on the elements of sets with even partition function
TRANSCRIPT
Ramanujan JDOI 10.1007/s11139-014-9618-z
On the elements of sets with even partition function
N. Baccar
Received: 2 April 2014 / Accepted: 20 July 2014© Springer Science+Business Media New York 2014
Abstract Let P ∈ F2[z] such that P(0) = 1 and degree(P) ≥ 1. Nicolas, Ruzsaand Sárközy proved that there exists a unique subset A = A(P) of N such that∑
n≥0 p(A, n)zn ≡ P(z) (mod 2), where p(A, n) is the number of partitions of nwith parts in A. Let m be an odd positive integer and let S(A, m) be the 2-adic integergiven by the expansion S(A, m) = χ(A, m) + 2χ(A, 2m) + 4χ(A, 4m) + · · · =∑∞
k=0 2kχ(A, 2km), where χ(A, ·) is the characteristic function of the set A. Byknowing the expansion S(A, m), one can obtain the elements of the set A of the form2km. In this paper, we prove that S(A, m) is an algebraic number.
Keywords Partitions · Periodic sequences · Order of a polynomial · Cyclotomicpolynomials · Resultant · 2-Adic integers · The Graeffe transformation
Mathematics Subject Classification 11P83 · 11B50 · 11D88
1 Introduction
Let N be the set of positive integers and let A be a non-empty subset of N. For n ∈ N,a partition of n with parts in A is a finite non-increasing sequence n1, n2, . . . , nk
belonging to A such that
n = n1 + n2 + · · · + nk .
N. Baccar (B)Dép. de Math Inf., Université de Sousse ISITCOM Hammam Sousse, 5 Bis Rue 1 Juin 1955,4011 Hammam Sousse, Tunisiae-mail: [email protected]
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N. Baccar
The number of partitions of n with parts in A is denoted by p(A, n) with p(A, 0) = 1.The set A is called an even partition set if the sequence (p(A, n))n≥0 is even from acertain point on.
Let F2 be the field with two elements and let P be a polynomial of positive degreein F2 [z] such that P(0) = 1. Nicolas, Ruzsa and Sárközy(see [16]) proved that thereis a unique even partition set A = A(P) satisfying the congruence
∑
n≥0
p(A, n)zn ≡ P(z) (mod 2). (1.1)
Two issues arise in the study of such sets: a qualitative aspect which deals with thequestion whether a positive integer n is in A or not? and a quantitative aspect, whosepurpose is the asymptotic study of the counting function of the set A = A(P); thatis, the function A(P, x) defined by
A(P, x) = |{n ∈ A : n ≤ x}|.
From now on, for any odd positive integer d, we write s(d) for the order of 2 modulod; that is, s(d) is the smallest positive integer such that 2s(d) ≡ 1 (mod d). We alsouse r(d) to denote the positive integer satisfying ϕ(d) = s(d)r(d), where ϕ is Euler’sfunction.
In [9], [10] and [17], the authors determined the elements of the set A(P), whenP(z) = 1+z+z3+z4+z5 and P(z) = 1+z+z3. They obtained asymptotic formulasfor the related counting function. This was generalized in [3] for irreducible P withprime order p. Here, the order of a polynomial P is the smallest positive integer w
such that P(z) divides 1 + zw in F2[z].Let the factorization of P into irreducible factors over F2[z] be
P = Qα11 Qα2
2 · · · Qαtt .
We denote by P∗ the squarefree kernel of P; that is, P∗ = Q1 Q2 · · · Qt the productof the distinct irreducible polynomials dividing P . In [11], F. Ben Saïd et al provedthat the counting function A(P, x) of the set A(P) satisfies
A(P, x) � x(log x)− 1
r(β) ,
where β is the order of P∗. In [8], Ben Saïd and Nicolas improved on the constantc = c(β) for which
A(P, x) � x(log x)−c.
Let m be an odd positive integer. By χ(A, n), we denote the characteristic functionof the set A = A(P); that is, we put χ(A, n) = 1 if n ∈ A and zero otherwise. We
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Set elements with even partition function
define the 2-adic integer S(A, m) by the expansion
S(A, m)=χ(A, m)+ 2χ(A, 2m)+4χ(A, 4m)+ · · ·=∞∑
k=0
2kχ(A, 2km). (1.2)
By knowing S(A, m) modulo 2k+1, we can deduce the values of χ(A, 2i m) for alli , i ≤ k. This may explain the importance of such expansion in the study of the setsA(P), since S(A, m) gives a complete description of the elements of the set A(P) ofthe form 2km.
