# On Some Sets with Even Valued Partition Function

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<ul><li><p>THE RAMANUJAN JOURNAL, 9, 6375, 2005c 2005 Springer Science + Business Media, Inc. Manufactured in the Netherlands.</p><p>On Some Sets with Even Valued Partition FunctionF. BEN SAID Fethi.Bensaid@fsm.rnu.tnFaculte des Sciences de Monastir, Avenue de lenvironnement, 5019, Monastir, Tunisie</p><p>Dedicated to Professor J.-L. Nicolas on the occasion of his 60th birthday</p><p>Received June 13, 2002; Accepted April 10, 2003</p><p>Abstract. Let IN be the set of positive integers, B = {b1 < < bh} IN , N IN and N bh .A = A0(B, N ) is the set (introduced by J.-L. Nicolas, I.Z. Ruzsa and A. Sarkozy) such that A {1, . . . , N } = Band p(A, n) 0 (mod 2) for n IN and n > N , where p(A, n) denotes the number of partitions of n with partsin A. Let us denote by (A, n) the sum of the divisors of n belonging to A. In a paper jointly written with J.-L.Nicolas, we have recently proved that, for all k 0, the sequence ( (A, 2kn))n1 mod 2k+1 is periodic with anodd period qk . In this paper, we will characterize for any fixed odd positive integer q, the sets B and the integersN such that q0 = q, and those for which qk = q for all k 0. Moreover, a setA = A0(B, N ) is constructed withthe property that its period, i.e. the period of ( (A, n))n1 mod 2, is 217, and for which the counting function isasymptotically equal to that of A0({1, 2, 3, 4, 5}, 5) which is a set of period 31.</p><p>Key words: even partition functions, periodic sequences, sets defined by partition parity</p><p>2000 Mathematics Subject Classification: Primary11P81, 11P83</p><p>1. Introduction</p><p>IN will denote the set of positive integers andP(IN ) is the subsets of IN . IfA = {a1, a2, . . .} IN (where a1 < a2 < ), then p(A, n) denotes the number of partitions of n with partsin A, i.e. the number of solutions of the equation</p><p>a1x1 + a2x2 + = n</p><p>in non negative integers x1, x2, . . . . As usual, we shall set p(A, 0) = 1.We shall use the generating function</p><p>fA(z) =</p><p>n=0p(A, n)zn =</p><p>aA</p><p>11 za . (1.1)</p><p>If B = {b1, . . . , bh} = (where b1 < < bh), B IN , N IN and N bh , then thereis (cf. [5]) a unique set A = A0(B, N ) such that</p><p>A {1, . . . , N } = B and p(A, n) 0 (mod 2) for n IN , n > N . (1.2)Research supported by MIRA 2002 program no 0203012701, Number Theory, Lyon-Monastir.</p></li><li><p>64 BEN SAID</p><p>Let us recall the construction as described in [5]. The setA = A0(B, N ) will be defined byrecursion. We write An = A {1, 2, . . . , n} so that</p><p>AN = A {1, 2, . . . , N } = B.</p><p>Assuming that n N + 1 andAn1 has been defined so that p(A, m) is even for N + 1 m n 1. Then set</p><p>n A if and only if p(An1, n) is odd.</p><p>It follows from the construction that for n N + 1 we have</p><p>if n A, p(A, n) = 1 + p(An1, n)if n / A, p(A, n) = p(An1, n)</p><p>which shows that p(A, n) is even for n N + 1.Note that in the same way, any finite set B = {b1, b2, . . . , bk} can be extended to an</p><p>infinite set A so that Abk = B and the parity of p(A, n) is given for n N + 1 (where Nis any integer such that N bk).