A63 INTEGERS 21 (2021) NEWTON SEQUENCES and DIRICHLET CONVOLUTION Klaudiusz Wójcik Faculty of Mathematics, Jagiellonian Univer

#A63 INTEGERS 21 (2021) NEWTON SEQUENCES AND DIRICHLET CONVOLUTION

Klaudiusz W´ojcik Faculty of Mathematics, Jagiellonian University, Krak´ow,Poland [email protected]

Received: 12/4/20, Revised: 4/20/21, Accepted: 5/29/21, Published: 6/3/21

Abstract An integer sequence a : N −→ Z is called a Newton sequence generated by the sequence of integers c : N −→ Z, if the following Newton identities hold a(n) = c(1) a(n − 1) + ... + c(n − 1) a(1) + n c(n). We show that a is a Newton sequence if and only if X f(d) a(n/d) ≡ 0 (mod n), n ≥ 1, d|n for every function f : N −→ Z satisfying the two conditions X f(d) ≡ 0 (mod n), f(1) = ±1. d|n In particular, f may be the M¨obiusfunction, the Euler totient function or the Jordan totient function.

1. Introduction

We recall some basic facts concerning Dirichlet involution (see [2]). Let A(C) be the set of all complex-valued arithmetical functions, i.e., the set of functions deﬁned on N with values in C. By A(Z) ⊂ A(C) we denote the set of all integer-valued sequences. The Dirichlet convolution of functions f, g ∈ A(C) is the function f ∗ g ∈ A(C) deﬁned by X X (f ∗ g)(n) = f(d)g(n/d) = f(d1)g(d2), n ≥ 1.

d|n d1d2=n

For any non-empty subsets A, B of A(C) we put A ∗ B := {a ∗ b : a ∈ A, b ∈ B}. We distinguish the following integer-valued sequences: INTEGERS: 21 (2021) 2

() the multiplicative identity with respect to Dirichlet convolution: ( 1 1, if n = 1 (n) = = n 0, if n > 1,

(µ) the M¨obiusfunction µ:  1, if n = 1  µ(n) = (−1)r, if n is a product of r diﬀerent primes 0, otherwise.

k (Ik) Ik(n) = n for k ≥ 0,

(Sf ) the divisor sum over f ∈ A(Z): X Sf (n) := f ∗ I0(n) = f(d), n ≥ 1, d|n

(ϕ) the Euler totient function ϕ: X ϕ(n) = µ ∗ I1(n) = 1, 1≤k≤n (k,n)=1

(Jk) the Jordan totient function Jk: X Jk(n) = µ ∗ Ik(n) = 1.

1≤a1,...,ak≤n (a1,...,ak,n)=1

The set of arithmetical functions A(C) is a commutative ring under pointwise ad- dition and Dirichlet convolution. The invertible elements of this ring are the arithmetical functions f with f(1) 6= 0. The following well-known facts will be useful later ([2]):

(1) Sµ = µ ∗ I0 = ,

(2) the M¨obiusinversion formula: Sf = f ∗ I0 if and only if f = µ ∗ Sf ,

(3) Sϕ = I1,

(4) SJk = Ik.

Deﬁnition 1. Let a ∈ A(Z) be a sequence of integers. We say that a ≡ 0, if a(n) ≡ 0 (mod n), for all n ≥ 1. INTEGERS: 21 (2021) 3

We put

∗ A (Z) := a ∈ A(Z): a(1) = ±1 , A0(Z) := a ∈ A(Z): a ≡ 0 , ∗ ∗ A0(Z) :=A (Z) ∩ A0(Z), ∗ M(Z) := f ∈ A(Z): Sf ∈ A0(Z) .

∗ Example 2. It follows that Ik ∈ A0(Z) and µ, ϕ, Jk ∈ M(Z). Corollary 3.

