On the Dirichlet Convolution of Completely Additive Functions

Isao Kiuchi and Makoto Minamide Department of Mathematical Sciences Yamaguchi University Yamaguchi 753-8512 Japan

                   kiuchi@yamaguchi-u.ac.jp
                  minamide@yamaguchi-u.ac.jp

Abstract Let k and l be non-negative integers. For two completely additive functions f and g, we consider various identities for the Dirichlet convolution of the kth powers of f and the lth powers of g. Furthermore, we derive some asymptotic formulas for sums of convolutions on the natural logarithms.

1 Statements of results

Let f and g be two arithmetical functions that are completely additive. That is, these functions satisfy f(mn) = f(m) + f(n) and g(mn) = g(m) + g(n) for all positive integers m and n. We shall consider the arithmetical function

D_{k,l}(n; f, g) := ∑_{d|n} f^{k}(d)g^{l} (\frac{n}{d}) ,  (1)

which represents the Dirichlet convolution of the kth power of f and the lth power of g for non-negative integers k and l. The above function provides a certain generalization of the classical number-of-divisors function d(n). In fact,

D_{0,0}(n; f, g) = d(n).

The first purpose of this study is to investigate some recurrence formulas for D_{k,l}(n; f, g) with respect to k and l. Since

∑_{d|n} f(d) = \frac{1}{2}d(n)f(n),  (2)

where f is a completely additive function, we have

D_{1,1}(n; f, g) = \frac{1}{2}d(n)f(n)g(n) − ∑_{d|n} f(d)g(d).  (3)

Similarly, as in (3), we use (1) for D_{k,l+1}(n; f, g) to obtain

D_{k,l+1}(n; f, g) = ∑_{d|n} f^{k}(d)g^{l} (\frac{n}{d}) g (\frac{n}{d})

= g(n) ∑_{d|n} f^{k}(d)g^{l} (\frac{n}{d}) − ∑_{d|n} f^{k}(d)g(d)g^{l} (\frac{n}{d}) .

Hence, we deduce the following two recurrence formulas.

Theorem 1. Let k and l be non-negative integers, and let f and g be completely additive functions. Then we have

D_{k,l+1}(n; f, g) + ∑_{d|n} f^{k}(d)g^{l} (\frac{n}{d}) g(d) = g(n)D_{k,l}(n; f, g),  (4)
D_{k+1,l}(n; f, g) + ∑_{d|n} f^{k}(d)f (\frac{n}{d}) g^{l} (\frac{n}{d}) = f(n)D_{k,l}(n; f, g).  (5)

Now, we put f = g in (4) (or (5)), and set D_{k,l}(n; f) := D_{k,l}(n; f, f). Then, we deduce the following corollary.

Corollary 2. Using the same notation given above, we have

D_{k,l+1}(n; f) + D_{k+1,l}(n; f) = f(n)D_{k,l}(n; f).  (6)

Particularly, if k = l, we have

D_{k+1,k}(n; f) = D_{k,k+1}(n; f) = \frac{1}{2}f(n)D_{k,k}(n; f).  (7)

Because the symmetric property D_{k,l}(n; f) = D_{l,k}(n; f), we only consider the function D_{k,k+j}(n, f) for j = 1, 2, . . ..

Example 3. The formulas (6) and (7) imply that

D_{k,k+2}(n; f) = \frac{1}{2}f^{2}(n)D_{k,k}(n; f) − D_{k+1,k+1}(n; f),
D_{k,k+3}(n; f) = \frac{1}{2}f^{3}(n)D_{k,k}(n; f) − \frac{3}{2}f(n)D_{k+1,k+1}(n; f),
D_{k,k+4}(n; f) = \frac{1}{2}f^{4}(n)D_{k,k}(n; f) − 2f^{2}(n)D_{k+1,k+1}(n; f) + D_{k+2,k+2}(n; f),
D_{k,k+5}(n; f) = \frac{1}{2}f^{5}(n)D_{k,k}(n; f) − \frac{5}{2}f^{3}(n)D_{k+1,k+1}(n; f) + \frac{5}{2}f(n)D_{k+2,k+2}(n; f),
D_{k,k+6}(n; f) = \frac{1}{2}f^{6}(n)D_{k,k}(n; f) − 3f^{4}(n)D_{k+1,k+1}(n; f) + \frac{9}{2}f^{2}(n)D_{k+2,k+2}(n; f)

