On the elements of sets with even partition function

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<ul><li><p>Ramanujan JDOI 10.1007/s11139-014-9618-z</p><p>On the elements of sets with even partition function</p><p>N. Baccar</p><p>Received: 2 April 2014 / Accepted: 20 July 2014 Springer Science+Business Media New York 2014</p><p>Abstract Let P F2[z] such that P(0) = 1 and degree(P) 1. Nicolas, Ruzsaand Srkzy proved that there exists a unique subset A = A(P) of N such that</p><p>n0 p(A, n)zn P(z) (mod 2), where p(A, n) is the number of partitions of nwith parts inA. Let m be an odd positive integer and let S(A,m) be the 2-adic integergiven by the expansion S(A,m) = (A,m) + 2(A, 2m) + 4(A, 4m) + =</p><p>k=0 2k(A, 2km), where (A, ) is the characteristic function of the set A. Byknowing the expansion S(A,m), one can obtain the elements of the setA of the form2km. In this paper, we prove that S(A,m) is an algebraic number.</p><p>Keywords Partitions Periodic sequences Order of a polynomial Cyclotomicpolynomials Resultant 2-Adic integers The Graeffe transformation</p><p>Mathematics Subject Classification 11P83 11B50 11D88</p><p>1 Introduction</p><p>Let N be the set of positive integers and letA 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, . . . , nkbelonging to A such that</p><p>n = n1 + n2 + + nk .</p><p>N. Baccar (B)Dp. de Math Inf., Universit de Sousse ISITCOM Hammam Sousse, 5 Bis Rue 1 Juin 1955,4011 Hammam Sousse, Tunisiae-mail: naceurbaccar@yahoo.fr</p><p>123</p></li><li><p>N. Baccar</p><p>The number of partitions of n with parts inA 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))n0 is even from acertain point on.</p><p>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 Srkzy(see [16]) proved that thereis a unique even partition set A = A(P) satisfying the congruence</p><p>n0p(A, n)zn P(z) (mod 2). (1.1)</p><p>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</p><p>A(P, x) = |{n A : n x}|.</p><p>From now on, for any odd positive integer d, wewrite s(d) for the order of 2modulod; 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 (d) = s(d)r(d), where is Eulersfunction.</p><p>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 wsuch that P(z) divides 1+ zw in F2[z].</p><p>Let the factorization of P into irreducible factors over F2[z] be</p><p>P = Q11 Q22 Qtt .</p><p>We denote by P the squarefree kernel of P; that is, P = Q1Q2 Qt the productof the distinct irreducible polynomials dividing P . In [11], F. Ben Sad et al provedthat the counting function A(P, x) of the set A(P) satisfies</p><p>A(P, x) x(log x) 1r() ,</p><p>where is the order of P. In [8], Ben Sad and Nicolas improved on the constantc = c() for which</p><p>A(P, x) x(log x)c.</p><p>Let m be an odd positive integer. By (A, n), we denote the characteristic functionof the set A = A(P); that is, we put (A, n) = 1 if n A and zero otherwise. We</p><p>123</p></li><li><p>Set elements with even partition function</p><p>define the 2-adic integer S(A,m) by the expansion</p><p>S(A,m)=(A,m)+ 2(A, 2m)+4(A, 4m)+ =</p><p>k=02k(A, 2km). (1.2)</p><p>By knowing S(A,m) modulo 2k+1, we can deduce the values of (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 setA(P) ofthe form 2km.