(1) Every a ∈ A∗(Z) admits an inverse a−1 ∈ A(Z),

(2) A0(Z) ∗ A0(Z) ⊂ A0(Z),

∗ ∗ ∗ (3) A0(Z) ∗ A0(Z) ⊂ A0(Z),

∗ (4) (A0(Z), ∗) is the commutative group,

∗ (5) M(Z) = µ ∗ g : g ∈ A0(Z) .

Proof. The conditions (1) and (5) are obvious. If a, b ∈ A0(Z), then X a ∗ b(n) = a(d)b(n/d) ≡ 0 (mod n), d|n

−1 ∗ hence (2) follows. We claim that a ≡ 0 for a ∈ A0(Z). We use the induction on n. It is trivial for n = 1. For n > 1, by [[2], Theorem 2.8] and the inductive step, we get that X a−1(n) = ± a(d)a−1(n/d) ≡ 0 (mod n). d|n d>1

Deﬁnition 4 (Newton sequence). A sequence of integers a ∈ A(Z) is called a Newton sequence generated by the sequence of integers c ∈ A(Z), if the following Newton identities hold: for all n ∈ N

a(n) = c(1) a(n − 1) + ... + c(n − 1) a(1) + n c(n).

Denote by AN (Z) the set of Newton sequences, i.e.,

AN (Z) = {a : a is a Newton sequence generated by a sequence of integers c}. INTEGERS: 21 (2021) 4

Example 5 (Sequence of traces). For a square integer matrix A ∈ Ml(Z) we deﬁne a sequence of traces tr [A] ∈ A(Z) by n (tr [A])n := tr A , n ≥ 1.

It follows by the classical Newton’s identities that a Newton sequence a ∈ A(Z) is generated by a ﬁnite sequence c = (c(1), . . . , c(l)) ∈ A(Z) (i.e., c(n) = 0 for n > l) if and only if a is the sequence of traces tr [A], where A is the companion matrix of the monic polynomial w(x) = xl − c(1)xl−1 − ... − c(l − 1)x − c(l) ∈ Z[x], i.e.,  0 0 ... 0 c(l)   1 0 ... 0 c(l − 1)     0 1 ... 0 c(l − 2)  A := A[c(1), . . . , c(l)] :=   ∈ Ml(Z).  . .. .   . 0 . 0 .  0 ... 0 1 c(1)

Since w(X) = det (XIl − A) ∈ Z[X], the eigenvalues of A are the roots of w. Deﬁnition 6 (Dold-Fermat sequence [6]). A sequence a ∈ A(Z) is called a Dold- Fermat sequence, if µ ∗ a ∈ A0(Z), i.e., X µ(d) a(n/d) ≡ 0 (mod n), n ≥ 1. d|n In combinatorics and number theory, Dold-Fermat sequences have also been called pre-realizable sequences, relatively realizable sequences, Gauss sequences and generalized Fermat sequences ([1, 3, 7, 8, 9, 10, 11, 13, 14]). We prove the following theorem.

Theorem 7. For a ∈ A(Z) the following conditions are equivalent: (1) a is a Newton sequence, (2) a is a Dold-Fermat sequence,

(3) f ∗ a ∈ A0(Z), i.e., X f(d) a(n/d) ≡ 0 (mod n), n ≥ 1, d|n

for every f ∈ M(Z). The equivalence of the conditions (1) and (2) was proved in [8]. On the other hand, it was observed in [1] that in the definition of the Dold-Fermat sequence the Möbiusfunction can be replaced by the Euler totient function ϕ. In the last section, we define the unitary analog of the Newton sequences. In particular, we show that a unitary Newton sequence is periodic if and only if it is bounded. We also prove that a p-periodic unitary sequence with prime period has to be constant. INTEGERS: 21 (2021) 5

2. Newton Sequences

Lemma 8. Let a ∈ A(Z) and f ∈ M(Z). The following conditions are equivalent:

(1) µ ∗ a ∈ A0(Z),

(2) f ∗ a ∈ A0(Z). Proof. We ﬁrst show that (1) implies (2). We have

f ∗ a = ∗ f ∗ a = µ ∗ I0 ∗ f ∗ a = (µ ∗ a) ∗ Sf ∈ A0(Z), by Corollary 3. Assume that (2) holds. We have

(µ ∗ a) ∗ Sf = f ∗ a ∈ A0(Z).

