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Arithmetic properties derived from coefficients of certain eta quotients
Journal of Inequalities and Applications volume 2020, Article number: 104 (2020)
Abstract
For a positive integer k, let
be the eta quotients. The coefficients \(\frak{a}_{1} (n)\) can be interpreted as a certain kind of restricted divisor sums. In this paper, we give the signs and modulo values for \(\frak{a}_{1} (n)\) and \(\frak{a}_{2} (m)\) and calculate several convolution sums involving \(\frak{a}_{k} (n)\).
1 Introduction
The study of arithmetical congruences is classical in elementary number theory, and such investigations have been carried out by several mathematicians including Ramanujan and Glaisher. For \(d,m,n\in \mathbb{N}\) and \(r,s\in \mathbb{N} \cup \{0\}\), we define some divisor functions for our further use:
We also use the following convention:
The exact evaluation of the basic convolution sum
first appeared in the letter of Besge to Liouville in 1862 [1]. The evaluation of such sums also appear in the works of Glaisher [5], Lahiri [11], Lehmer [12], Ramanujan [18], Skoruppa [19], and Williams [20]. For instance, Ramanujan obtained the identities
and
using only elementary arguments. For \(a,b,n \in \mathbb{N}\), Ramanujan showed that the sum
can be evaluated in terms of the quantities
for the nine pairs \((a,b) \in \mathbb{N}^{2}\) satisfying
Let
and
Here q denotes a fixed complex number with \(|q| <1\), so that we may write \(q=e^{\pi i \tau }\), where \(\operatorname{Im} (\tau ) >0\). For \(k=2\), the right side of (4) becomes
So we note that
More precisely, we prove the following theorems.
Theorem 1
Let\(n=2^{a} m\)be a positive integer with\((2,m)=1\). Then
In particular, we get:
- (1)
If\(n\equiv 3 \ (\mathrm{mod }\ 4)\), then\(\frak{a}_{1} (n)\equiv 0 \ (\mathrm{mod }\ 16)\).
- (2)
If\(n\equiv 1 \ (\mathrm{mod }\ 4)\), then
$$ \begin{aligned}[b] \frak{a}_{1} (n)\equiv \textstyle\begin{cases} 4 \ (\mathrm{mod }\ 8) & \textit{if } n \textit{ is square}, \\ 0 \ (\mathrm{mod }\ 8) & \textit{otherwise}. \end{cases}\displaystyle \end{aligned} $$ - (3)
If\(a\geq 2\), then
$$ \begin{aligned}[b] \frak{a}_{1} (n)\equiv \textstyle\begin{cases} 8 \ (\mathrm{mod }\ 16) & \textit{if } a=2 \textit{ and } n \textit{ is square}, \\ 0 \ (\mathrm{mod }\ 16) & \textit{if } a=2 \textit{ and } n \textit{ is non-square}, \\ 24 \ (\mathrm{mod }\ 48) & \textit{if } a> 2 \textit{ and } n \textit{ is square}, \\ 0 \ (\mathrm{mod }\ 48) & \textit{otherwise}. \end{cases}\displaystyle \end{aligned} $$
Remark 1
Using computer program, L. Pehlivan and K.S. Williams found exact formula for \(\frak{a}_{1} (n)\) in [17, Theorem 1.2 (ii)] and [21, Table 1]. On the other hand, we will prove Theorem 1 using basic arithmetic tools.
Theorem 2
Let\(2n=2^{k} m\)be an even positive integer with\((2,m)=1\). Then
and\(\frak{a}_{2} (2n) \equiv 0 \ (\mathrm{mod }\ 16\sigma _{3} (m))\). In particular, \(\frak{a}_{2} (4l+2)=16\sigma _{3} (2l+1)\), \(\frak{a}_{2} (8l+4)=-144\sigma _{3} (2l+1)\), and\(\frak{a}_{2} (8l)=-16(\sigma _{3} (2l)-25\sigma _{3} (l))\)for\(l\in \mathbb{N}\).