The aim of the present paper is to prove the following theorem:
Theorem 1.1 Let P ∈ F2[z] be a polynomial of positive degree with P(0) = 1 andA(P) be the set satisfying (1.1). Then, for all odd positive integer m, the 2-adic integerS(A(P), m) defined by (1.2) is algebraic.
Moreover, the proof of Theorem 1.1 gives a way to compute an explicit polynomialfor whose S(A(P), m) is a root. Theorem 1.1 extends a result proved in [1] whenP is irreducible of prime order p in F2[z] ( in this case, the minimal polynomialof S(A(P), m) is obtained explicitly when s(p) = p−1
2 (see [1] and [3]) and when
s(p) = p−13 (see [2])). As an example, we consider the case p = 7 (see [9]): here,
the only irreducible polynomials in F2[z] of order 7 are P1(z) = 1 + z + z3 andP2(z) = 1 + z2 + z3 since in F2[z],
1 + z7 = (1 + z)(1 + z + z3)(1 + z2 + z3).
In this case, we have
y2 − y + 2 = (y − S(A(P1), 1)
)(y − S(A(P2), 1)),
where
S(A(P1), 1) = 1 + 2 + 23 + 24 + 26 + 213 + · · · + 2997 + · · ·
and
S(A(P2), 1) = 2 + 22 + 25 + 27 + 28 + 29 + · · · + 2999 + · · ·,
and so the elements of A(P1) (resp. A(P2)) of the form 2k are 1, 2, 23, 24, 26, 213,
. . . , 2997, . . . (resp. 2, 22, 25, 27, 28, 29, . . . , 2999, . . .).As described in [16], the set A(P) is constructed by recursion, and this enables
to give an algorithm that allows us to compute only the first elements of A(P). Onthe other hand, as it can be seen in the last example, solving the equation satisfiedby S(A(P), m) allows us to obtain the elements of A(P) of the form 2km for largevalues of k.
In Sect. 2, we recall some important facts and give some preliminary results oneven partition sets. In Sect. 3, the proof of Theorem 1.1 is given.
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N. Baccar
2 Some results on even partition sets
Let P be a polynomial in F2 [z] of positive degree such that P(0) = 1 and let β bethe order of P∗.
Definition 2.1 Let Z2 be the 2-adic ring. We shall say that two formal power seriesf (z) and g(z) with coefficients in Z2 are congruent modulo a 2-adic integer M andwrite f (z) ≡ g(z) (mod M), if their coefficients of the same power of z are congruentmodulo M .
Congruences of formal power series may be added, multiplied and derivated. More-over, we have the following:
if f (z) ≡ g(z) (mod M) and f (0) = g(0) = 1, then1
f (z)≡ 1
g(z)(mod M).
Definition 2.2 The Graeffe transformation. Let K be some field and K[[z]] be thering of formal power series with coefficients in K. For an element
f (z) = a0 + a1z + a2z2 + . . . + anzn + . . .
of this ring, the product
f (z) f (−z) = b0 + b1z2 + b2z4 + . . . + bnz2n + . . .
is an even power series. We shall call G( f ) (Graeffe transformation) the series
G( f )(z) = b0 + b1z + b2z2 + . . . + bnzn + . . . .
It follows immediately that for f, g ∈ K[[z]],
G( f g) = G( f )G(g). (2.1)
We shall use the following notation for the iterates of f by the transformation G:
f(0) = f, f(1) = G( f ), . . . , f(k) = G( f(k−1)) = G(k)( f ).