∗ −1 Since Sf belongs to A0(Z), Sf does and by Corollary 3, one has

−1 µ ∗ a = Sf ∗ (f ∗ a) ∈ A0(Z).

Remark 9. Observe that in the proof of implication (1) =⇒ (2) we need only that Sf ∈ A0(Z) and f does not have to be invertible. In particular, one can take f of the form f = µ ∗ g for some g ∈ A0(Z). As a consequence, one has the following result.

Theorem 10. Assume that a ∈ A(Z). The following conditions are equivalent:

(1) a ∈ AN (Z),

(2) µ ∗ a ∈ A0(Z),

(3) f ∗ a ∈ A0(Z) for every f ∈ M(Z). Proof. The equivalence of the conditions (1) and (2) was proved in [8].

Lemma 11. Let f ∈ M(Z). Then

−1 F : A0(Z) 3 b 7−→ f ∗ b ∈ AN (Z) is a bijective map.

−1 −1 Proof. If b ∈ A0(Z) then f ∗ f ∗ b = b ∈ A0(Z), hence f ∗ b ∈ AN (Z) by Theorem 10, so F is well-deﬁned. Since F is obviously injective, it is suﬃcient to show its surjectivity. If a ∈ AN (Z) then f ∗ a ∈ A0(Z), hence f ∗ a = b for some −1 b ∈ A0(Z), so a = f ∗ b. INTEGERS: 21 (2021) 6

Corollary 12. The following maps are bijective:

−1 (a) A0(Z) 3 b 7−→ µ ∗ b = Sb ∈ AN (Z),

−1 (b) A0(Z) 3 b 7−→ ϕ ∗ b = Sµ·I1 ∗ b ∈ AN (Z). Example 13 (Newton sequences). The following well-known sequences can be interpreted as Newton sequences: P (1) σ1(n) = SI1 (n) = d|n d,

P k (2) σk(n) = SIk (n) = d|n d , P (3) Sµ·I1 ∗ (n!) = d|n dµ(d)(n/d)!,

P k P nk (4) Sµ·I1 ∗ Ik(n) = d|n dµ(d)(n/d) = d|n µ(d) dk−1 .

Lemma 14. If a ∈ AN (Z) and b ∈ A0(Z), then a ∗ b ∈ AN (Z).

Proof. Since b, µ ∗ a ∈ A0(Z), we have

µ ∗ (a ∗ b) = b ∗ (µ ∗ a) ∈ A0(Z), so a ∗ b ∈ AN (Z). Example 15 (Newton sequences). The following sequences are Newton sequences:

P k (1) a ∗ Ik(n) = d|n d a(n/d), P (2) b(n) = d|n a(n/d) d!.

Corollary 16. If a, b ∈ AN (Z), then µ ∗ a ∗ b ∈ AN (Z). Corollary 17. The map

∗ A0(Z) × AN (Z) 3 (b, a) 7−→ b ∗ a ∈ AN (Z)

∗ is an action of the group A0(Z) on the set AN (Z). We put

AN∗ (Z) = {a ∈ AN (Z): a(1) 6= 0}, AN1 (Z) = {a ∈ AN∗ (Z): a(1) = ±1}, so AN∗ (Z) is the set of all invertible Newton sequences and for every a ∈ AN1 (Z) the inverse of a belongs to A(Z). −1 Example 18. We have I0 ∈ AN1 (Z), but I0 = µ∈ / AN1 (Z). INTEGERS: 21 (2021) 7

Lemma 19. The map ∗ A0(Z) × AN∗ (Z) 3 (b, a) 7−→ b ∗ a ∈ AN∗ (Z) ∗ is a free but not transitive action of the group A0(Z) on the set AN∗ (Z).

