0 (−1) n q (an 2 +bn)/2 and how fast it grows. Moreover, we prove there is a special sign pattern for the coefficients of q-expansion ∞ n=1 1 1 − q n


Introduction
Let a 1 a 2 • • • a m be an ordinary partition [1].In a recent paper [9], the first author defines a subpartition of an ordinary partition as follows.Let us fix a positive integer d.Then, for a given partition, a subpartition with gap d is defined as the longest sequence satisfying a 1 > a 2 > • • • > a s and a s > a s+1 , where a i − a j d for all i < j s. a s+1 must be understood as a zero if it comes after the final part.This is a generalization of L. Kolitsch's Rogers-Ramanujan subpartition [10], which is the case d = 2.We call the first condition involving d a gap condition and the second condition a s > a s+1 a tail condition.For convenience, we define the subpartition of the empty partition as the empty partition.We define the length of the subpartition with gap d as the number of parts in the subpartition.When the gap d is clear in the context, we will write "the subpartition" instead of "the subpartition with gap d".In [9], the author uses subpartitions to find combinatorial proofs of entries in Ramanujan's lost notebook [11].Moreover, these subpartitions play a crucial role in obtaining an asymptotic formula for certain q-series involving partial theta functions [8].
Define p(n) to be the number of partitions of n and p(n, , d) to be the number of partitions of n having a subpartition of length with gap d.In [9], by finding a generating function via a case by case argument, the first author proved that Theorem 1.For all nonnegative integers n and and a positive integer d, where, for each nonnegative integer k, In this note, by employing a combinatorial argument, we give a simpler proof.Now we further generalize the notion of subpartitions as follows.We introduce a new parameter t and replace the tail condition by a s − a s+1 t.The case t = 1 is the original definition of a subpartition with gap d.Now we define p(n, , d, t) to be the number of partitions of n having a subpartition of length with gap d and tail condition t.Then, by employing essentially same argument, we can prove the following theorem.where, for each nonnegative integer k, By summing even 's, we see that, for a positive integer a and an integer b with a + b > 0 and a ≡ b (mod 2), we find that where (q; q) ∞ = ∞ n=1 (1 − q n ) and p e (n, d, t) is the number of partitions of n having subpartitions of even length with gap d and tail condition t.Here the assumption on a and b is for the positive integrality of (an 2 + bn)/2 for all positive integers n.From the representation of the partial theta function on the left side of (1), it is not clear at all the positivity of its q-expansion and what it counts.Since n copies of 1 is always counted by p e (n, a, (a + b)/2), the positivity of q-expansion is now clear from the combinatorial description.The case a = b = 1 appears Andrews [2] as a generating function for the number of partitions of n in which the first non-occurrence number as a part is odd, which is a conjugation of partition with subpartition of even length with gap d as noted in [8].When a = 1 and b = 3, we have Among 7 partitions of 5, there are 5 partitions having subpartitions of even length with gap 1 with tail condition 2 as follows: Moreover, by adopting the argument in [8], we can prove the following theorem.Remark 6.The case a = 2 and b = 0 was discussed in [8, Theorem 2].In this case, the sign of p e (n, 2, 1) − p o (n, 2, 1) is alternating.This difference is due to that the generating function is essentially modular in this case.The more precise statement in the second part is also due to that 1 − 2 ∞ n=0 (−1) n q (an 2 +bn)/2 becomes a theta function in these cases.Remark 7. The conditions on a and b, i.e. a > 0, a + b > 0, and a ≡ b (mod 2), are needed just for having non-negative integer exponents in the q-expansion.
This paper is organized as follows.In Section 2, we prove the combinatorial results.By adopting the circle method and elementary q-series manipulation, we will prove Theorem 5 in Section 3.