− D_{k+3,k+3}(n; f),

D_{k,k+7}(n; f) = \frac{1}{2}f^{7}(n)D_{k,k}(n; f) − \frac{7}{2}f^{5}(n)D_{k+1,k+1}(n; f) + 7f^{3}(n)D_{k+2,k+2}(n; f)

− \frac{7}{2}f(n)D_{k+3,k+3}(n; f),

D_{k,k+8}(n; f) = \frac{1}{2}f^{8}(n)D_{k,k}(n; f) − 4f^{6}(n)D_{k+1,k+1}(n; f) + 10f^{4}(n)D_{k+2,k+2}(n; f)

− 8f^{2}(n)D_{k+3,k+3}(n; f) + D_{k+4,k+4}(n; f).

Next, we shall demonstrate that the explicit evaluation of the function D_{k,k+m}(n; f) (m = 2, 3, …) can be expressed as a combination of the functions D_{k,k}(n; f), D_{k+1,k+1}(n; f), D_{k+2,k+2}(n; f), …, D_{k+⌊ \frac{m}{2} ⌋,k+⌊ \frac{m}{2} ⌋}(n; f). Hence, we shall give a recurrence formula between D_{k,k}(n; f), …, D_{k+⌊ \frac{m}{2} ⌋,k+⌊ \frac{m}{2} ⌋}(n; f) and D_{k,k+m}(n; f).

Theorem 4. Let k and m be positive integers, and let D_{k,k+m}(n; f) be the function defined by the above formula. Then we have

D_{k,k+m}(n; f) = ∑_{j=0}^{⌊ \frac{m}{2} ⌋} c_{k,j}^{(m)}f^{m−2j}(n)D_{k+j,k+j}(n; f),  (8)

where

c_{k,j}^{(m)} = \begin{cases}\frac{1}{2}, & if j = 0; \\ −\frac{m}{2}, & if j = 1; \\ (−1)^{j} \frac{m}{2 · j!} ∏_{i=1}^{j−1} (m − (j + i)) , & if 2 ≤ j ≤ ⌊\frac{m}{2}⌋. \end{cases}

Proof. By (7) in Corollary 2, the equality (8) holds for m = 1 and all k ∈ ℕ. Now, we assume that (8) is true for m = 1, 2, …, l and k ∈ ℕ. Using this assumption and (6) in Corollary 2, we observe that

D_{k,k+l+1}(n; f) = ∑_{j=0}^{⌊ \frac{l}{2} ⌋} c_{k,j}^{(l)}f^{l+1−2j}(n)D_{k+j,k+j}(n; f)

− ∑_{j=0}^{⌊ \frac{l−1}{2} ⌋} c_{k+1,j}^{(l−1)}f^{l−1−2j}(n)D_{k+1+j,k+1+j}(n; f).

For even l = 2q, we have

D_{k,k+2q+1}(n; f) = \frac{1}{2}f^{2q+1}(n)D_{k,k}(n; f)

+ ∑_{j=1}^{q} (c_{k,j}^{(2q)} − c_{k+1,j−1}^{(2q−1)}) f^{2q+1−2j}(n)D_{k+j,k+j}(n; f)

and

c_{k,j}^{(2q)} − c_{k+1,j−1}^{(2q−1)} = (−1)^{j} \frac{2q + 1}{2 · j!} ∏_{i=1}^{j−2} (2q − (j + i)) (2q − j) = c_{k,j}^{(2q+1)}.