</p><p>As observed in [6], by the uniqueness of the construction of A = A0(B, N ), if M is anyinteger N and B = A {1, . . . , M}, then</p><p>A = A0(B, N ) = A0(B, M). (1.3)</p><p>If A IN , let (A, n) denotes the characteristic function of A, i.e.,</p><p> (A, n) ={1 if n A</p><p>0 if n / A,and for n 1,</p><p> (A, n) =</p><p>d|n (A, d)d =</p><p>d|n,dAd. (1.4)</p><p>It was recently proved (cf. [3]) that for all k 0 and all A = A0(B, N ), the sequence( (A, 2kn))n1 mod 2k+1 is periodic with an odd period qk . This property allowed one in[1] to determine the elements of the setA0({1, 2, 3}, 3) which satisfies qk = 7 for all k 0,and then to obtain an asymptotic for the counting function of this set. Although the methodgiven there can be applied to any setA = A0(B, N ), the calculations may prove difficult forlarger q0. In this paper, we will show out the general nature of the setsA = A0(B, N ), andwe will give a new method capable of solving more easily the above mentioned problemsfor some sets A = A0(B, N ) with large q0.</p><p>If A = A0(B, N ), let us define the polynomial PB,N (already considered in [6])</p><p>PB,N (z) =</p><p>0nJnz</p><p>n, (1.5)</p></li><li><p>ON SOME SETS WITH EVEN VALUED PARTITION FUNCTION 65</p><p>where J is the largest integer such that p(A, J ) is odd (such a J does exist since p(A, 0) =1), and n is defined by</p><p>p(A, n) n (mod 2), n {0, 1}.</p><p>We shall say that PB,N is the characteristic polynomial ofA = A0(B, N ) and we will write</p><p>PB,N = charA0(B, N ). (1.6)</p><p>Note that PB,N is of degree at most N and if IF2[z] is the ring of polynomials with coefficientsin {0, 1}, then PB,N IF2[z] and PB,N (0) = 1. Let the factorization of PB,N into pairwiserelatively prime irreducible polynomials over IF2[z] be</p><p>PB,N = P11 P22 Pss . (1.7)</p><p>We denote by i the order of Pi , i.e., the smallest integer such that Pi divides 1 + zi inIF2[z], and for all k 0, we set</p><p>Ik = { j : 1 j s, j 2k(mod 2k+1)},Jk = I0 I1 Ik = { j, 1 j s, j 0(mod 2k+1)},</p><p>and Tk = lcm jJk j (with Tk = 1 if Jk = ).In [3], it is proved that the period qk of the sequence ( (A, 2kn))n1 mod 2k+1 is odd</p><p>and</p><p>qk = Tk . (1.8)</p><p>In this paper, we shall prove (Proposition 2) that given any fixed polynomial P IF2[z], ofpositive degree N , and such that P(0) = 1, there is a unique set B {1, . . . , N } such thatP = PB,N = charA0(B, N ). Next, we will give (Theorem 1) a method which determinesBwhen P = PB,N is of the form (1.7) and when the sets Bi satisfying Pi = charA0(Bi , Ni ),with Ni = degree(Pi ), are known. As an application of the last result, a set A = A0(B, N )is constructed with the property that its period, i.e. the period of ( (A, n))n1 mod 2, is 217,and for which the counting function is asymptotically equal to that of A0({1, 2, 3, 4, 5}, 5)which is a set of period 31. In Section 3, we will characterize the setsA = A0(B, N ) whengiven any positive odd integer q , the period qk of ( (A, 2kn))n1 mod 2k+1 is q for allk 0; the sets A = A0(B, N ) for which q0 = q are also characterized.</p><p>All polynomials and series as well as sums and products that will be considered in thepaper are to be taken with coefficients in {0, 1}.</p><p>2. Sets A = A0(B, N) with fixed characteristic polynomial</p><p>Let us consider the ring of formal power series IF2[[z]]. For an element of this ring</p><p>f (z) = a0 + a1z + a2z2 + .