If G( f ) = f , then for all k, G(k)( f ) = f , and we will say that f is a fixed point ofthe Graeffe transformation (for example f (z) = 1 − zq , where q is an odd integer).Let us assume now that K = Z2. As in [7], it is easy to see that if f and g are inZ2[[z]], then
f(1) ≡ f (mod 2), (2.2)
and for any 2-adic integer M ,
f ≡ g (mod 2M) ⇒ f(1) ≡ g(1) (mod 4M). (2.3)
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Moreover, the last property is generalized (cf. [7]) to
f ≡ g (mod 2) ⇒ f(k) ≡ g(k) (mod 2k+1), ∀k ≥ 0. (2.4)
Using the elementary theory of finite fields (see [14]), we know that
1 − zβ =∏
d | βφd(z), (2.5)
where φd is the cyclotomic polynomial of index d. Since β is odd, then φβ factors inF2[z] into r(β) distinct irreducible polynomials, each of degree s(β) and of order β. Itwill also be of use to include here the following basic identities concerning cyclotomicpolynomials:
φβ(z) =∏
d | β(1 − z
βd )μ(d). (2.6)
In [1] (see also [3]), it is proved that the polynomial P can be extended in a 2-adicpolynomial P such that
P ≡ P(k) (mod 2k+1), ∀k ≥ 0. (2.7)
Let Q is a polynomial of F2[z] of positive degree such that Q(0) = 1. Therefore, from(2.1), it follows that
P Q = P Q. (2.8)
Let A = A(P) be the even partition set defined by (1.1). Let m be an odd positiveinteger and let σ(A, n) denote the sum of those divisors of n belonging to A; that is,
σ(A, n) =∑
d | n, d∈Ad. (2.9)
It is proved (see [4]) that for all k ≥ 0, β is a period of the sequence(σ (A, 2kn) mod 2k+1)n≥1. Moreover, if P is irreducible in F2, then for all k ≥ 0,
σ(A, 2kβ) ≡ −s(β) (mod 2k+1). (2.10)
Let (Z/βZ)∗ denote the group of invertible residues modulo β and let < 2 > be itssubgroup generated by 2. Consider the action � of < 2 > on the set Z/βZ given bya �n = an for all a ∈< 2 > and n ∈ Z/βZ. Then, denote the orbit of some n ∈ Z/βZ
by Oβ(n); that is,
Oβ(0) = {0},
andOβ(n) = {2 j n mod β, 0 ≤ j ≤ s(β/δ) − 1}, (2.11)
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N. Baccar
where δ := δ(n) = gcd(β, n). It is easy to see that for every non-negative integer h,we have Oβ(2hn) = Oβ(n). Moreover, for all positive integer i , one has
{2 j n mod β, is(β/δ) ≤ j ≤ (i + 1)s(β/δ) − 1} = Oβ(n). (2.12)
Such action allows one to write
(Z/βZ)/<2> = {Oβ(x1), . . . , Oβ(xr(β))} ∪ {Oβ(xr(β)+1), . . . , Oβ(x f ), Oβ(0)},(2.13)
where Oβ(x1), . . . , Oβ(xr(β)) are the invertible orbits, while Oβ(xr(β)+1), . . . ,
Oβ(x f ), Oβ(0) are the non-invertible ones. We will always set xr(β) = 1. In [4],it is proved that if q mod β and n mod β are in the same orbit, then for all k ≥ 0,
σ(A, 2kq) ≡ σ(A, 2kn) (mod 2k+1). (2.14)
In particular (taking q = 2n),
σ(A, 2k+1n) ≡ σ(A, 2kn) (mod 2k+1)
Hence, we shall consider the 2-adic integers ρ(A, n) defined by
ρ(A, n) ≡ σ(A, 2kn) (mod 2k+1), ∀k ≥ 0. (2.15)
One may easily verify that the sequence (ρ(A, n))n≥1 is periodic with period β.Moreover, it turns out (see [1] and [3]) that
+∞∑
n=1
ρ(A, n)zn = zP ′(z)P(z)
. (2.16)
The last equality with the properties of the logarithmic derivative gives
ρ(A(P Q), n) = ρ(A(P), n) + ρ(A(Q), n). (2.17)
Let S(A, m) be the 2-adic integer given by (1.2). Using (2.9) and (2.15), one cansee that
ρ(A, m) =∑
d | m
d S(A, d). (2.18)
Applying Möbius inversion formula on the last equality, we obtain
mS(A, m) =∑
d | m
μ(d)ρ(A,
m
d
)=
∑
d | m
μ(d)ρ(A,
m
d
), (2.19)
where m = ∏
p | m, p primep denotes the radical of m with 1 = 1, and where μ is the
Möbius’s function.
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Set elements with even partition function
Lemma 2.1 Let α be a positive integer and let S(A(Pα), m) be the 2-adic integerdefined by (1.2). Then
S(A(Pα), m) = αS(A(P), m) (2.20)
Proof From (2.17), we see that
ρ(A(Pα), n) = αρ(A(P), n), (2.21)
which when combined with (2.19) gives the desired result. �Remark 2.1 According to what has been said in the previous lemma, in the case α isa power of 2, the set A(Pα) can be expressed in terms of the set A(P), in other words
A(P2q) = 2q · A(P) := {2qn, n ∈ A(P)}.
Let K be some field, H(z) = a0+a1z+· · ·+anzn and L(z) = b0+b1z+· · ·+bk zk
be polynomials in K[z] of degrees n and k, respectively. We denote the resultant ofH and L by resz(H(z), L(z)) and recall the following well-known results (see, forinstance, [15]).