</p><p>The aim of the present paper is to prove the following theorem:</p><p>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) = p12 (see [1] and [3]) and whens(p) = p13 (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],</p><p>1+ z7 = (1+ z)(1+ z + z3)(1+ z2 + z3).</p><p>In this case, we have</p><p>y2 y + 2 = (y S(A(P1), 1))(y S(A(P2), 1)),</p><p>where</p><p>S(A(P1), 1) = 1+ 2+ 23 + 24 + 26 + 213 + + 2997 + </p><p>and</p><p>S(A(P2), 1) = 2+ 22 + 25 + 27 + 28 + 29 + + 2999 + ,</p><p>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, . . .).</p><p>As described in [16], the set A(P) is constructed by recursion, and this enablesto 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.</p><p>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.</p><p>123</p></li><li><p>N. Baccar</p><p>2 Some results on even partition sets</p><p>Let P be a polynomial in F2 [z] of positive degree such that P(0) = 1 and let bethe order of P.</p><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 .</p><p>Congruences of formal power series may be added, multiplied and derivated. More-over, we have the following:</p><p>if f (z) g(z) (mod M) and f (0) = g(0) = 1, then 1f (z)</p><p> 1g(z)</p><p>(mod M).</p><p>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</p><p>f (z) = a0 + a1z + a2z2 + . . .+ anzn + . . .</p><p>of this ring, the product</p><p>f (z) f (z) = b0 + b1z2 + b2z4 + . . .+ bnz2n + . . .</p><p>is an even power series. We shall call G( f ) (Graeffe transformation) the series</p><p>G( f )(z) = b0 + b1z + b2z2 + . . .+ bnzn + . . . .</p><p>It follows immediately that for f, g K[[z]],</p><p>G( f g) = G( f )G(g). (2.1)</p><p>We shall use the following notation for the iterates of f by the transformation G:</p><p>f(0) = f, f(1) = G( f ), . . . , f(k) = G( f(k1)) = G(k)( f ).</p><p>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</p><p>f(1) f (mod 2), (2.2)and for any 2-adic integer M ,</p><p>f g (mod 2M) f(1) g(1) (mod 4M). (2.3)</p><p>123</p></li><li><p>Set elements with even partition function</p><p>Moreover, the last property is generalized (cf. [7]) to</p><p>f g (mod 2) f(k) g(k) (mod 2k+1), k 0. (2.4)</p><p>Using the elementary theory of finite fields (see [14]), we know that</p><p>1 z =</p><p>d |d(z), (2.5)</p><p>where d is the cyclotomic polynomial of index d. Since is odd, then factors inF2[z] into r() distinct irreducible polynomials, each of degree s() and of order . Itwill also be of use to include here the following basic identities concerning cyclotomicpolynomials:</p><p>(z) =</p><p>d |(1 z d )(d). (2.6)</p><p>In [1] (see also [3]), it is proved that the polynomial P can be extended in a 2-adicpolynomial P such that</p><p>P P(k) (mod 2k+1), k 0. (2.7)</p><p>Let Q is a polynomial of F2[z] of positive degree such that Q(0) = 1. Therefore, from(2.1), it follows that</p><p>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 (A, n) denote the sum of those divisors of n belonging to A; that is,</p><p>(A, n) =</p><p>d | n, dAd. (2.9)</p><p>It is proved (see [4]) that for all k 0, is a period of the sequence( (A, 2kn)mod 2k+1)n1. Moreover, if P is irreducible in F2, then for all k 0,</p><p>(A, 2k) s() (mod 2k+1). (2.10)</p><p>Let (Z/Z) denote the group of invertible residues modulo and let&lt; 2 &gt; be itssubgroup generated by 2. Consider the action of &lt; 2 &gt; on the set Z/Z given