Theorem 3
Letmandrbe positive integers. Then
Here\(\epsilon (t_{1}, a_{1} -t_{1} ,\ldots, t_{r} ,a_{r} -t_{r}):=\# \{ t_{i}=0\ \textit{or}\ a_{j} -t_{j} =0 | 1\leq i,j\leq r \}\)and\(\bar{\sigma }(0):=1\).
Similarly, we get
Using the theory of modular forms, we can also get several convolution sums.
Theorem 4
Let\(n\geq 2\)be an integer. Then we have
In particular, if\(4\nmid n\), then
whereχis the nontrivial Dirichlet character modulus 4.
To state the next theorems, for \(i=1,2,\ldots ,7\), let
and put
Theorem 5
Let\(n\geq 3\)be an integer. Then we have
Theorem 6
Let\(n\geq 4\)be an integer. Then we have
Remark 2
Comparing old results on convolution sums with Theorem 4, 5, and 6, we have the Table 1.
2 Proof of Theorem 1
In [4, p. 23], we find the curious identity
Putting \(u=\frac{\pi }{2}\) into (13), we get
We obtain three cases:
Therefore we get
Lemma 1
Let\(n\equiv 2 \ (\mathrm{mod }\ 4)\)be a positive integer. Then\(\frak{a}_{1} (n)=0\).
Proof
Let d be a positive divisor of n. If \(d\equiv 1\) (resp., 0) (mod 2), then \(\frac{n}{d}\equiv 0\) (resp., 1) (mod 2). Thus we obtain \(d-\frac{n}{d}\equiv 1 \ (\mathrm{mod }\ 2)\). We easily check that \(\bar{\sigma }(n)=0\) and \(\frak{a}_{1} (n)=-4\bar{\sigma }(n)=0\). This completes the proof of Lemma 1. □
Lemma 2
Let\(n\equiv 3 \ (\mathrm{mod }\ 4)\)be a positive integer. Then\(\frak{a}_{1} (n)=4\sigma (n) > 0\)and\(\frak{a}_{1} (n)\equiv 0 \ (\mathrm{mod }\ 16)\).
Proof
If \(d\equiv 1\) (resp., 3) (mod 4), then \(\frac{n}{d}\equiv 3\) (resp., 1) (mod 4). Thus we obtain that \(d-\frac{n}{d}\equiv 2 \ (\mathrm{mod }\ 4)\). Furthermore, there does not exist a pair \(( d,\frac{n}{d} )\) satisfying \(d\equiv \frac{n}{d} \ (\mathrm{mod }\ 4)\). Hence
and \(\frak{a}_{1} (n) = - 4 \bar{\sigma }(n)=4 \sigma (n) > 0\). Note that
and
By (17) and (18) we get \(\frak{a}_{1} (n) = - 4 \bar{\sigma }(n)=4 \sigma (n) \equiv 0 \ (\mathrm{mod }\ 16)\). This completes the proof of Lemma 2. □
Lemma 3
Let\(n\equiv 1 \ (\mathrm{mod }\ 4)\)be a positive integer. Then\(\frak{a}_{1} (n)=-4\sigma (n) < 0\). In particular,
Proof
First, we consider the case where n is a nonsquare integer. If \(n\equiv 1 \ (\mathrm{mod }\ 4)\) is a nonsquare integer, then we write \(n=p_{1}^{f_{1}} \cdots p_{r}^{f_{r}}\) for some \(f_{i} \equiv 1 \ (\mathrm{mod }\ 2)\) with \(1\leq i \leq r\), where, \(p_{i}\) are odd distinct prime integers. Since \(d \equiv \frac{n}{d} \ (\mathrm{mod }\ 4)\), we have
On the other hand, we obtain \(d \neq \frac{n}{d}\) when \(d|n\). Since \(d+\frac{n}{d} \equiv 0 \ (\mathrm{mod }\ 2)\), we note that
Therefore \(\frak{a}_{1} (n)= -4 \bar{\sigma }(n)= -4 \sigma (n)<0\) and \(\frak{a}_{1} (n)\equiv 0 \ (\mathrm{mod }\ 8)\). Second, we consider the case where n is a square integer. Let \(n = p_{1}^{2e_{1}} \cdots p_{r}^{2e_{r}}\). In this case, all factors of n have their pairs \((d, \frac{n}{d})\) satisfying \(d\neq \frac{n}{d}\) except for \(p_{1}^{f_{1}} \cdots p_{r}^{f_{r}}\). So,