Lemma 2.2 (i) The resultant resz(H(z), L(z)) is a homogeneous multivariate poly-nomial, with integer coefficients, and of degree n + k in the n + k + 2 variables ai
and b j .(ii) If H(z) is written as H(z) = an(z − z1)(z − z2) · · · (z − zn) in the splitting field
of H over K, then
resz(H(z), L(z)) = akn
n∏
i=1
L(zi ). (2.22)
(iii) If z1, z2, . . . , zn and q1, q2, . . . , qν are the roots of H and L, respectively, in analgebraic closure K of K , then the nν roots of the polynomial resz(H(y−z), L(z))in K are zi + q j , 1 ≤ i ≤ n, 1 ≤ j ≤ ν.
3 Elements of even partition sets
Let P be a polynomial in F2 [z] of positive degree such that P(0) = 1 and let β be theorder of P∗. Let A = A(P) be the even partition set obtained from (1.1) and let mbe an odd positive integer. The goal of this section was to prove that the elements ofthe even partition set A(P) satisfying (1.1), of the form 2km, are given by the 2-adicexpansion of some root of a polynomial Rm(y) ∈ Z[y]. In other words, we prove thatthe 2-adic integer S(A, m) defined by (1.2) is an algebraic number and explain how toobtain a polynomial Rm(y) ∈ Z[y] having S(A, m) as a root. For this, we proceed inthree steps: we will start with the case where P is an irreducible polynomial, then weconsider the case P is a product of power of irreducible polynomials of same orderand we end by the general case. In what follows, we set s = s(β) and r = r(β) andrecall that
Z/βZ = Oβ(x1) ∪ Oβ(x2) ∪ · · · Oβ(x f ) ∪ Oβ(0). (3.1)
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N. Baccar
3.1 Case P is an irreducible polynomial
Let φβ be (cf. Sect. 2) the cyclotomic polynomial of index β and write
φβ(z) ≡r∏
�=1
P�(z) (mod 2), (3.2)
where the P�’s are the only irreducible polynomials over F2[z] of the same degree sand all of which are of order β. It follows from (2.6) and (2.1) that φβ is a fixed pointof the Graeffe transformation. Therefore, from (2.8), we have
φβ = P1 P2 · · · Pr = P1 P2 · · · Pr . (3.3)
Let ζ be a β-th primitive root of unity over the 2-adic field Q2 and such that Pr (ζ ) = 0.Since for all k ≥ 0, (Pr )(k) = Pr , it follows that ζ, ζ 2, . . . , ζ 2s−1
are the roots of Pr
and the polynomials P� can be arranged so that (see (2.13))
P�(z) = (−1)s(z − ζ x� )(z − ζ 2x� ) · · · (z − ζ 2s−1x� ), 1 ≤ � ≤ r. (3.4)
For all �, 1 ≤ � ≤ r , let A� = A(P�) be the even partition set satisfying (1.1).
Lemma 3.1 Let v be a positive integer and let ρ(A�, v) be the 2-adic integer definedby (2.15). Then
ρ(A�, v) = −D(θvx� ), 1 ≤ � ≤ r, (3.5)
where θ = ζ−1 and D(z) is the polynomial given by
D(z) = z + z2 + · · · + z2s−1 mod β. (3.6)
Proof Using the fact that β is a period of (ρ(Ar , n))n≥1 and that θ is β-th root ofunity, it suffices to prove the last lemma when v ∈ {1, 2, . . . , β}. Since
∞∑
n=1n≡v (mod β)
ρ(Ar , n)zn = ρ(Ar , v)
∞∑
u=0
zv+uβ,
it follows that
∞∑
n=1
ρ(Ar , n)zn =β−1∑
v=1
⎛
⎜⎜⎝
∞∑
n=1n≡v (mod β)
ρ(Ar , n)zn
⎞
⎟⎟⎠ +
∞∑
n=1n≡0 (mod β)
ρ(Ar , n)zn .