For odd l = 2q − 1, we observe that

D_{k,k+2q}(n; f) = \frac{1}{2}f^{2q}(n)D_{k,k}(n; f)

+ ∑_{j=1}^{q−1} (c_{k,j}^{(2q−1)} − c_{k+1,j−1}^{(2q−2)}) f^{2q−2j}(n)D_{k+j}(n; f)

+ (−1)^{⌊ \frac{2q}{2} ⌋}D_{k+⌊ \frac{2q}{2} ⌋,k+⌊ \frac{2q}{2} ⌋}(n; f).

By our assumption, since

c_{k,j}^{(2q−1)} − c_{k+1,j−1}^{(2q−2)} = (−1)^{j} \frac{2q}{2 · j!} ∏_{i=1}^{j−2} (2q − 1 − (j + i)) (2q − 1 − j) = c_{k,j}^{(2q)},

we obtain the assertion (8) for all k and m ∈ ℕ.

Now, we consider another expression for D_{k,l}(n; f, g) using the arithmetical function

H_{k,m}(n; f, g) := ∑_{d|n} f^{k}(d)g^{m}(d).  (9)

If f = g, we set H_{k+m}(n; f) = H_{k,m}(n; f, f). The right-hand side of (9) implies the Dirichlet convolution of 1 and f^{k}g^{m}. Since g is a completely additive function, we have

D_{k,l}(n; f, g) = ∑_{d|n} f^{k}(d) (g(n) − g(d))^{l}

= ∑_{d|n} f^{k}(d) ∑_{m=0}^{l}(−1)^{m} (\begin{matrix}l \\ m \end{matrix}) g^{l−m}(n)g^{m}(d).

From (9) and the above, we obtain the following theorem.

Theorem 5. Let k and l be non-negative integers, and let f and g be completely additive functions. Then we have

D_{k,l}(n; f, g) = ∑_{m=0}^{l}(−1)^{m}\binom{l}{m}g^{l−m}(n)H_{k,m}(n; f, g),

where the function H_{k,m}(n; f, g) is defined by (9).

We immediately obtain the following corollary.

Corollary 6. Let k and l be non-negative integers, and let f and g be completely additive functions. Then we have

D_{k,l}(n; f) = ∑_{m=0}^{l}(−1)^{m}\binom{l}{m}f^{l−m}(n)H_{k+m}(n; f).  (10)

Note that

H_{k,m}(n; f, g) = ∑_{d|n} f^{k} (\frac{n}{d}) g^{m} (\frac{n}{d})

= ∑_{i=0}^{k} ∑_{j=0}^{m}(−1)^{i+j}\binom{k}{i}\binom{m}{j}f^{k−i}(n)g^{m−j}(n) ∑_{d|n} f^{i}(d)g^{j}(d).  (11)

Applying (11) to Theorem 5, we have the following theorem.

Theorem 7. Let k and l be non-negative integers, and let f and g be completely additive functions. Then we have

D_{k,l}(n; f, g)

= ∑_{m=0}^{l} ∑_{i=0}^{k} ∑_{j=0}^{m}(−1)^{m+i+j}\binom{l}{m}\binom{k}{i}\binom{m}{j}f^{k−i}(n)g^{l−j}(n) ∑_{d|n} f^{i}(d)g^{j}(d).

In the case where f = g, note that

H_{k+m}(n; f) = ∑_{d|n} (f(n) − f(d))^{k+m}

= ∑_{j=0}^{k+m}(−1)^{j}\binom{k + m}{j}f^{k+m−j}(n) ∑_{d|n} f^{j}(d).

From (10) and the above, we obtain the following corollary.

Corollary 8. Let k and l be non-negative integers, and let f be a completely additive function. Then we have

D_{k,l}(n; f) = ∑_{m=0}^{l} ∑_{j=0}^{k+m}(−1)^{m+j}\binom{l}{m}\binom{k + m}{j}f^{k+l−j}(n) ∑_{d|n} f^{j}(d).  (12)

2 Recurrence formula connecting D_{k,l}(n; f) with ∑_{d|n} f^{j}(d)

The second purpose of this study is to derive another expression for D_{k,l}(n; f) that involves the divisor function d(n). Before stating Theorem 10, we prepare the following lemma.