Thus \(\frak{a}_{1} (n) = -4 \bar{\sigma }(n) =-4 \sigma (n) < 0\) and \(\frak{a}_{1} (n) \equiv 4 \ (\mathrm{mod }\ 8)\). These complete the proof of Lemma 3. □
Lemma 4
Let\(n=4m\)be a positive integer with\((2,m)=1\). Then\(\frak{a}_{1} (n)=-8 \sigma (\frac{n}{4}) < 0\)and
Proof
Let \(n=4p_{1}^{e_{1}} \cdots p_{r}^{e_{r}}\) be a positive integer with odd distinct primes \(p_{i}\). All odd divisors d of n satisfy \(d -\frac{n}{d} \equiv 1\ (\mathrm{mod }\ 2)\), so we do not consider them. Hence we only consider the divisor d of n satisfying \(d \equiv \frac{n}{d}\equiv 2 \ (\mathrm{mod }\ 4)\), that is, we can choose \(d=2S_{1}\) and \(\frac{n}{d}=2S_{2}\) with \(4S_{1} S_{2}=n\), where, \(S_{1} \equiv S_{2} \equiv 1 \ (\mathrm{mod }\ 2)\). Thus \(d \equiv \frac{n}{d} \ (\mathrm{mod }\ 4)\). So,
If \(\frac{n}{4}\) is a square integer, then \(\sigma (\frac{n}{4}) \equiv 1 \ (\mathrm{mod }\ 2)\) by [20, p. 28]. So, by (20), \(\frak{a}_{1} (n)=-4 \bar{\sigma }(n) = -8 \sigma (\frac{n}{4})<0\) with \((8,\sigma (\frac{n}{4}))=1\). Therefore \(\frak{a}_{1} (n) \equiv 8 \ (\mathrm{mod }\ 16)\).
On the other hand, if \(\frac{n}{4}\) is not a square, then \(S_{1} \neq \frac{n}{4S_{1}}\) for all \(S_{1} |\frac{n}{4}\). It is obvious that
We have \(\frak{a}_{1} (n)=-8 \sigma (\frac{n}{4}) \equiv 0 \ (\mathrm{mod }\ 16)\) by (20) and (21). These complete the proof of Lemma 4. □
Using (20) and (21), we obtain a more general congruence of the result in Lemma 4.
Corollary 1
If\(n=4p_{1}^{f_{1}} \cdots p_{r}^{f_{r}}\)is a nonsquare integer, then\(\sigma (\frac{n}{4} )\equiv 0 \ (\mathrm{mod }\ (e_{1}+1) \cdots (e_{r}+1))\)and\(\frak{a}_{1} (n) \equiv 0 \ (\mathrm{mod }\ 8(e_{1}+1) \cdots (e_{r}+1))\). Here\(p_{i}\)are distinct odd prime integers.
Lemma 5
Let\(n=2^{a}p_{1}^{e_{1}} \cdots p_{r}^{e_{r}}\)be a positive integer with\(a\geq 3\). Then\(\frak{a}_{1} (n) > 0\). In particular,
Proof
Let \(n=2^{a}m\) be an integer with \((m,2)=1\) and \(a \geq 3\). If \(d\not \equiv \frac{n}{d} \ (\mathrm{mod }\ 2)\), then we do not consider these divisors d of n. Putting \(n= 2^{k}\cdot 2^{a-k}m\) with \(1 \leq k \leq a-1\), assume that \(S_{1} | m\). Then we get
and
and
Here we easily check that \(\sigma (m)=\sigma _{1,1}(\frac{n}{2^{j}};2)\) with \(0\leq j\leq a\). Therefore we obtain that
On the other hand, by [20, p. 28] we obtain
This completes the proof of Lemma 5. □
Proof of Theorem 1
Using Lemmas 1, 2, 3, 4, and 5, we can get the proof of Theorem 1. □
3 The proof of Theorem 2 and Theorem 3
Glaisher [3, p. 300] proved that
In [20, p. 192], a more general formula for natural numbers n is given:
To prove Lemma 7 and 8, we need a Glaisher’s result in [5, p. 11] and [9]:
Lemma 6
Let\(n\in \mathbb{N}\cup \{0 \}\). Then we have\(\frak{a}_{2} (4n+2)>0\). In particular, \(\frak{a}_{2} (4n+2)=16\sigma _{3}{(2n+1)}\).