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Set elements with even partition function
Therefore,
∞∑
n=1
ρ(Ar , n)zn = 1
1 − zβ
β−1∑
v=1
ρ(Ar , v)zv + ρ(Ar , β)zβ
1 − zβ. (3.7)
On the other hand, from (3.4), we get
zP ′
r (z)
Pr (z)=
s−1∑
j=0
z
z − ζ 2 j = −s∑
j=0
∞∑
n=1
(z/ζ 2 j)n
= −s−1∑
j=0
⎛
⎜⎜⎝
β−1∑
v=1
∞∑
n=1n≡v (mod β)
(z/ζ 2 j)n +
∞∑
n=1n≡0 (mod β)
(z/ζ 2 j)n
⎞
⎟⎟⎠
= −s−1∑
j=0
⎛
⎝β−1∑
v=1
zv
ζ 2 j v
+∞∑
u=0
zuβ + zβ
1 − zβ
⎞
⎠
= − 1
1 − zβ
β−1∑
v=1
zv
⎛
⎝s−1∑
j=0
θ2 j v
⎞
⎠ − szβ
1 − zβ,
where θ = 1ζ
. Using (2.16) and comparing the last equality with (3.7), we obtain (3.5)for � = r . To get (3.5) for all �, 1 ≤ � ≤ r , we just have to replace in the above proofr by � and ζ by ζ x� . �
For �, 1 ≤ � ≤ r , let S(A�, m) be the 2-adic integer defined by (1.2). We definethe polynomial Dm by
Dm(z) =f∑
h=1
λ(m, h)Bh(z) + sγ (m), (3.8)
where Bh(z) is the polynomial given by
Bh(z) =s−1∑
j=0
z2 j xh mod β (3.9)
and
λ(m, h) =∑
d | mmd mod β ∈Oβ(xh)
μ(d), (3.10)
γ (m) =∑
d | mmd ≡0 (mod β)
μ(d). (3.11)
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N. Baccar
Clearly, γ (m) = 0 if β | m while if β � m, we have
γ (m) =∑
d | gcd(m, mβ
)
μ(d) ={
1 if m = β
0 if m �= β
We now have the tools necessary to prove the following
Proposition 3.1 Under the above notation, S(A1, m), S(A2, m), . . . , S(Ar , m) arethe roots of the polynomial Rm(y) ∈ Z[y] given by
Rm(y) = resz(φβ(z), my + Dm(z)
), (3.12)
where φβ(z) is the cyclotomic polynomial of index β.
Proof It follows from (2.19) that
mS(A�, m)=f∑
h=1
⎛
⎜⎜⎜⎝
∑
d | mmd mod β ∈Oβ(xh)
μ(d)ρ(A�,m
d)
⎞
⎟⎟⎟⎠
+∑
d | mmd ≡0 (mod β)
μ(d)ρ(A�,m
d),
which, by (2.14) and (2.15), gives
mS(A�, m) =f∑
h=1
ρ(A�, xh)
⎛
⎜⎜⎜⎝
∑
d | mmd mod β ∈Oβ(xh)
μ(d)
⎞
⎟⎟⎟⎠
− s∑
d | mmd ≡0 (mod β)
μ(d).
Consequently,
mS(A�, m) =f∑
h=1
λ(m, h)ρ(A�, xh) − sγ (m),
which, by (3.5) and (3.9), gives
mS(A�, m) = −f∑
h=1
λ(m, h)Bh(θ x� ) − sγ (m).
From (3.8) and the last equality, we have
Dm(θ x� ) = −mS(A�, m). (3.13)
From (3.3), (2.22) and (3.4), it follows that the polynomial defined by (3.12) can bewritten as
Rm(y) =r∏
�=1
s−1∏
i=0
(my + Dm(θ2i x� )
).
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Set elements with even partition function
Now recall that for all i, 0 ≤ i ≤ s − 1,
Bh(θ2i x� ) =s−1∑
j=0
θ2i+ j x�xh , 1 ≤ h ≤ f.
Let t = |Oβ(x�xh)|. We claim that the last sum does not depend on i since from (2.11)and (2.12),
s−1∑
j=0
θ2i+ j x�xh =st −1∑
l=0
(l+1)t−1∑
j=lt
θ2i+ j x�xh
=st −1∑
l=0
(l+1)t−1∑
j=lt
θ2 j x�xh
=s−1∑
j=0
θ2 j x�xh .
Hence, Dm(θ2i x� ) = Dm(θ x� ) and so
Rm(y) =r∏
�=1
(my + Dm(θ x� )
)s.
Finally, from (3.13), it follows that
Rm(y) = mϕ(β)r∏
�=1
(y − S(A�, m))s,
which gives the desired result. �Remark 3.1 Let M(y) be the polynomial of Q[z] defined by
M(y) := M(P)m(y) = 1
mϕ(β)(Rm(y))
1s. (3.14)
Taking m = 1, we have
M(y) = (y − D(θ))(y − D(θ x1)) · · · (y − D(θ xr−1)).
We note that D(θ), D(θ x1), . . . , D(θ xr−1) are the Gaussian periods corresponding tothe subgroup < 2 > which are known to be conjugate and distinct (see [12]). So, onecan deduce that M(y) can be regarded as the minimal polynomial of the algebraicnumbers S(A1, 1), S(A2, 1), . . . , S(Ar , 1). Hence, it is naturel to ask whether similardeduction can be given when m �= 1?