Proof. Since b ∗ a(1) = b(1)a(1) = ±a(1) 6= 0, b ∗ a ∈ AN∗ (Z). We ﬁrst show that the action is free. Let b ∗ a = a. This implies b = because a is invertible. Next, we claim that the action is not transitive. It is obvious that I0, 2 · I0 ∈ AN∗ (Z). ∗ Suppose that b ∗ I0 = 2 · I0 for some b ∈ A0(Z). Then b = 2 · and b(1) = 2 6= ±1. ∗ This creates a contradiction because b ∈ A0(Z). Lemma 20. The map ∗ A0(Z) × AN1 (Z) 3 (b, a) 7−→ b ∗ a ∈ AN1 (Z) ∗ is a free and transitive action of the group A0(Z) on AN1 (Z).

Proof. Since b ∗ a(1) = b(1)a(1) = ±1, b ∗ a ∈ AN1 (Z). The action is free by Lemma 19. We show the transitivity of the action. For this, we consider two −1 diﬀerent elements a1 and a2 of AN1 (Z). Let b = a1 ∗ a2. We have b ∗ a1 = a2 and b(1) = ±1, so it is suﬃcient to show that b ∈ A0(Z). Since µ ∗ a1, µ ∗ a2 ∈ A0(Z), we have −1 −1 b = a1 ∗ I0 ∗ µ ∗ a2 = (µ ∗ a1) ∗ (µ ∗ a2) ∈ A0(Z).

−1 ∗ Corollary 21. If a1, a2 ∈ AN1 (Z), then a1 ∗ a2 ∈ A0(Z). −1 Example 22. For σ1 ∈ AN1 (Z), we have σ1 = (µ · I1) ∗ µ, hence (µ · I1) ∗ µ ∗ a ∈ ∗ A0(Z) for all a ∈ AN1 (Z). ∞ Proposition 23. A0(Z) ∩ AN (Z) = {0 }. Proof. We use the induction with respect to n. Assume that n = 1. Since a ∈ A0(Z) ∩ AN (Z), for prime p we have a(p) ≡ 0 (mod p) and a(p) ≡ a(1) (mod p). It follows that a(1) ≡ 0 (mod p) for every prime p, i.e., a(1) = 0. Let n ≥ 2 and assume that a(1) = ... = a(n − 1) = 0. We show that a(n) = 0. Let p > n be a prime number. We have X pn X pn X n µ(d)a = µ(d)a + µ(pd)a ≡ 0 (mod pn), d d d d|pn d|n d|n and consequently, X pn µ(d)a − a(n) ≡ 0 (mod pn). d d|n pn pn On the other hand, a ∈ A0(Z) implies that a d ≡ 0 (mod d ) for all d|n, hence pn a d ≡ 0 (mod p). It follows that a(n) ≡ 0 (mod p) for every prime p > n, and a(n) = 0. INTEGERS: 21 (2021) 8

3. Unitary Newton Sequences

We recall that d is a unitary divisor of the positive integer n if d|n and (d, n/d) = 1. It will be denoted by dkn. Obviously, 1kn and nkn. For f, g ∈ A(Z) we deﬁne their unitary convolution f ◦ g by the formula X f ◦ g(n) = f(d)g(n/d), n ≥ 1. dkn

Notice that f ◦ g(n) = f ∗ g(n) for all square-free n ≥ 1. We summarize some well-known facts concerning unitary convolution.

• The Dirichlet and the unitary convolution, have the same identity element . −1 • If f ∈ A(Z) satisﬁes f(1) = ±1, then f has the unitary inverse f ∈ A(Z).