Proof
By (8) we note that
If \(2m\equiv 2 \) (resp., 0) (mod 4), then \((4n+2-2m)\equiv 0\) (resp., 2) (mod 4). It is easy to check that \(\frak{a}_{1} (2m)=0\) or \(\frak{a}_{1} (4n+2-2m)=0\) by (1). Then we have \(\sum_{m=0}^{2n+1}\frak{a}_{1} (2m)\frak{a}_{1} (4n+2-2m)=0\).
If \(2m-1\equiv 1\) (resp., 3) (mod 4), then \((4n+3-2m)\equiv 1\) (resp., 3) (mod 4).
This completes the proof of Lemma 6. □
Lemma 7
Let\(n\in \mathbb{N}\). Then we have\(\frak{a}_{2} (4n) < 0\)with\((n,2)=1\). In particular, \(\frak{a}_{2} (4n)=-144 \sigma _{3} (n)\).
Proof
By (8) we note that
First, from Lemma 4 we find
Second, we can consider \(\sum_{\mathrm{odd}}\frak{a}_{1} (2m-1)\frak{a}_{1}(4n-(2m-1))\).
If \(2m-1 \equiv 1\) (resp., 3) (mod 4), then \(4n-(2m-1)\equiv 3\) (resp., 1) (mod 4). So, \(\frak{a}_{1}(2m-1)\frak{a}_{1} (4n-(2m-1))=-16\sigma (2m-1)\sigma (4n-(2m-1))\). Thus we note that
by (30).
Third, by Lemma 1 we obtain
Finally, by (16) we have to check that
If \(m\equiv 1\) (resp., 0) (mod 2), then \(4m\equiv 4\) (resp., 0) (mod 8) and \(4n-4m\equiv 0\) (resp., 4) (mod 8). Thus by (20) and (26) we obtain
By (32), (38), and (39) we obtain
It is well known that \(\sigma _{1,1}(n;2)=\sigma _{1}(n)\) for odd n. Therefore the proof of Lemma 7 is completed by (34), (35), (36), (37), and (40). □
To simplify Lemma 8, we introduce a useful formula from [20, p. 26]. Let p be a prime. For k, \(n\in \mathbb{N}\), we have
Lemma 8
Let\(n\in \mathbb{N}\). Then we have\(\frak{a}_{2} (8n) =-16(\sigma _{3} (2n) -25\sigma _{3} (n))> 0\).
Proof
By (8) we note that
First, by (27) we find
Second, using a similar method as for (36), we obtain
Third, by Lemma 1 we obtain
Note that \(8m-4\equiv 8n -8m +4 \equiv 4 \ (\mathrm{mod }\ 8)\) and \(\frak{a}_{1} (8m-4)\frak{a}_{1} (8n-8m+4)=64\sigma (2m-1)\sigma (2n-2m+1)\) by Lemma 4.
Forth, by (30) we get
Fifth, by (27) and (32) we have
It is well known that
by (41). By (42)–(47) we obtain
Let \(n=2^{t} u\) be an integer with \((u,2)=1\). Then
If \(t\geq 1\) (resp., =0), then \(\sigma _{3} (2n) -25 \sigma _{3} ( n)\leq -152\sigma _{3} (u)\) (resp., \(-16\sigma _{3}(n)\)). Therefore \(\frak{a}_{2} (8n) >0\). This completes the proof of Lemma 8. □
Corollary 2
Ifnis an odd integer, then\(\frak{a}_{2} (8n) =256\sigma _{3} ( n)\).