Lemma 9. Let f be a completely additive function. There exist the constants e_{q,q}, e_{q,q−1}, …, e_{q,1} (q = 1, 2, …) that satisfy the equation

∑_{d|n} f^{2q−1}(d) = e_{q,q}d(n)f^{2q−1}(n) + ∑_{j=1}^{q−1} e_{q,q−j}f^{2q−2j−1}(n) ∑_{d|n} f^{2j}(d).  (13)

Moreover, the relations among sequences (e_{q,q−j})_{j=1}^{q} are as follows.

e_{q,q} = \frac{1}{2} (1 − ∑_{j=1}^{q−1} \binom{2q − 1}{2j − 1}e_{j,j}) = \frac{(2^{2q} − 1) B_{2q}}{q} ,  (14)

e_{q,q−j} = \frac{1}{2} (\binom{2q − 1}{2j} − ∑_{{i=2 \atop i−j≥1}}^{q−1} \binom{2q − 1}{2i − 1}e_{i,i−j}) ,

where B_{n} denotes the nth Bernoulli number, which is defined by the Taylor expansion

\frac{z}{e^{z} − 1} = ∑_{n=1}^{∞} \frac{B_{n}}{n!} z^{n}, (|z| < 2π).

Proof. By (2), the case q = 1 in (13) is trivial. Assume that there exist e_{p,p}, e_{p,p−1}, …, e_{p,1} (p ≤ q) such that

∑_{d|n} f^{2p−1}(d) = e_{p,p}d(n)f^{2p−1}(n) + ∑_{j=1}^{p−1} e_{p,p−j}f^{2p−2j−1}(n) ∑_{d|n} f^{2j}(d).  (15)

Since

∑_{d|n} f^{2q+1}(d) = ∑_{j=0}^{2q+1}(−1)^{j}\binom{2q + 1}{j}f^{2q+1−j}(n) ∑_{d|n} f^{j}(d),

we have

∑_{d|n} f^{2q+1}(d) = \frac{1}{2}d(n)f^{2q+1}(n) + \frac{1}{2} ∑_{j=1}^{q} \binom{2q + 1}{2j}f^{2q+1−2j}(n) ∑_{d|n} f^{2j}(d)

− \frac{1}{2} ∑_{j=1}^{q} \binom{2q + 1}{2j − 1}f^{2q+2−2j}(n) ∑_{d|n} f^{2j−1}(d).  (16)

Applying (15) to (16), we obtain

∑_{d|n} f^{2q+1}(d) =\frac{1}{2} (1 − ∑_{j=1}^{q} \binom{2q + 1}{2j − 1}e_{j,j}) d(n)f^{2q+1}(n)

+ \frac{1}{2} ∑_{j=1}^{q} (\binom{2q + 1}{2j} − ∑_{{i=2 \atop i−j≥1}}^{q} \binom{2q + 1}{2i − 1}e_{i,i−j}) f^{2q−2j+1}(n) ∑_{d|n} f^{2j}(d).

By induction, this completes the proof, except for the second term on the right-hand side of (14).

The first term on the right-hand side of (14) implies

e_{q,q} = 1 − ∑_{k=1}^{q} \binom{2q − 1}{2k − 1}e_{k,k} = 1 − ∑_{k=1}^{q} \binom{2q}{2k}\frac{k}{q} e_{k,k}.

Here we put a(k) = ke_{k,k}. Then we have

a(q) = q − ∑_{k=1}^{q} \binom{2q}{2k}a(k).  (17)

Since a(1) = e_{1,1} = 1/2 and (2^{2} − 1) B_{2} = 1/2, we only need to show that (2^{2k} − 1) B_{2k} (k = 1, …, q) satisfies the recurrence formula (17). Consider the nth Bernoulli polynomial B_{n}(x), which is defined by the following Taylor expansion:

\frac{ze^{xz}}{e^{z} − 1} = ∑_{n=0}^{∞} \frac{B_{n}(x)}{n!}z^{n} (|z| < 2π).