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N. Baccar
3.2 If P is a product of powers of irreducible polynomials of the same order
Let us assume that
P = Qα11 Qα2
2 · · · Qαtt
is the irreducible factorization of P over F2, where Q1, Q2, . . . and Qt are of sameorder β. It follows immediately from (2.17) that
ρ(A(P), n) = α1ρ(A(Q1), n) + α2ρ(A(Q2), n) + · · · + αtρ(A(Qt ), n).
Therefore, from (2.19), we have
S(A(P), m) = α1S(A(Q1), m) + α2S(A(Q2), m) + · · · + αt S(A(Qt ), m).
Let ζ be a β-th primitive root of unity over the 2-adic field Q2. It follows that thereexist a1, a2, . . . , at distinct in {x1, x2, . . . , xr } (see (3.4)) so that
Qi (z) = (−1)s(z − ζ ai )(z − ζ 2ai ) · · · (z − ζ 2s−1ai ), 1 ≤ i ≤ t, (3.15)
where the Qi ’s are the polynomial defined by (2.7). Consequently, by (3.5), for allv ≥ 1,
ρ(A(Qi ), v) = −D(θvai ), 1 ≤ i ≤ t, (3.16)
where θ = ζ−1 and D(z) is the polynomial given by (3.6). In this case, it turns out tobe fruitful to consider the polynomials T� := T�(P) of F2[z] given by
T� = Hα11 Hα2
2 · · · Hαtt , 1 ≤ � ≤ r,
where H1, H2, . . . , and Ht are irreducible polynomials of F2 of order β such that
Hi (z) = (−1)s(z − ζ ai x� )(z − ζ 2ai x� ) · · · (z − ζ 2s−1ai x� ), 1 ≤ i ≤ t.
One can easily see that P = Tr . We may now state the following
Proposition 3.2 The 2-adic integers S(A(T1), m), S(A(T2), m), . . .and S(A(Tr ), m)
are the roots of the polynomial Rm given by
Rm(y) = resz(φβ(z), my + Cm(z)
), (3.17)
where
Cm(z) =f∑
h=1
λ(m, h)
(t∑
i=1
αi Bh,i (z)
)
+ sγ (m)
t∑
i=1
αi , (3.18)
with
Bh,i (z) =s−1∑
j=0
z2 j xhai mod β, (3.19)
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Set elements with even partition function
where λ(m, h) and γ (m) are, respectively, defined by (3.10) and (3.11).
Proof The concept of proof is similar to the proof of Proposition 3.1. �Remark 3.2 By following the same steps as in the proof of Proposition 3.1, we obtain
Rm(y) =r∏
�=1
(my + Cm(θ x� )
)s, (3.20)
with
Cm(θ x� ) = −mS(A(T�), m), 1 ≤ � ≤ r.
Unlike the previous case, the polynomial M(y) = 1mϕ(β) (Rm(y))
1s is not necessarily
the minimal polynomial of S(A(P), m), since for some � �= �′, it can happen thatS(A(T�), m) = S(A(T�′), m). However, M(y) can be seen as a power of the minimalpolynomial of S(A(P), m).This can be illustrated by the example where P is the product of two irreduciblepolynomials of order β = 113. In this case, we have s = 28, r = 4,
Z/113Z = O113(3) ∪ O113(9) ∪ O113(27) ∪ O113(1) ∪ O113(0),
and we take x1 = 3, x2 = 9, x3 = 27 and x4 = 1. Let ζ be a 113-th primitive rootof unity over Q2 and take θ = ζ−1. We assume that P = P1 P2 where P1 and P2 areirreducible polynomials of order 113 such that
P1(z) = (z − ζ )(z − ζ 2) · · · (z − ζ 227)
and
P2(z) = (z − ζ 9)(z − ζ 18) · · · (z − ζ 9.227).
Recall that T� = H1 H2 (1 ≤ � ≤ 4), where
H1(z) = (z − ζ x� )(z − ζ 2x� ) · · · (z − ζ 227x� )
and
H2(z) = (z − ζ 9x� )(z − ζ 18x� ) · · · (z − ζ 9.227x� ).
We remark that 3 · O113(9) = O113(27), 27 · O113(9) = O113(3) and 9 · O113(9) =O113(1); therefore, T1 = T3 and T2 = T4. Consequently, when taking m = 1, weobtain
M(y) =[(
y − S(A(T1), 1))(
y − S(A(T2), 1))]2
,
123
N. Baccar
and so M(y) is the square of the minimal polynomial of S(A(P), 1). On the otherhand,
M(y) =(
resz(φ113(z), y +27∑
j=0
z2 j mod 113 +27∑
j=0
z2 j 9 mod 113)) 1
28 = (y2 − y − 28)2.