Proof of Theorem 2
We can get the proof of the theorem by using Lemmas 6–8 and Corollary 2, □
Corollary 3
Let\(n,M,N\in \mathbb{N}\)with\(N\equiv 1 \ (\mathrm{mod }\ 2)\)and\(M\geq 3\).
- (1)
\(\sum_{m=1}^{4n+1}m \frak{a}_{1} (m)\frak{a}_{1} (4n+2-m) =(2n+1) \sigma _{3} (2n+1)\).
- (2)
\(\sum_{m=1}^{8n-1}m \frak{a}_{1} (m)\frak{a}_{1} (8n-m) =-64n( \sigma _{3} (2n) -25\sigma _{3} (n) +6\sigma (n) -12\sigma (\frac{n}{2}))\).
- (3)
\(\sum_{m=1}^{4N-1} m \frak{a}_{1} (m)\frak{a}_{1} (4N-m) =-32N( 9 \sigma _{3} (N) -\sigma (N) )\).
- (4)
\(\sum_{k_{1} +k_{2} +k_{3} =M} \frak{a}_{1} (8k_{1})\frak{a}_{1} (8k_{2}) \frak{a}_{1} (8k_{3}) =-18(29 \sigma _{5} (M) -\sigma _{5} (2M)+44 \sigma _{3} (M)-4 \sigma _{3} (2M)-4 \sigma (M) +8 \sigma ( \frac{M}{2}))\).
- (5)
\(\sum_{k_{1} +\cdots k_{l} =N} \frak{a}_{1} (2k_{1})\cdots \frak{a}_{1} (2k_{l})=0\).
Proof
(1) We easily check that
Thus it is clear by Lemma 6.
(2) This is easily proved by using the same method as for (49) and then Lemma 8 and (42).
(3) It is obtained by using the same method as for (49) and then Lemma 7 and (35).
(4) By (27) we obtain
It is obtained by [9, Theorem 3.7].
(5) If \(k_{i} \equiv 0 \ (\mathrm{mod }\ 2)\) for all \(1\leq i\leq l\), then \(N\equiv 0 \ (\mathrm{mod }\ 2)\). This contradicts the fact that N is an odd integer. Thus at least one odd integer \(k_{i}\) exists, which is clear by Lemma 1. □
To prove Theorem 3, we need the following lemma.
Lemma 9
Let\(n\in \mathbb{N}\). Then we have\(\frak{a}_{2} (2n-1)=-8\frak{e}_{1} (2n-1)\).
Proof
Let
To prove Lemma 9, we have to prove that \(f(q)\) is an even function, that is, \(f(q)-f(-q) =0\). By the Jacobi product identity [6, Theorem 3.9] we obtain that
Thus we obtain that \(\frak{a}_{2} (2n-1)=v(2n-1)=-8\frak{e}_{1} (2n-1)\). □
Proof of Theorem 3
Since
we have
From Lemma 9 we have \(\frak{a}_{2} (m) = -8 \frak{e}_{1}(m)\) for m odd. Thus we get
Recall that \(\frak{a}_{1} (0)=1\) and \(\frak{a}_{1} (n) = -4 \bar{\sigma } (n)\) for \(n \ge 1\) from (16). Then we have
where \(\epsilon (t_{1}, a_{1} -t_{1} ,\ldots, t_{r} ,a_{r} -t_{r}):=\# \{ t_{i}=0 \text{ or } a_{j} -t_{j} =0 | 1\leq i,j\leq r \}\). Also, we can obtain another expression:
□
4 The proof of Theorem 4
Proof of Theorem 4
We will use the theory of modular forms. In fact, \(F(q)^{2}= [4]\sum_{m \ge 0} \frak{a}_{2} (m)q^{m}\) is in the space \(M_{4}(\varGamma _{0}(8))\), which is a five-dimensional vector space; see [8, Theorem 3.8]. Let
Then the space \(M_{4}(\varGamma _{0}(8))\) is spanned by \(E_{4}(q)\), \(E_{4}(q^{2})\), \(E_{4}(q^{4})\), \(E_{4}(q^{8})\), and \(Y (q)\). Comparing the Fourier coefficients, we have
and then \(\frak{a}_{2} (n)=16\sigma _{3}(\frac{n}{2}) -288\sigma _{3}( \frac{n}{4})+512\sigma _{3}(\frac{n}{8})-8\frak{e}_{1} (n)\) for \(n \ge 1\). In particular, \(\frak{a}_{2} (n)=-8\frak{e}_{1} (n)\) for n odd. The same result is in Lemma 9.