The following relations are known among B_{n}(1), B_{n}(1/2) and B_{n},

B_{n}(1) = B_{n}, B_{n} (\frac{1}{2}) = − (1 − 2^{1−n}) B_{n}.

By the formula [1, Thm. 12.12, p. 264]

B_{n}(y) = ∑_{k=0}^{n} \binom{n}{k}B_{k}y^{n−k},  (18)

we observe that

B_{2q}(y) = y^{2q} − qy^{2q−1} + ∑_{k=2}^{2q} \binom{2q}{k}B_{k}y^{2q−k}.

In this equation, we consider y = 1 and y = 1/2; then

B_{2q} = 1 − q + ∑_{k=1}^{q} \binom{2q}{2k}B_{2k}  (19)

and

2^{2q}B_{2q} (\frac{1}{2}) = (2 − 2^{2q}) B_{2q}
= 1 − 2q + ∑_{k=1}^{q} \binom{2q}{2k}B_{2k}2^{2k}.  (20)

Subtracting (20) from (19), we obtain

(2^{2q} − 1) B_{2q} = q − ∑_{k=1}^{q} \binom{2q}{2k} (2^{2k} − 1) B_{2k}.

This recurrence formula for (2^{2k} − 1) B_{2k}’s is equivalent to (17). This completes the proof of (13).

Applying Lemma 9 to (12) in Corollary 8, we have

D_{k,l}(n; f) = ∑_{m=0}^{l}(−1)^{m}\binom{l}{m} ∑_{j=0}^{⌊ \frac{k+m}{2} ⌋} \binom{k + m}{2j}f^{l+k−2j}(n) ∑_{d|n} f^{2j}(d)

− ∑_{m=0}^{l}(−1)^{m}\binom{l}{m} ∑_{j=1}^{⌊ \frac{k+m+1}{2} ⌋} \binom{k + m}{2j − 1}f^{l+k+1−2j}(n) ∑_{d|n} f^{2j−1}(d).  (21)

The second term on the right-hand side of (21) gives us

− (∑_{m=0}^{l}(−1)^{m}\binom{l}{m} ∑_{j=1}^{⌊ \frac{k+m}{2} ⌋} e_{j,j}\binom{k + m}{2j − 1}) f^{l+k}(n)d(n)

− ∑_{m=0}^{l} ∑_{j=1}^{⌊ \frac{k+m+1}{2} ⌋} ∑_{i=1}^{j−1}(−1)^{m}\binom{l}{m}\binom{k + m}{2j − 1}e_{j,j−i}f^{l+k−2i}(n) ∑_{d|n} f^{2i}(d)  (22)

using (13). From (14), (21) and (22), we have the following theorem.

Theorem 10. Let k and l be non-negative integers, and let f be a completely additive function. There exist the constants e_{j,j}, e_{j,j−i} (j = 1, 2, …, ⌊\frac{k+m+1}{2}⌋, 1 ≤ j − i < j) and A_{k,l} such that

D_{k,l}(n; f) =A_{k,l}f^{l+k}(n)d(n)
+ ∑_{m=0}^{l} ∑_{j=0}^{⌊ \frac{k+m}{2} ⌋}(−1)^{m}\binom{l}{m}\binom{k + m}{2j}f^{l+k−2j}(n) ∑_{d|n} f^{2j}(d)  (23)

− ∑_{m=0}^{l} ∑_{j=1}^{⌊ \frac{k+m}{2} ⌋} ∑_{i=1}^{j−1}(−1)^{m}\binom{l}{m}\binom{k + m}{2j − 1}e_{j,j−i}f^{l+k−2i}(n) ∑_{d|n} f^{2i}(d),

where

A_{k,l} = ∑_{m=0}^{l}(−1)^{m−1}\binom{l}{m} ∑_{j=1}^{⌊ \frac{k+m+1}{2} ⌋} \frac{(2^{2j} − 1)B_{2j}}{j} \binom{k + m}{2j − 1}

= 2 ∑_{m=0}^{l}(−1)^{m}\binom{l}{m}\frac{2^{k+m+1} − 1}{k + m + 1}B_{k+m+1}.  (24)

Proof. We only need to show (24) to complete the proof of Theorem 10. We set

A_{k,l} = ∑_{m=0}^{l}(−1)^{m−1}\binom{l}{m}J_{k,m},  J_{k,m} = ∑_{j=1}^{⌊ \frac{k+m+1}{2} ⌋} \frac{(2^{2j} − 1)B_{2j}}{j} \binom{k + m}{2j − 1}.