Proof of Combinatorial Results
For a given partition λ, we always write it in the form λ 1 λ 2 • • • λ m , and for convenience, we define λ s = 0 for all integer s > m.It is well known [1] that Now we define p(n, t, d) to be the number of partitions of n having subpartitions of length m with gap d.Then, the following lemma immediately implies Theorem 1.
Lemma 8.For all nonnegative integers m, By definition, it is clear that the above holds when t = 0, and thus we may assume that t 1.We first observe that q S m,d generates the partition π = (1 + (t − 1)d, 1 + (t − 2)d, . . ., 1).
By employing the same argument, we can easily see that q T m,d,t (q;q)∞ is a generating function for the number of partitions of n having subpartitions of length m with gap d and tail condition t, which implies Theorem 3. Now we turn to the proof of Theorem 4. To this end, we are going to employ Ingham's Tauberian theorem ([7, Theorem 1] and [4, Theorem 5.3]).To apply the Tauberian theorem, we have to show that p e (n, d, t) is weakly increasing.To see this, suppose that λ is a partition of n with subpartition of even length.If there is no subpartition in λ, we add a part 1 to the partition λ.Then, the resulting partition is a partition of n + 1, and since the size of the first two parts remains the same, the length of the subpartition is 0. If λ contains the subpartition, we increase the largest part of λ by 1.Then, the resulting partition is a partition of n + 1 and this operation does not affect the length of the subpartition.It is clear that this map is an injection, thus we observe that p e (n, d, t) p e (n + 1, d, t).Theorem 4 now immediately follows from [3, Theorem 1] and Ingham's Tauberian theorem.

Proof of Theorem 5
For a positive integer a and an integer b with a + b > 0 and a ≡ b (mod 2), define Note that α a,b (N ) = p e (N, a, (a + b)/2) − p o (N, a, (a + b)/2).Therefore, to prove Theorem 5, it suffices to see the sign of α a,b (N ).We are going to get an asymptotic formula for α a,b (N ) by similar argument of Bringmann and Mahlburg [4].Main idea of the proof is that we can get an asymptotic formula by focusing on asymptotic behavior of S a,b (q) near q = 1.Set q = e 2πiτ with τ = x + iy.The following proposition describes an asymptotic behavior of S a,b (q) near q = 1.
the electronic journal of combinatorics 21(4) (2014), #P4.21 To prove this result, we need the following Zagier's result on asymptotic expansions for series (the first generalization of Proposition 3 in [12] with a correction on the sign), Lemma 10.Suppose that h has the asymptotic expansion as t → 0+ and that h and all of its derivatives are of rapid decay at infinity, i.e. ∞ l |h (k) (x)| dx converges for some l > 0.Then, for a > 0, as t → 0+, We can rewrite f a,b (τ ) as follows: where We will find asymptotic formulas for the real and imaginary parts of g a,b (τ ).The real part of g a,b (τ ) can be written as The imaginary part can be treated similarly.
where v s (t) = e −4πat 2 sin(4πast 2 ) and Together with the assumption |x| y, we get Therefore, by considering the Taylor expansion of e − b 2 4a πiτ , as y → 0+ with |x| y.
Next, we consider the behavior of S a,b (q) away from q = 1.
(2) Since (q k ;q k ) k ∞ (q;q)∞ is a generating function for the number of k-core partition, it is clear that q-expansion of the above infinite product is nonnegative.By the result of A. Granvile and K. Ono [6], we know that there is a k-core partition of n if k 4. Thus, when k 4, the q-expansion of (2) has positive coefficients.When k = 3, since the partitions 1 and 1 + 1 are 3-core partitions and since 1 (q 3 ;q 3 )∞ = ∞ n=0 p(n)q 3n , the coefficients of q-expansion of (2) are always positive.Finally, when k = 2, we see that ( 2) is (q 2 ; q 2 ) 2 ∞ (q; q) ∞ (q 4 ; q 4 ) ∞ .

Theorem 4 .
As n tends to infinity, for positive integers d and t, p e (n, d, t) ∼ 1 2 p(n).This is a generalization of [8, Theorem 1] and says that asymptotically half of the partitions of n have subpartitions of even length.Much less obviously, there are inequalities between p e (n, a, (a + b)/2) and p o (n, a, (a + b)/2), where p o (n, a, (a + b)/2) = p(n) − p e (n, a, (a + b)/2), i.e. the number of partitions of n having subpartitions of odd length with gap a and tail condition (a + b)/2.These inequalities are unexpected since both p e (n, a, (a + b)/2) and p o (n, a, (a + b)/2) are asymptotically p(n)/2.Theorem 5.For integers a and b satisfying a > 0, a + b > 0, and a ≡ b (mod 2), we have p e (n, a, (a + b)/2) > p o (n, a, (a + b)/2), if b > 0, p e (n, a, (a + b)/2) < p o (n, a, (a + b)/2), if b < 0, for large enough integers n.Moreover, for b = 0 and even integers a > 2, we have p e (n, a, a/2) > p o (n, a, a/2), for all positive integers n except that the equality holds when n = 2 and a = 4.