• The unitary sum function Sf is deﬁned as X Sf (n) = f ◦ I0(n) = f(d), n ≥ 1. dkn

• The unitary M¨obiusfunction µ is deﬁned as the unitary inverse of I0, i.e., µ ∗ I0 = . Since µ is multiplicative, it is determined by its values at the powers of primes. We have X 0 = µ(d) = µ(pk) + µ(1), dkpk

hence µ(pk) = −1 and µ(n) = (−1)ω(n), where ω(n) is the number of distinct prime factors of n.

Deﬁnition 24 (Unitary Newton sequence). Let a ∈ A(Z). We say that a is a unitary Newton sequence, if µ ◦ a ∈ A0(Z), i.e., X (−1)ω(d) a(n/d) ≡ 0 (mod n), n ≥ 1. dkn

By AUN (Z) we denote the set of all unitary Newton sequences.

Corollary 25. For a ∈ A(Z) the following conditions are equivalent:

(1) a ∈ AUN (Z),

(2) a = Sb = b ◦ I0 for some b ∈ A0(Z).

Proof. Obviously, µ ◦ a = b for some b ∈ A0(Z) is equivalent to a = b ◦ I0 = Sb for some b ∈ A0(Z). INTEGERS: 21 (2021) 9

Example 26. The sum of unitary divisors function σ1 given by X σ1(n) = d = SI1 (n). dkn

It follows that σ1 ∈ AUN (Z), but σ1 ∈/ AN (Z). Indeed, for prime p we have

2 2 σ1(p ) = 1 + p , σ1(p) = 1 + p, hence X 2 2 2 2 µ(d) σ1(p /d) = σ1(p ) − σ1(p) = p − p 6≡ 0 (mod p ). d|p2 Example 27. The sum of k-th powers of the unitary divisors of n, i.e.,

X k σk(n) = d = SIk (n), k ≥ 1, dkn is a unitary Newton sequence.

Corollary 28. For a ∈ AUN (Z) the following conditions hold: (1) a(pk) ≡ a(1) (mod pk) for all prime p and k ≥ 1,

(2) if a is bounded and p is prime, then a(pk) = a(1) for almost all k ≥ 1,

(3) if n is square-free, then X X µ ◦ a(n) = (−1)ω(d) a(n/d) = µ(d) a(n/d) ≡ 0 (mod n), dkn d|n

k−1 k (4) if a ∈ AN (Z)∩AUN (Z), then a(p ) ≡ a(1) (mod p ) for prime p and k ≥ 1.

Proposition 29. If a ∈ AUN (Z) is p-periodic with a prime p, then a is constant.

Proof. Since µ ◦ a ∈ A0(Z), a(p) ≡ a(1) (mod p). We first show that a(1) = ... = a(p − 1). Let l ∈ {2, . . . , p − 1}. We consider the arithmetic progression pk = l + kp. Then, by the Dirichlet theorem, pk is a prime number for infinitely many k ≥ 1. If pk is prime, then a(1) ≡ a(pk) = a(l) (mod pk). Consequently, a(1) ≡ a(l) (mod pk) for infinitely many primes pk. This shows that a(1) = a(l). It remains to prove that a(1) = a(p). We have that a(pk) = a(p) because a is p-periodic. It follows by Corollary 28 that a(p) ≡ a(1) (mod pk) for all k ≥ 1. This means that a(p) = a(1).

Example 30. Let k, l ∈ Z be such that k ≡ l (mod p) for prime p. Let a be a p-periodic sequence such that a(1) = ... = a(p − 1) = l and a(p) = k. If l 6= k, then a ∈ AN (Z) and a∈ / AUN (Z). INTEGERS: 21 (2021) 10

n Example 31. Let a(n) = 2 . Then a ∈ AN (Z) and a∈ / AUN (Z). More generally, n let a(n) = m for some positive integer m ≥ 1. One can check that, a ∈ AUN (Z) if and only if m = 1.

P ◦ P Let τ(n) = d|n 1 be the number of divisors of n ≥ 1 and τ (n) = dkn 1 be the number of unitary divisors of n. Since one of two complementary divisors of n √ is always not greater than n, one has √ τ ◦(n) ≤ τ(n) ≤ 2 n.