From the identity \frac{1}{j} \binom{k+m}{2j−1} = \frac{2}{k+m+1} \binom{k+m+1}{2j}, we have

J_{k,m} = \frac{2}{k + m + 1} ∑_{j=1}^{⌊ \frac{k+m+1}{2} ⌋}(2^{2j} − 1)B_{2j}\binom{k + m + 1}{2j}.

Since B_{2j+1} = 0 if j ≥ 1 and B_{1} = −\frac{1}{2}, we have

J_{k,m} = \frac{2}{k + m + 1} ∑_{j=0}^{k+m+1}(2^{2j} − 1)B_{j}\binom{k + m + 1}{j} + 1.  (25)

Taking y = \frac{1}{2} and y = 1 in (18), we get

2^{n}B_{n} (\frac{1}{2}) = ∑_{j=0}^{n} 2^{j}\binom{n}{j}B_{j} and B_{n} = ∑_{j=0}^{n} \binom{n}{j}B_{j},

respectively. Hence we have

2^{n}B_{n} (\frac{1}{2}) − B_{n} = ∑_{j=0}^{n} (2^{j} − 1) \binom{n}{j}B_{j}.  (26)

On the other hand, it is known that

B_{n}(mx) = m^{n−1} ∑_{j=0}^{m−1} B_{n} (x + \frac{j}{m}) .  (27)

See [1, p. 275]. Taking x = 0 and m = 2 in (27), we obtain

2^{n}B_{n} (\frac{1}{2}) − B_{n} = − (2^{n} − 1) B_{n}.

Then we have, from (26) and the above,

− (2^{n} − 1) B_{n} = ∑_{j=0}^{n} (2^{j} − 1) B_{j}\binom{n}{j}.

Substituting this relation in (25), we have

J_{k,m} = −\frac{2}{k + m + 1}(2^{k+m+1} − 1)B_{k+m+1} + 1.

Hence we have

A_{k,l} = 2 ∑_{m=0}^{l}(−1)^{m}\binom{l}{m}\frac{2^{k+m+1} − 1}{k + m + 1}B_{k+m+1}.

From (23), we obtain the following corollary.

Corollary 11. For every positive integer k, there are constants c_{k}, c_{k−1}, …, c_{0} such that

D_{k,k}(n; f) = c_{k}d(n)f^{2k}(n) + ∑_{j=1}^{k} c_{k−j}f^{2k−2j}(n) ∑_{d|n} f^{2j}(d).  (28)

From (8) and (28), we obtain the following corollary.

Corollary 12. For every positive integer k and m (≥ 2), there are constants c_{0}, c_{1}, …, c_{k}, …, c_{k+⌊ \frac{m}{2} ⌋} as in Corollary 11 and c_{k,1}^{(m)} = −\frac{m}{2}, c_{k,2}^{(m)} = \frac{m(m−3)}{4}, c_{k,3}^{(m)}, …, c_{k,⌊ \frac{m}{2} ⌋}^{(m)} as in Theorem 4 such that

D_{k,k+m}(n; f) = (\frac{1}{2}c_{k} + ∑_{p=1}^{⌊ \frac{m}{2} ⌋} c_{k,p}^{(m)}c_{k+p}) d(n)f^{2k+m}(n)

+ \frac{1}{2} ∑_{j=1}^{k} c_{k−j}f^{2k+m−2j}(n) ∑_{d|n} f^{2j}(d)

+ ∑_{p=1}^{⌊ \frac{m}{2} ⌋} c_{k,p}^{(m)} ∑_{j=1}^{k+p} c_{k+p−j}f^{2k+m−2j}(n) ∑_{d|n} f^{2j}(d).