bya n = an for all a &lt; 2 &gt; and n Z/Z. Then, denote the orbit of some n Z/Zby O(n); that is,</p><p>O(0) = {0},</p><p>andO(n) = {2 j n mod , 0 j s(/) 1}, (2.11)</p><p>123</p></li><li><p>N. Baccar</p><p>where := (n) = gcd(, n). It is easy to see that for every non-negative integer h,we have O(2hn) = O(n). Moreover, for all positive integer i , one has</p><p>{2 j n mod , is(/) j (i + 1)s(/) 1} = O(n). (2.12)</p><p>Such action allows one to write</p><p>(Z/Z)/ = {O(x1), . . . , O(xr())} {O(xr()+1), . . . , O(x f ), O(0)},(2.13)</p><p>where O(x1), . . . , O(xr()) are the invertible orbits, while O(xr()+1), . . . ,O(x f ), O(0) are the non-invertible ones. We will always set xr() = 1. In [4],it is proved that if q mod and n mod are in the same orbit, then for all k 0,</p><p>(A, 2kq) (A, 2kn) (mod 2k+1). (2.14)</p><p>In particular (taking q = 2n),</p><p>(A, 2k+1n) (A, 2kn) (mod 2k+1)</p><p>Hence, we shall consider the 2-adic integers (A, n) defined by</p><p>(A, n) (A, 2kn) (mod 2k+1), k 0. (2.15)</p><p>One may easily verify that the sequence ((A, n))n1 is periodic with period .Moreover, it turns out (see [1] and [3]) that</p><p>+</p><p>n=1(A, n)zn = z P</p><p>(z)P(z)</p><p>. (2.16)</p><p>The last equality with the properties of the logarithmic derivative gives</p><p>(A(P Q), n) = (A(P), n)+ (A(Q), n). (2.17)</p><p>Let S(A,m) be the 2-adic integer given by (1.2). Using (2.9) and (2.15), one cansee that</p><p>(A,m) =</p><p>d |md S(A, d). (2.18)</p><p>Applying Mbius inversion formula on the last equality, we obtain</p><p>mS(A,m) =</p><p>d |m(d)</p><p>(A, m</p><p>d</p><p>)=</p><p>d | m(d)</p><p>(A, m</p><p>d</p><p>), (2.19)</p><p>where m = p |m, p prime</p><p>p denotes the radical of m with 1 = 1, and where is theMbiuss function.</p><p>123</p></li><li><p>Set elements with even partition function</p><p>Lemma 2.1 Let be a positive integer and let S(A(P),m) be the 2-adic integerdefined by (1.2). Then</p><p>S(A(P),m) = S(A(P),m) (2.20)Proof From (2.17), we see that</p><p>(A(P), n) = (A(P), n), (2.21)</p><p>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 isa power of 2, the setA(P) can be expressed in terms of the setA(P), in other words</p><p>A(P2q ) = 2q A(P) := {2qn, n A(P)}.</p><p>LetK be some field, H(z) = a0+a1z+ +anzn and L(z) = b0+b1z+ +bk zkbe 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]).</p><p>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 aiand b j .</p><p>(ii) If H(z) is written as H(z) = an(z z1)(z z2) (z zn) in the splitting fieldof H over K, then</p><p>resz(H(z), L(z)) = aknn</p><p>i=1L(zi ). (2.22)</p><p>(iii) If z1, z2, . . . , zn and q1, q2, . . . , q are the roots of H and L, respectively, in analgebraic closure K of K , then the n roots of the polynomial resz(H(yz), L(z))in K are zi + q j , 1 i n, 1 j .</p><p>3 Elements of even partition sets</p><p>Let P be a polynomial in F2 [z] of positive degree such that P(0) = 1 and let 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() and r = r() andrecall that</p><p>Z/Z = O(x1) O(x2) O(x f ) O(0). (3.1)</p><p>123</p></li><li><p>N. Baccar</p><p>3.1 Case P is an irreducible polynomial</p><p>Let be (cf. Sect. 2) the cyclotomic polynomial of index and write</p><p>(z) r</p><p>=1P(z) (mod 2), (3.2)</p><p>where the Ps are the only irreducible polynomials over