</p></li><li><p>66 BEN SAID</p><p>and for n 0, we define the polynomial f Mod zn+1 by( f Mod zn+1)(z) =</p><p>0inai z</p><p>i . (2.1)</p><p>Note that if f and g are in IF2[[z]], then( f g) Mod zn+1 = (( f Mod zn+1)(g Mod zn+1)) Mod zn+1. (2.2)</p><p>Let B = {b1, . . . , bh} IN , B = , b1 < < bh , N bh , and consider the setA = A0(B, N ) defined by (1.2). Denote by PB,N the characteristic polynomial of A givenby (1.5). The following result defines PB,N in another way.</p><p>Proposition 1. Under the above notations, we have</p><p>PB,N = char A0(B, N ) = fB(z) Mod zN+1. (2.3)</p><p>Proof: Recall that if A = A0(B, N ) and fA(z) =</p><p>n=0 p(A, n)zn is its generatingfunction considered as an element of IF2[[z]], then from (1.1), we have</p><p>fA(z) =</p><p>aA</p><p>11 za =</p><p>aB</p><p>11 za</p><p>aAa>N</p><p>11 za , (2.4)</p><p>since by (1.2), A {1, . . . , N } = B.But, from the definition given to PB,N = charA0(B, N ) in (1.5), we deduce immediately</p><p>that</p><p>PB,N = fA(z) Mod zN+1,which together with (2.4) and (2.2) prove (2.3).</p><p>Now, let</p><p>E = {(B, N ) : N IN ,B {1, . . . , N },B = }. (2.5)By (1.3), it turns out that the relation defined on E by</p><p>(B, N ) (B, N ) A0(B, N ) = A0(B, N ), (2.6)is an equivalence relation. Moreover, if (B, N ) is the class of (B, N ) E , it is obvious by(1.3) that</p><p>(B, N ) = {(B, N ) E : N N and B = A0(B, N ) {1, . . . , N }} {(B, N ) E : N N and B = A0(B, N ) {1, . . . , N }}</p><p>Denote by E/ the quotient set, and let</p><p>F = {P : P IF2[z], P(0) = 1 and degree (P) 1}. (2.7)</p></li><li><p>ON SOME SETS WITH EVEN VALUED PARTITION FUNCTION 67</p><p>From the definition of the characteristic polynomial ofA = A0(B, N ), we can prove withoutdifficulty that the mapping</p><p>E/ F(B, N ) PB,N = char A0(B, N )}</p><p>is injective. In fact, we will prove in the following result that it is a one to one correspondence.</p><p>Proposition 2. Let N be a positive integer, P IF2[z], P(0) = 1 and degree (P) = N.There exists a unique set B {1, . . . , N }, B = , such that</p><p>P = PB,N = char A0(B, N ).</p><p>Note that if (B, N ) E and P is the characteristic polynomial of A = A0(B, N ) definedby (1.5), then degree(P) = M N . Thus Proposition 2 implies that there is a uniqueset B such that the couple (B, M) represents the class (B, N ) according to the equivalencerelation given by (2.6). Moreover, with the conventionA0(, 0) = , the conditions B = and degree(P) 1 in (2.5) and (2.7) can be removed.</p><p>Proof of Proposition 2: We shall proceed by induction on N . For if N = 1, then P = 1+z,so that by setting B = {1}, we get from (2.3),</p><p>PB,1 = charA0(B, 1) = 11 z Mod z2 = 1 + z = P,</p><p>which proves that the result is true for N = 1. Suppose that the result holds up to anyN 1, and take P IF2[z] such that P(0) = 1 and degree (P) = N + 1. If we set P1 = PMod zN+1, then 0 degree (P1) N .</p><p> When degree (P1) = 0, we have P = 1 + zN+1, so that by setting B = {N + 1}, we getas above</p><p>PB,N+1 = charA0(B, N + 1) = 11 zN+1 Mod zN+2 = 1 + zN+1 = P.</p><p> If 1 degree(P1) = M N , then from induction hypothesis, there is some B1 {1, . . . , M} satisfying</p><p>P1 = PB1,M = charA0(B1, M).