Example 13. Corollaries 11 and 12 give us

D_{1,1}(n; f) = \frac{1}{2}d(n)f^{2}(n) − ∑_{d|n} f^{2}(d),

D_{2,1}(n; f) = \frac{1}{4}d(n)f^{3}(n) − \frac{1}{2}f(n) ∑_{d|n} f^{2}(d),

D_{3,1}(n; f) = −\frac{1}{4}d(n)f^{4}(n) + \frac{3}{2}f^{2}(n) ∑_{d|n} f^{2}(d) − ∑_{d|n} f^{4}(d),

D_{2,2}(n; f) = \frac{1}{2}d(n)f^{4}(n) − 2f^{2}(n) ∑_{d|n} f^{2}(d) + ∑_{d|n} f^{4}(d),

D_{4,1}(n; f) = −\frac{1}{2}d(n)f^{5}(n) + \frac{5}{2}f^{3}(n) ∑_{d|n} f^{2}(d) − \frac{3}{2}f(n) ∑_{d|n} f^{4}(d),

D_{3,2}(n; f) = \frac{1}{4}d(n)f^{5}(d) − f^{2}(n) ∑_{d|n} f^{2}(d) + \frac{1}{2}f(n) ∑_{d|n} f^{4}(d).

3 Applications

The second author [2, p. 330] showed an asymptotic formula for ∑_{n≤x} D_{1,1}(n; log)

∑_{n≤x} D_{1,1}(n; log) = \frac{1}{6}x log^{3} x − \frac{1}{2}x log^{2} x + (1 − 2A_{1})x log x
+ (2A_{1} − 4A_{2} − 1)x + O_{ε} (x^{\frac{1}{3} +ε})  (29)

for x > 2 and all ε > 0. Here the constants A_{1} and A_{2} are coefficients of the Laurent expansion of the Riemann zeta-function ζ(s) in the neighbourhood s = 1:

ζ(s) = \frac{1}{s − 1} + A_{0} + A_{1}(s − 1) + A_{2}(s − 1)^{2} + A_{3}(s − 1)^{3} + · · · .

We use (7), (29) and Abel’s identity [1, Thm. 4.2, p. 77] to obtain

∑_{n≤x} D_{2,1}(n; log) = \frac{1}{12}x log^{4} x − \frac{1}{3}x log^{3} x + (1 + A_{1})x log^{2} x
− 2(1 + A_{2})x log x + 2(1 + A_{2})x + O_{ε} (x^{\frac{1}{3} +ε}) .

Furthermore, a generalization of (29) for the partial sums of D_{k,k}(n; log) for positive integers k was considered by the second author [2, Thm. 1.2, p. 326], who demonstrated that there exists a polynomial P_{2k+1} of degree 2k + 1 such that

∑_{n≤x} D_{k,k}(n; log) = xP_{2k+1}(log x) + O_{k,ε} (x^{\frac{1}{3} +ε})  (30)

for every ε > 0. Applying Theorem 4 and the above formula (30), we have the following theorem.

Theorem 14. There exists a polynomial U_{2k+m+1} of degree 2k + m + 1 such that

∑_{n≤x} D_{k,k+m}(n; log) = xU_{2k+m+1}(log x) + O_{k,m,ε} (x^{\frac{1}{3} +ε})

for m = 2, 3, … and every ε > 0.

4 Acknowledgment

The authors deeply thank the referee for carefully reading this paper and indicating some mistakes.

References

[1] T. M. Apostol. Introduction to Analytic Number Theory. Springer-Verlag, 1976.

[2] M. Minamide. The truncated Voronoï formula for the derivative of the Riemann zetafunction. Indian J. Math. 55 (2013), 325–352.

2010 Mathematics Subject Classification: Primary 11A25; Secondary 11P99. Keywords: completely additive function, Dirichlet convolution.

Received April 23 2014; revised versions received May 15 2014; July 9 2014; July 20 2014; July 31 2014. Published in Journal of Integer Sequences, August 5 2014.

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