F2[z] of the same degree sand all of which are of order . It follows from (2.6) and (2.1) that is a fixed pointof the Graeffe transformation. Therefore, from (2.8), we have</p><p> = P1P2 Pr = P1 P2 Pr . (3.3)</p><p>Let be a -th primitive root of unity over the 2-adic fieldQ2 and such that Pr ( ) = 0.Since for all k 0, (Pr )(k) = Pr , it follows that , 2, . . . , 2s1 are the roots of Prand the polynomials P can be arranged so that (see (2.13))</p><p>P(z) = (1)s(z x )(z 2x ) (z 2s1x ), 1 r. (3.4)</p><p>For all , 1 r , let A = A(P) be the even partition set satisfying (1.1).</p><p>Lemma 3.1 Let v be a positive integer and let (A, v) be the 2-adic integer definedby (2.15). Then</p><p>(A, v) = D(vx ), 1 r, (3.5)where = 1 and D(z) is the polynomial given by</p><p>D(z) = z + z2 + + z2s1 mod . (3.6)</p><p>Proof Using the fact that is a period of ((Ar , n))n1 and that is -th root ofunity, it suffices to prove the last lemma when v {1, 2, . . . , }. Since</p><p>n=1nv (mod )</p><p>(Ar , n)zn = (Ar , v)</p><p>u=0zv+u,</p><p>it follows that</p><p>n=1(Ar , n)zn =</p><p>1</p><p>v=1</p><p>n=1nv (mod )</p><p>(Ar , n)zn</p><p>+</p><p>n=1n0 (mod )</p><p>(Ar , n)zn .</p><p>123</p></li><li><p>Set elements with even partition function</p><p>Therefore,</p><p>n=1(Ar , n)zn = 1</p><p>1 z1</p><p>v=1(Ar , v)zv + (Ar , ) z</p><p>1 z . (3.7)</p><p>On the other hand, from (3.4), we get</p><p>zP r (z)Pr (z)</p><p>=s1</p><p>j=0</p><p>z</p><p>z 2 j = s</p><p>j=0</p><p>n=1(z/ 2</p><p>j)n</p><p>= s1</p><p>j=0</p><p>1</p><p>v=1</p><p>n=1nv (mod )</p><p>(z/ 2j)n +</p><p>n=1n0 (mod )</p><p>(z/ 2j)n</p><p>= s1</p><p>j=0</p><p>1</p><p>v=1</p><p>zv</p><p> 2jv</p><p>+</p><p>u=0zu + z</p><p>1 z</p><p>= 11 z</p><p>1</p><p>v=1zv</p><p>s1</p><p>j=02</p><p>jv</p><p> s z</p><p>1 z ,</p><p>where = 1. Using (2.16) and comparing the last equality with (3.7), we obtain (3.5)</p><p>for = r . To get (3.5) for all , 1 r , we just have to replace in the above proofr by and by x . </p><p>For , 1 r , let S(A,m) be the 2-adic integer defined by (1.2). We definethe polynomial Dm by</p><p>Dm(z) =f</p><p>h=1(m, h)Bh(z)+ s (m), (3.8)</p><p>where Bh(z) is the polynomial given by</p><p>Bh(z) =s1</p><p>j=0z2</p><p>j xh mod (3.9)</p><p>and</p><p>(m, h) =</p><p>d | mmd mod O(xh)</p><p>(d), (3.10)</p><p> (m) =</p><p>d | mmd 0 (mod )</p><p>(d). (3.11)</p><p>123</p></li><li><p>N. Baccar</p><p>Clearly, (m) = 0 if |m while if m, we have</p><p> (m) =</p><p>d | gcd(m,m)</p><p>(d) ={1 if m = 0 if m = </p><p>We now have the tools necessary to prove the following</p><p>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</p><p>Rm(y) = resz((z),my + Dm(z)</p><p>), (3.12)</p><p>where (z) is the cyclotomic polynomial of index .</p><p>Proof It follows from (2.19) that</p><p>mS(A,m)=f</p><p>h=1</p><p>d | mmd mod O(xh)</p><p>(d)(A, md)</p><p>+</p><p>d | mmd 0 (mod )</p><p>(d)(A, md),</p><p>which, by (2.14) and (2.15), gives</p><p>mS(A,m) =f</p><p>h=1(A, xh)</p><p>d | mmd mod O(xh)</p><p>(d)</p><p> s</p><p>d | mmd 0 (mod )</p><p>(d).</p><p>Consequently,</p><p>mS(A,m) =f</p><p>h=1(m, h)(A, xh) s (m),</p><p>which, by (3.5) and (3.9), gives</p><p>mS(A,m) = f</p><p>h=1(m, h)Bh(</p><p>x ) s (m).</p><p>From (3.8) and the last equality, we have</p><p>Dm(x ) = mS(A,m). (3.13)</p><p>From (3.3), (2.22) and...</p></li></ul>