</p><p>But if B2 = A0(B, M) {1, . . . , N }, we have from (1.3), A0(B1, M) = A0(B2, N ), andso</p><p>P1 = PB1,M = PB2,N .</p></li><li><p>68 BEN SAID</p><p>Suppose for instance that</p><p>fB2 (z) Mod zN+1 = fB2 (z) Mod zN+2,</p><p>and set B = B2. We have from (2.3)</p><p>charA0(B, N + 1) = fB2 (z) Mod zN+2 = fB2 (z) Mod zN+1 + zN+1= PB2,N + zN+1 = P1 + zN+1 = P.</p><p>Now, if</p><p>fB2 (z) Mod zN+1 = fB2 (z) Mod zN+2,</p><p>then by setting B = B2 {N + 1}, we obtain as above</p><p>charA0(B, N + 1) =(</p><p>fB2 (z).1</p><p>1 zN+1)</p><p>Mod zN+2</p><p>= ( fB2 (z)(1 + zN+1)) Mod zN+2 = fB2 (z) Mod zN+1 + zN+1,</p><p>since fB2 (0) = 1. Hence</p><p>charA0(B, N + 1) = P1 + zN+1 = P.</p><p>This completes the proof of the existence of B. The uniqueness of such a B follows fromthe fact that if B {1 . . . , N } is another set satisfying</p><p>fB(z) Mod zN+1 = fB (z) Mod zN+1,</p><p>then for all i , 1 i N , we have</p><p>fB(z) Mod zi+1 = fB (z) Mod zi+1.</p><p>Now, what can one say about the set B defined by the relation P = charA0(B, N ), whenP is of degree N and is the product of two polynomials, P1, P2 F? In particular, can Bbe expressed in terms of the sets B1,B2 obtained from the relations</p><p>P1 = charA0(B1, N1), P2 = charA0(B2, N2),</p><p>where N1 and N2 are respectively the degrees of P1 and P2?Let us define the following application</p><p>g : P(IN ) P(IN ) P(IN ) P(IN )(2.8)</p><p>(A, B) g(A, B) = (g1(A, B), g2(A, B)),</p></li><li><p>ON SOME SETS WITH EVEN VALUED PARTITION FUNCTION 69</p><p>where g1(A, B) = AB = (A B)\(A B) and g2(A, B) = 2(A B) = 2A 2B. Weshall write gk for the k composite of g , i.e.,</p><p>gk = ((gk)1, (gk)2). (2.9)</p><p>Let f A be the generating function of A defined as in (1.1). The following result gives arecursive congruence mod 2 for f A. fB .</p><p>Proposition 3. Under the above notations. For all k 1 and all A, B P(IN ), we have</p><p>f A. fB f(gk )1(A,B). f(gk )2(A,B) mod 2. (2.10)</p><p>Proof: We shall proceed by induction. For if k = 1, then</p><p>f A. fB = f AB f 2AB f AB . f2(AB) mod 2,</p><p>since for any C P(IN ), f 2C f2C mod 2. Hence</p><p>f A. fB fg1(A,B). fg2(A,B) mod 2,</p><p>and the result is true for k = 1. Assuming that the result holds up to any k 1 then we have</p><p>f A. fB f(gk )1(A,B). f(gk )2(A,B) mod 2, (2.11)</p><p>and by induction hypothesis</p><p>f(gk )1(A,B). f(gk )2(A,B) f(gk )1(A,B)(gk )2(A,B). f2((gk )1(A,B)(gk )2(A,B))= f(gk+1)1(A,B). f(gk+1)2(A,B),</p><p>which with (2.11) complete the proof.</p><p>Note that if A B = , then f A. fB = f AB . f = f AB , and thus</p><p>(gk)2(A, B) = , (gk)1(A, B) = A B, for all k IN . (2.12)</p><p>We also have another interesting property of the application g.</p></li><li><p>70 BEN SAID</p><p>Proposition 4. Let A and B be in P(IN ), and for k IN , gk = ((gk)1, (gk)2) is theapplication defined by (2.8) and (2.9). Then</p><p>(gk)2(A, B) 2k(A B), (2.13)</p><p>(AB) </p><p>i=12i (A B) (gk)1(A, B) (AB) </p><p>i=12i (A B). (2.14)</p><p>Proof: The relation (2.13) is immediate by induction since</p><p>(g1)2(A, B) = g2(A, B) = 2(A B),(gk+1)2(A, B) = 2((gk)1(A, B) (gk)2(A, B)).</p><p>Let us prove (2.14). Since</p><p>(g1)1(A, B) = g1(A, B) = AB,</p><p>the case k = 1 is true. Suppose that the result holds up to any k 1, then</p><p>(gk+1)1(A, B) = (gk)1(A, B)(gk)2(A, B) (gk)1(A, B) (gk)2(A, B)</p><p>(</p><p>(AB) </p><p>i=12i (A B)</p><p>) 2k(A B)</p><p>= (AB) </p><p>i=12i (A B).