For the five remaining possibilities of P represented as a product of two distinctirreducible polynomials of order 113, the corresponding polynomial M(y) will haveone of the following forms
•⎛
⎝resz
⎛
⎝φ113(z), y +27∑
j=0
z2 j mod 113 +27∑
j=0
z2 j 3 mod 113
⎞
⎠
⎞
⎠
128
= y4 − 2y3 − 55y2
+ 56y + 332
•⎛
⎝resz
⎛
⎝φ113(z), y +27∑
j=0
z2 j mod 113 +27∑
j=0
z2 j 27 mod 113
⎞
⎠
⎞
⎠
128
= y4 − 2y3
− 55y2 + 56y + 332
•⎛
⎝resz
⎛
⎝φ113(z), y +27∑
j=0
z2 j 27 mod 113 +27∑
j=0
z2 j 9 mod 113
⎞
⎠
⎞
⎠
128
= y4 − 2y3
− 55y2 + 56y + 332
•⎛
⎝resz
⎛
⎝φ113(z), y +27∑
j=0
z2 j 9 mod 113 +27∑
j=0
z2 j 3 mod 113
⎞
⎠
⎞
⎠
128
= y4 − 2y3
− 55y2 + 56y + 332
•⎛
⎝resz
⎛
⎝φ113(z), y +27∑
j=0
z2 j 27 mod 113 +27∑
j=0
z2 j 3 mod 113
⎞
⎠
⎞
⎠
128
= (y2 − y − 28)2,
giving the minimal polynomial only in the four first cases.
Unfortunately, the expressions of P1 and P2 do not allow us to determine thecorresponding polynomials P1 and P2. However, this can be done by computing thefirst few elements of the sets A(P) for the six possibilities of P represented as a productof two distinct irreducible polynomials of order 113 and then comparing S(A(P), 1)
with the roots in Z2 of y2 − y − 28. Doing so, we obtain 2, 3, 4, 5, and 6 for thedifferent possible values of S(A(P), 1) mod 23. Since the roots of y2 − y − 28 arey1 = 22 + 24 + 25 + 26 + 210 + 212 + 214 + 218 + 219 + 220 + 221 + 225 + · · ·and y2 = 1 + 22 + 23 + 27 + 28 + 29 + 211 + 213 + 215 + 216 + 217 + 222 + 223 +· · · ,then the corresponding P = P1 P2 is such that S(A(P), 1) mod 23 = 4 or 5 whichyieldsP1(z) = 1 + z3 + z4 + z6 + z7 + z13 + z14 + z15 + z21 + z22 + z24 + z25 + z28
andP2(z) = 1+z5 +z6+z8 +z11+z12 +z13+z14 +z15+z16+z17+z20 +z22 +z23+z28
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Set elements with even partition function
orP1(z) = 1 + z2 + z6 + z8 + z9 + z10 + z14 + z18 + z19 + z20 + z22 + z26 + z28
andP2(z) = 1 + z3 + z4 + z5 + z6 + z8 + z9 + z10 + z13 + z14 + z15 + z18 + z19 + z20 +z22 + z23 + z24 + z25 + z27 + z28.
3.3 Proof of Theorem 1.1
Now, we consider the general case and write P = ∏i Pαi
i for the irreducible factor-ization of P over F2. Using the fact that
S(A(P), m) =∑
i
S(A(Pαii ), m)
and Lemma 2.2 (iii), one can deduce that S(A(P), m) is algebraic. It follows that bygrouping the Pi having the same order and taking into account the above cases, weobtain a polynomial M(P)(y) with integer coefficients having S(A(P), m) as a root.
An example: Let Q = Pa1 Pb
2 Pc3 where P1(z) = 1 + z + z2, P2(z) = 1 + z + z3 and
P3(z) = 1 + z + z2 + z4 + z6 + z7 + z8 are irreducible of order, respectively, 3, 7 and17. Taking m = 1, we obtain
M(Pa1 )(y) = y − a, M(Pb
2 )(y) = y2 − by + 2b2 and M(Pc3 )(y) = y2 − cy − 4c2.