It follows that τ(n) lim = 0. n→∞ n

Lemma 32. Let a ∈ AN (Z) (a ∈ AUN (Z)) be a bounded sequence and let kak∞ = 2 max{|ai| : i ≥ 1}. If n > 4kak∞, then (µ ∗ a)(n) = 0 ((µ ◦ a)(n) = 0).

Proof. We have X X √ |(µ ∗ a)(n)| ≤ |a(n/d)| ≤ kak∞ 1 ≤ 2kak∞ n, d|n d|n

2 hence |(µ ∗ a)(n)| < n for n > 4kak∞. Since (µ ∗ a)(n) ≡ 0 (mod n), (µ ∗ a)(n) = 0 2 for n > 4kak∞. The proof in the unitary case is exactly the same because √ τ ◦(n) ≤ τ(n) ≤ 2 n.

k Lemma 33. Let a ∈ AUN (Z) be p -periodic for some prime p and k ≥ 1. Then a(pk) = a(1).

Proof. We have a(pk) = a(plpk) = a(pl+k) = a(1) for l suﬁciently large.

Deﬁnition 34. Let k ≥ 1. We deﬁne bk ∈ A0(Z) by ( k, if n = k bk(n) = 0, otherwise.

It follows that S ∈ A ( ) and bk UN Z ( X k, if kkn S (n) = bk(d) = bk 0, otherwise. dkn

Let us observe that S is k2-periodic. Indeed, it is easy to check that kkn if and bk only if kk(n + k2). This shows that S (n) = S (n + k2). bk bk INTEGERS: 21 (2021) 11

Proposition 35. Let a ∈ AUN (Z). The following conditions are equivalent: (1) a is bounded,

(2) there exist l ≥ 1 and integers xk ∈ Z for 1 ≤ k ≤ l such that

l X a = x S , k bk k=1

(3) a is periodic.

Proof. Obviously, (2) implies (3) and (3) implies (1). We show that (1) implies (2). Assume that a is bounded and |a(n)| ≤ M for n ≥ 1. Since a = Sb with b ∈ A0(Z), a = SI·˜b for some b ∈ A(Z). We have I · b = µ ∗ a, so 1 X M X τ(n) |b(n)| = µ(d) a(n/d) ≤ 1 ≤ M . n n n dkn dkn

τ(n) Since n → 0, b(n) → 0. On the other hand, b is a sequence of integers, so there exists l ≥ 1 such that b(n) = 0 for n > l. It follows that

l X b(k) b = (b(1), . . . , b(l), 0,...) = b , k k k=1 and consequently, l X b(k) a = Sb = xk S , xk = . bk k k=1

Remark 36. For k ≥ 1 we consider the sequence bk ∈ A0(Z) given as ( k, if n = k bk(n) = 0, otherwise.

It follows that Sbk ∈ AN (Z) and ( X k, if k|n Sb (n) = bk(d) = k 0, otherwise. d|n The proof of Proposition 35 shows the well-known fact that a Newton sequence a ∈ AN (Z) is bounded if and only if there are l ≥ 1 and integers x1, . . . , xl such that l X a = xk Sbk . k=1 INTEGERS: 21 (2021) 12

Lemma 37. If a, f ∈ A(Z) are such that µ ◦ a, Sf ∈ A0(Z), then f ◦ a ∈ A0(Z). In particular, a ∈ AUN (Z) implies f ◦ a ∈ A0(Z). Proof. Corollary 3 implies that

f ◦ a = ◦ f ◦ a = µ ◦ I0 ◦ f ◦ a = (µ ◦ a) ◦ Sf ∈ A0(Z).