Meanwhile, since
for \(n \ge 2\), we have
□
5 Several convolution sums
To prove Lemmas 10, 11, and 13, we need the following propositions.
Proposition 1
([9, Lemma 4.1, Corollary 4.7])
LetN (≥3) be an integer.
- (1)
\(\sum_{m=1}^{N} \sigma _{3} (2m-1)\sigma (2N-2m+1) =\frac{1}{32}( \sigma _{5} (2N) -\sigma _{5} (N) )\).
- (2)
\(\sum_{m=1}^{N-1} \sigma _{3}(m) \sigma _{1,1}(N-m;2) = \frac{1}{240} (11\sigma _{5} (N)- 32\sigma _{5} (\frac{N}{2})- 10 \sigma _{3} (N) -\sigma (N)+2\sigma (\frac{N}{2}))\).
- (3)
\(\sum_{m=1}^{m<\frac{N}{2}} \sigma _{3}(m) \sigma _{3}(N-2m) = \frac{1}{2040}\sigma _{7} (N)+ \frac{2}{255}\sigma _{7} (\frac{N}{2})- \frac{1}{240}\sigma _{3} (N) -\frac{1}{240}\sigma _{3} (\frac{N}{2}) + \frac{1}{272}\frak{d} (N)\).
- (4)
\(\sum_{m=1}^{m<\frac{N}{4}} \sigma _{3}(m) \sigma _{3}(N-4m) = \frac{1}{32\text{,}640}\sigma _{7} (N)+ \frac{1}{2176}\sigma _{7} ( \frac{N}{2})+ \frac{2}{255}\sigma _{7} (\frac{N}{4}) - \frac{1}{240} \sigma _{3} (N) -\frac{1}{240}\sigma _{3} (\frac{N}{4}) + \frac{9}{2176}\frak{d} (N) +\frac{9}{136}\frak{d} (\frac{N}{2})\).
Proposition 2
LetN (≥3) be an odd integer.
- (1)
\(\sum_{m=1}^{N} \sigma _{3} (2m-1)\sigma (2N-2m+1) =\sigma _{5} (N)\).
- (2)
\(\sum_{m=1}^{N-1} (\sigma _{3}(2m)-\sigma _{3}(m)) \sigma _{1,1}(N-m;2) = \frac{1}{3} (\sigma _{5} (N)-\sigma _{3} (N))\).
- (3)
\(\sum_{m=1}^{N-1} \sigma _{3}(m) \sigma _{1,1}(N;2) = \frac{1}{240} (11 \sigma _{5} (N)-10\sigma _{3} (N) -\sigma (N))\).
- (4)
\(\sum_{m< \frac{N}{2}} \sigma _{3}(2m) \sigma (N-2m) = \frac{17}{480}\sigma _{5} (N) -\frac{1}{240}\sigma (N) -\frac{1}{32} \frak{c} (N)\).
Proof
(1)–(3) See [7, Corollary 3] and [9, Proposition 3.11, 3.12].
(4) In 1997, Melfi [13], [14] proved that
In 2005, Cheng and Williams [2, Theorem 4.2 (iii)] showed that
By (41) we have
From (53), (54), and (55) we obtain the desired result. □
Lemma 10
IfNis an integer, then\(\sum_{m=1}^{4N-1} \frak{a}_{2} (2m)\frak{a}_{1} (8N-2m) = \frac{16}{5} (22\sigma _{5} (2N) -274\sigma _{5} (N) +5 \sigma _{3} (2N) -125\sigma _{3} (N) -12 \sigma (2N) +24\sigma (N))\).