Hence, we define M(Pa1 Pb
2 )(y) and M(Q)(y) by
M(Pa1 Pb
2 )(y) = resz
(M(Pa
1 )(y − z),M(Pb2 )(z)
),
M(Q)(y) = resz
(M(Pa
1 Pb2 )(y − z),M(Pc
3 )(z))
,
which give
M(Pa1 Pb
2 )(y) = y2 − (2a + b)y + a2 + ab + 2b2
and
M(Q)(y) = y4 + (−4a − 2b − 2c)y3 + (5b2 + 3bc − 7c2 + 6ab + 6a2 + 6ac)y2
+ (−6a2c + 7bc2 + 14ac2 + 8c3 − 4a3 − 10ab2 − 5b2c − 6abc
− 6a2b − 4b3)y + a4 + 3a2bc + 5a2b2 + 5ab2c + 2a3b − 7abc2
+ 2a3c − 7a2c2 + 4ab3 − 8ac3 + 2b3c + 16c4 − 4bc3 + 4b4 + 14b2c2
Let e be the degree of S(A(P), m) as an algebraic number. Let us write P =∏j N j , where the irreducible factors of each of the polynomials N j over F2 are of
123
N. Baccar
the same order β j . Since each of the polynomials M(N j ) is of degree r(β j ), then byconstructing M(P) as we did in the last example, one can deduce that e ≤ ∏
j r(β j ).
Remark 3.3 Let us consider the case where m is divisible by some prime p belongingto Oβ(1). Then, for all d dividing m
p , we have, for all h, 1 ≤ h ≤ f ,
m
pdmod β ∈ Oβ(xh) ⇐⇒ m
dmod β ∈ Oβ(xh).
So, we get
λ(m, h) =∑
d | mp
md mod β ∈Oβ(xh)
μ(d) −∑
d | mp
mpd mod β ∈Oβ(xh)
μ(d) = 0.
By the same way, we obtain γ (m) = 0. This leads to S(A(P), m) = 0, and that theset A(P) does not contain elements of the form 2km. This result has already beenproved in [11], Theorem 1, (i).
Remark 3.4 The same things happen when ββ divides m, since in such case, forall d dividing m, we have m
d ≡ 0 (mod β), so that λ(m, h) = 0 and γ (m) =∑d | m μ(d) = 0.
References
1. Baccar, N.: Sets with even partition function and 2-adic integers. Period. Math. Hung. 55(2), 177–193(2007)
2. Baccar, N., Zekraoui, A.: Sets with even partition function and 2-adic integers II. J. Int. Seq. 13 (2010),Article 10.1.3
3. Baccar, N., Ben Saïd, F.: On sets such that the partition function is even from a certain point on. Int.J. Number Theory 5(3), 1–22 (2009)
4. Baccar, N., Ben Saïd, F., Zekraoui, A.: On the divisor function of sets with even partition functions.Acta Math. Hung. 112(1–2), 25–37 (2006)
5. Ben Saïd, F.: On a conjecture of Nicolas-Sárközy about partitions. J. Number Theory 95, 209–226(2002)
6. Ben Saïd, F.: On some sets with even valued partition function. Ramanujan J. 9, 63–75 (2005)7. Ben Saïd, F., Nicolas, J.-L.: Sets of parts such that the partition function is even. Acta Arith. 106,
183–196 (2003)8. Ben Saïd, F., Nicolas, J.-L.: On the counting function of sets with even partition function. Publ. Math.
Debr. 79(3–4), 36 (2011)9. Ben Saïd, F., Nicolas, J.-L.: Even partition functions, Séminaire Lotharingien de Combinatoire 46,
B46i (2002). http://www.mat.univie.ac.at/~slc/10. Ben Saïd, F., Nicolas, J.-L., Zekraoui, A.: On the parity of generalized partition function III. Journal
de Théorie des Nombres de Bordeaux 22(1), 51–78 (2010)11. Ben Saïd, F., Lahouar, H., Nicolas, J.L.: On the counting function of the sets of parts such that the
partition function takes even values for n large enough. Discret. Math. 306, 1115–1125 (2006)12. Gurak, S.: On the minimal polynomial of gauss periods for prime powers. Math. Comput. 75(256),
2021–2035 (2006)13. Hardy, G.H., Wright, E.M.: An Introduction to the Theory of Numbers. Clarendon, Oxford (1979)14. Lidl, R., Niederreiter, H.: Introduction to Finite Fields and Their Applications. Cambridge University
Press, New York (1994)
123
Set elements with even partition function
15. Mignotte, M.: Mathématiques pour le calcul formel. Presses Universitaires de France, Paris (1989)16. Nicolas, J.-L., Ruzsa, I.Z., Sárközy, A.: On the parity of additive representation functions. J. Number
Theory 73, 292–317 (1998)17. Nicolas, J.-L.: On the parity of generalized partition functions II. Period. Math. Hung. 43, 177–189
(2001)
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