∗ Lemma 38. If a, f ∈ A(Z) are such that f ◦ a ∈ A0(Z), Sf ∈ A0(Z), then µ ◦ a ∈ A0(Z). In particular, f ◦ a ∈ A0(Z) implies that a ∈ AUN (Z) provided ∗ that Sf ∈ A0(Z). Proof. We have (µ ◦ a) ◦ Sf = f ◦ a ∈ A0(Z). −1 ∗ ∗ Since Sf ∈ A0(Z), Sf ∈ A0(Z). Consequently, Corollary 3 shows that

−1 µ ◦ a = Sf ◦ (f ◦ a) ∈ A0(Z).

∗ Corollary 39. Assume that a ∈ A(Z) and Sf ∈ A0(Z). The following conditions are equivalent:

(1) a ∈ AUN (Z),

(2) f ◦ a ∈ A0(Z). Proof. It follows by Lemma 37 and Lemma 38.

Example 40. The unitary Euler totient function is deﬁned by φ := µ ◦ I1, i.e., X n φ(n) = µ(d) . d dkn

Obviously, ∗ Sφ = φ ◦ I0 = µ ◦ I1 ◦ I0 = I1 ∈ A0(Z). ∗ Similarly, S ∈ A ( ) for φ := µ ◦ Ik. φk 0 Z k

It was proved in [7] that a ∈ AN (Z) if and only if

a(n) ≡ a(n/p) (mod pk), for any n ∈ N and for any prime p with pkkn (i.e., pk|n and n 6≡ 0 (mod pk+1)).

Proposition 41. Let a ∈ A(Z). The following conditions are equivalent: INTEGERS: 21 (2021) 13

(1) a ∈ AUN (Z), (2) a(n) ≡ a(n/pk) (mod pk) for any n ≥ 1 and any prime p with pkkn.

Proof. First, we show that (1) implies (2). We use the induction with respect to ω(n), i.e., the number of prime factors of n. If ω(n) = 1, then n = pk and the result follows by Corollary 28. Assume that formula holds for ω(n) = r. We prove k1 k2 kr+1 it for ω(n) = r + 1. Let n = p1 p2 . . . pr+1 with diﬀerent primes pi and ki ≥ 1. k1 k1 By symmetry, it is suﬃcient to show that a(n) ≡ a(n/p1 ) (mod p1 ). We consider the set k1 S = {d : dkn, d 6≡ 0 (mod p1 )}. We have X µ ◦ a(n) = (−1)ω(d) a(n/d) dkn X k1 ω(d) ω(p1 d) k1 = (−1) a(n/d) + (−1) a(n/p1 d) d∈S

k1 X ω(d) k1 = a(n) − a(n/p1 ) + (−1) (a(n/d) − a(n/p1 d)). d∈S k1 d6=1, d6=p1

k1 Obviously, µ ◦ a(n) ≡ 0 (mod p1 ) and by the inductive step

X ω(d) k1 k1 (−1) (a(n/d) − a(n/p1 d)) ≡ 0 (mod p1 ). d∈S k1 d6=1, d6=p1

k1 k1 It follows that a(n) − a(n/p1 ) ≡ 0 (mod p1 ). Next, we show that (2) implies (1). We have to show that µ ◦ a(n) ≡ 0 (mod n). We proceed by induction on ω(n). If ω(n) = 1, then n = pk and the result follows by (2). Assume that formula holds for ω(n) = r. We will prove it for ω(n) = r + 1. k1 k2 kr+1 Let n = p1 p2 . . . pr+1 with diﬀerent primes pi and ki ≥ 1. We set

ki Si = {d : dkn, d 6≡ 0 (mod pi )}. We have

ki X ω(d) ki µ ◦ a(n) = a(n) − a(n/pi ) + (−1) (a(n/d) − a(n/pi d)). d∈Si ki d6=1, d6=pi By the inductive step

X ω(d) ki ki (−1) (a(n/d) − a(n/pi d)) ≡ 0 (mod pi ), d∈Si ki d6=1, d6=pi INTEGERS: 21 (2021) 14

ki ki ki and a(n)−a(n/pi ) ≡ 0 (mod pi ) by (2). We conclude that µ◦a(n) ≡ 0 (mod pi ) for all i = 1, . . . , r + 1, and the result follows.