In particular, ifNis an odd integer, then\(\sum_{m=1}^{4N-1} \frak{a}_{2} (2m)\frak{a}_{1} (8N-2m) = \frac{64}{5}(113\sigma _{5} (N) -25\sigma _{3} (n)-3\sigma (n))\).
Proof
By Lemma 1, (41), and Proposition 1 we obtain
Similarly, by Lemma 1 and Proposition 2, for odd integers N, we obtain
In fact, even if Lemma 1 and Proposition 2 are not used, this equation is easily induced by \(\sigma _{5} (2) =33\), \(\sigma _{3} (2) =9\), and \(\sigma (2) =3\). □
Lemma 11
IfNis an integer, then\(\sum_{m=1}^{4N+1} \frak{a}_{2} (2m)\frak{a}_{1} (8N+4-2m) =-16 (2 \sigma _{5} (2N+1) -9 \sigma _{3} (2N+1) +7 \frak{c} (2N+1))\).
Proof
Let
By Lemma 1, Lemma 4, and (27) we have
Comparing [2, (1.7)] with Proposition 2 (4), we obtain the formula
From [9, Proposition 4.5] we see that integer N satisfy
By Proposition 2 (4) we have
Using [2, \(T_{3,1}(n)\)], we can rewrite \(T_{4}\) as
Finally, we apply (56)–(59) to get the result. □
Now we change our direction to modular forms to see \(T(n)\) defined in (12).
Lemma 12
If\(n \equiv 0 \ (\mathrm{mod }\ 8)\), then\(T(n)= 0\).
Proof
The space of cusp forms \(S_{6}(\varGamma _{0}(16))\) is a seven-dimensional vector space. Explicitly, this space is spanned by the following seven eta quotients:
By the definition of \(T(n)\) the generating function \(\sum_{n=1}^{\infty } T(n)q^{n}\) is in \(S_{6}(\varGamma _{0}(16))\). The space of cusp forms breaks into the spaces of newforms and oldforms. We write
By [15, Theorem 2.27], if \(h(q)=q+\sum_{n=2}^{\infty } a(n)q^{n} \in S_{k}^{\mathrm{new}}(\varGamma _{0}(4N))\) is a normalized newform, then \(a(2)=0\). Applying the Hecke operator \(T_{2}\) to \(h(q)\) (see [15, Definition 2.1]), we get
Since newforms are eigenforms for all Hecke operators (see [8, §4.3]), we have \(T_{2}(h) = \lambda h\) for some constant λ, but since \(a(2)=0\), λ must be zero. So, \(T_{2}(h)\) is identically zero, and thus \(a(2n)=0\) for \(n=1,2,\ldots \) . Thus all forms in \(S_{k}^{\mathrm{new}}(\varGamma _{0}(4))\), \(S_{k}^{\mathrm{new}}(\varGamma _{0}(8))\), and \(S_{k}^{\mathrm{new}}(\varGamma _{0}(16))\) have Fourier expansions of the form \(a(1)q+a(3)q^{3}+a(5)q^{5}+\cdots \) . In particular, \(a(8n)=0\). Now we consider the space \(S_{k}^{\mathrm{old}}(\varGamma _{0}(16))\). For a modular form \(h(q)\), \(V(d)\) is defined by \(h(q)|V(d) = h(q^{d})\). By [15, (2.16)] we have
Note that there are no forms in \(S_{6}(\varGamma _{0}(1))\) and \(S_{6}(\varGamma _{0}(2))\). Thus \(S_{6}^{\mathrm{old}}(\varGamma _{0}(16))\) is obtained from \(S_{6}(\varGamma _{0}(4))\) and \(S_{6}(\varGamma _{0}(8))\). Indeed, \(S_{6}(\varGamma _{0}(4))=S_{6}^{\mathrm{new}}(\varGamma _{0}(4))\) and \(S_{6}(\varGamma _{0}(8))=S_{6}^{\mathrm{new}}(\varGamma _{0}(8)) \oplus S_{6}( \varGamma _{0}(4)) \oplus S_{6}(\varGamma _{0}(4)) \mid V(2)\). By the definition of the V-operators the space \(S_{6}^{\mathrm{old}}(\varGamma _{0}(16))\) is spanned by forms
We can observe that \(b(n),c(n),d(n)=0\) for all \(n \equiv 0 \ (\mathrm{mod }\ 8)\). Therefore, for any form in \(S_{6}(\varGamma _{0}(16))\), its 8nth coefficients vanish. In particular, \(T(n)=0\) for all \(n \equiv 0 \ (\mathrm{mod }\ 8)\). □