</p><p>On the other hand,</p><p>(gk+1)1(A, B) = (gk)1(A, B)(gk)2(A, B) (gk)1(A, B) \ (gk)2(A, B)</p><p> (AB) </p><p>i=12i (A B).</p><p>This completes the proof of Proposition 4.</p><p>Now, we can assert the main result of this paper.</p><p>Theorem 1. Let P1, P2 IF2[z] ; Pi (0) = 1, degree (Pi ) = Ni 1. Denote by Bi the setsatisfying (cf. Proposition 2),</p><p>Bi {1, . . . , Ni } and Pi = charA0(Bi , Ni ).</p></li><li><p>ON SOME SETS WITH EVEN VALUED PARTITION FUNCTION 71</p><p>Moreover, let N = N1 + N2, and set</p><p>B1 = A0(B1, N1) {1, . . . , N }, (2.15)B2 = A0(B2, N2) {1, . . . , N }. (2.16)</p><p>Then, the unique set B {1, . . . , N } such that</p><p>P = P1 P2 = charA0(B, N ),</p><p>is given by</p><p>B = limk</p><p>(gk)1(B1,B2) {1, . . . , N }, (2.17)</p><p>where (gk)1 is the first coordinate of the application gk defined previously by (2.8) and(2.9). (Note that by (2.12), the above sequence in (2.17) is finite).</p><p>Moreover, if A = A0(B1, N1), A = A0(B2, N2) and A = A0(B, N ), then</p><p>A = limk</p><p>(gk)1(A,A), (2.18)</p><p>where the sequence here is infinite.</p><p>Proof: We have from (2.2),( fB1 . fB2</p><p>)Mod zN+1 = ( fB1 Mod zN+1</p><p>)( fB2 Mod zN+1)</p><p>Mod zN+1,</p><p>so that by (2.15), (2.16),(2.3) and (1.3), we get( fB1 . fB2</p><p>)Mod zN+1 = ( fB1 Mod zN1+1</p><p>)( fB2 Mod zN2+1))</p><p>Mod zN+1</p><p>= (P1 P2) Mod zN+1 = P1 P2.</p><p>Thus, by applying Proposition 3 with k such that</p><p>(gk)2(B1,B2) {1, . . . , N } = ,</p><p>(2.17) follows then from (2.12) and Proposition 2.(2.18) follows from (2.17) and the uniqueness of the sets of the form A0(B, N ), as</p><p>mentioned in (1.2).</p><p>Let as apply Theorem 1 on a concrete example. Consider the polynomials P1 and P2 ofdegrees N1 = 3 , N2 = 5 respectively, and given by</p><p>P1(z) = 1 + z + z3,P2(z) = 1 + z + z3 + z4 + z5.</p><p>By the use of Propositions 1 and 2, an easy computation gives B1 = {1, 2, 3} and B2 ={1, 2, 3, 4, 5}. Besides, with a slightly further calculation, we obtain (as already</p></li><li><p>72 BEN SAID</p><p>given in [6])</p><p>B1 = A0(B1, 3) {1, . . . , 8} = {1, 2, 3, 5, 8},B2 = A0(B2, 5) {1, . . . , 8} = {1, 2, 3, 4, 5, 7, 8}.</p><p>Hence the unique set B {1, . . . , 8} such thatP = P1 P2 = charA0(B, 8),</p><p>is obtained by the following procedure mod 2:</p><p>fB Mod z9 fB1 . fB2 Mod z9= f{1,2,3,5,8}. f{1,2,3,4,5,7,8} Mod z9 f{4,7}. f{2,4,6} Mod z9 f{2,6,7}. f{8} Mod z9 f{2,6,7,8} Mod z9,</p><p>which gives B = {2, 6, 7, 8}.Let us consider the setA = A0({2, 6, 7, 8}, 8). Since the period ofA = A0({1, 2, 3}, 3)</p><p>is 7 (cf. [6]), and that of A = A0({1, 2, 3, 4, 5}, 5) is 31 (cf. [4]), (these results can alsobe deduced from (1.8) ), then by the use of (1.8), we can prove without difficulty that theperiod ofA is 217. Moreover, for real x 1, let A(x), A(x) and A(x) be respectively thecounting functions of A, A and A then, when x tends to infinity, we have</p><p>A(x) A(x). (2.19)Proof of (2.19): From (2.18) and (1.3), we obtain</p><p>A(x) = limk</p><p>(gk)1(A,A) {1, . . . , x},</p><p>where A(x) (and so for A(x), A(x)) is defined here byA(x)...</p></li></ul>