Corollary 42. Let a ∈ A(Z). The following conditions are equivalent:

(1) a ∈ AN (Z) ∩ AUN (Z), (2) a(n) ≡ a(n/p) ≡ a(n/pk) (mod pk) for any n ≥ 1 and all prime p with pkkn.

In particular, if a ∈ AN (Z) ∩ AUN (Z), then

a(pk) ≡ a(1) (mod pk+1), for prime p and k ≥ 1.

Example 43. If a ∈ A(Z) is constant, then a ∈ AN (Z) ∩ AUN (Z). Example 44. The sequence ( 2k+1, if 2kkn for some k ≥ 1 a2(n) = 0, if n ≡ 1 (mod 2), belongs to AN (Z) ∩ AUN (Z). Indeed, we have

s s (i) a2(n) = a2(n/p) = a2(n/p ) = 0 if n is odd and p kn,

s k+1 k (ii) a2(n) = a2(n/p) = a2(n/p ) = 2 if n = 2 m (k ≥ 1) with m odd and pskm,

k k+1 k k (iii) for m odd and k ≥ 1 we have a2(2 m) = 2 , a2(2 m/2 ) = a2(m) = 0 and ( k k−1 0, if k = 1 a2(2 m/2) = a2(2 m) = 2k, if k > 1.

More generally, let p be a ﬁxed prime. Then ap ∈ AN (Z) ∩ AUN (Z), where ( pk+1, if pkkn for some k ≥ 1 ap(n) = 0, if n 6≡ 0 (mod p).

Example 45. We show that the sequence a2 deﬁned in Example 44 is not a sequence of traces. Let c ∈ A(Z) be a sequence of integers generating a2, i.e.,

a2(n) = c(1)a2(n − 1) + ... + c(n − 1)a2(1) + nc(n), n ≥ 1.

It is sufficient to show that c is infinite, i.e., there is no l ≥ 1 such that c(n) = 0 for n > l. We first claim that c(m) = 0 for all odd m ≥ 1. Obviously, we have INTEGERS: 21 (2021) 15

0 = a2(1) = c(1). Assume that a(s) = 0 for all odd s less than some odd m > 1. Then 0 = a2(m) = c(1)a2(m − 1) + ... + c(m − 1)a2(1) + mc(m). Since m is odd, i and m − i have diﬀerent parity for all i = 1, . . . , m − 1. By the inductive step c(i)a2(m − i) = 0. Consequently, c(m) = 0. Next, suppose that c is ﬁnite, i.e, c = (0, c(2), 0, c(4),..., 0, c(2s), 0, 0,...) for some s ≥ 1. Then 2s = 2tm for some t ≥ 1 and odd m. Let r ≥ 1 be such that |c(2i)| ≤ 2r for i = 1, . . . , s. For all k > 2s, we have

k+1 k k k 2 = a2(2 ) = c(2)a2(2 − 2) + ... + c(2s)a2(2 − 2s) k−1 t k−t = c(2)a2(2(2 − 1)) + ... + c(2s)a2(2 (2 − m)) = c(2)22 + ... + c(2s)2t+1.

Since

k k k |a2(2 )| = |c(2)a2(2 − 2) + ... + c(2s)a2(2 − 2s)| ≤ 2r(22 + ... + 2t+1) ≤ s2r+t+1, we get a contradiction by taking k suﬃciently large.

Problem 46. The following natural questions are worth of consideration:

• describe AN (Z) ∩ AUN (Z),

• characterize the integer matrices A ∈ Ml(Z) such that tr [A] ∈ AN (Z) ∩ AUN (Z) (obviously, tr [I] ∈ AN (Z) ∩ AUN (Z)),

• characterize the generating sequences c ∈ A(Z) for a ∈ AN (Z) ∩ AUN (Z).

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