Proof Theorem 5
We consider \(F(q)^{3}\). \(F(q)^{3}=\sum_{m \ge 0} \frak{a}_{3} (m) q^{m}\) is a modular form in \(M_{6}(\varGamma _{0}(16))\). This space is 13-dimensional, so \(F(q)^{3}\) can be expressed as a linear combination of thirteen linearly independent modular forms in \(M_{6}(\varGamma _{0}(16))\). We take seven cusp forms \(\sum_{n \ge 1} t_{i}(n)q^{n}\) for \(i=1,2,\ldots ,7\) and choose
in the Eisenstein subspace. Then we obtain the following formula:
Since
we have
Note that
where \(\epsilon (a_{1},a_{2},a_{3})=\#\{a_{i}=0 |i=1,2,3 \}\). Combining this with the previous result, we get
Observe that for \(n \ge 3\),
where the last equality holds by (52). Finally, we have
□
Lemma 13
IfNis an integer, then\(\sum_{m=1}^{N-1} \frak{a}_{2} (8m )\frak{a}_{2} (8(N-m)) = \frac{16}{255} (21\mbox{,}167\sigma _{7} (2N) -1327\sigma _{7} (N) -256 \sigma _{7} (\frac{N}{2}) -20\mbox{,}417 \sigma _{3} (2N) +697 \sigma _{3} (N) +136 \sigma _{3} (\frac{N}{2}) +750 \frak{d} (2N) -2160\frak{d} (N) )\).
Proof
Let
By Lemma 8 we obtain
By (3) and Proposition 1 we deduce Lemma 13. □
Proof of Theorem 6
We can regard \(F(q)^{4}\) as a modular form in the five-dimensional space \(M_{8}(\varGamma _{0}(4))\), which is spanned by \(E_{8}(q)\), \(E_{8}(q^{2})\), \(E_{8}(q^{4})\), \(H(q)\), and \(Y(q)^{2}\), where
Then we get the following formula:
Thus we have
or, equivalently,
since \(H(q)=Y(q^{1/2})^{2}\), so that \(\frak{d}(n)=\frak{e}_{2}(2n)\). In particular, if n is odd, then \(\frak{a}_{4}(n)=-16\frak{d}(n)=-16 \frak{e}_{2}(2n)\). Also, we get
Observing that
we obtain
□
6 Conclusion
Although many other research papers about divisor functions, restricted divisor functions, and the coefficients of modular functions have been written in recent years, active, productive, and applied approaches are still continuing in these areas. For this reason, arithmetic properties for new identities of special numbers and polynomials involving eta quotients and modular forms and combinatorial numbers are constructed. By considering these coefficients with their modular equations, difference equations, and combinatorial equations, we obtained and studied various properties for divisor functions, restricted divisor functions, and some combinatorial numbers. The use of the convolution sums of these divisor functions is helpful in the theory of convolution sums of various restricted divisor functions and also helpful in theories of modular forms, elliptic curves, and partitions.
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The corresponding author was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (NRF-2018R1D1A1B07041132).
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Hwang, J., Li, Y. & Kim, D. Arithmetic properties derived from coefficients of certain eta quotients. J Inequal Appl 2020, 104 (2020). https://doi.org/10.1186/s13660-020-02368-y
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DOI: https://doi.org/10.1186/s13660-020-02368-y