A crank-based approach to the theory of 3-core partitions
aa r X i v : . [ m a t h . N T ] J a n A CRANK-BASED APPROACH TO THE THEORY OF -COREPARTITIONS OLIVIER BRUNAT AND RISHI NATH
Abstract.
This note is concerned with the set of integral solutions of theequation x + 3 y = 12 n + 4 , where n is a positive integer. We will describea parametrization of this set using the -core partitions of n . In particularwe construct a crank using the action of a suitable subgroup of the isometricgroup of the plane that we connect with the unit group of the ring of Eisensteinintegers. We also show that the process goes in the reverse direction: from thesolutions of the equation and the crank, we can describe the -core partitions of n . As a consequence we describe an explicit bijection between -core partitionsand ideals of the ring of Eisenstein integers, explaining a result of G. Han andK. Ono obtained using modular forms. Introduction
Let t be a positive integer. The set of t -core partitions is a subset of the set ofinteger partitions which is important in the representation theory of the symmetricgroups S n . Indeed, the famous Nakayama conjecture (since proven) says that,when t is a prime number, t -core partitions label the t -blocks of the symmetricgroups. In particular, defect zero characters of the symmetric group are labeled by t -core partitions of n . On the other hand, t -core partitions also appear cruciallyin number theory. For example, F. Garvan, D. Kim and D. Stanton give in [8] avery elegant proof of Ramanujan’s congruences by constructing cranks using t -corepartitions. Following Dyson [6], a crank over a finite set A is a map c : A → B toanother set B , such that all fibers of c have the same cardinality. In this paper, weconsider the equation over integers x + 3 y = k, (1)where k is a nonnegative integer. Denote by U k the set of tuples ( x, y ) ∈ Z that aresolution of (1), and set u ( k ) = |U k | . Equation (1) was first considered by Fermatwho conjectured in 1654 [5, p. 8] that if k is a prime number congruent to modulo , then u ( k ) = 0 . This conjecture was proved by Euler [5, p. 12] in 1759.In this note, we consider Equation (1) for k = 12 n + 4 , where n is a nonnegativeinteger. We will construct a crank c : U n +4 → ( Z / Z ) connected with the set E n of -core partitions of n . In particular, we obtain a parametrization of theelements of U n +4 that depends on E n . Moreover, this process goes in the reversedirection. From the set of solutions of (1), we can recover E n as a fiber of c . In fact,such a connection is not surprising. Indeed, in 2014, N.D. Baruah and K. Nathshowed in [2, Theorem 3.1] using the theory of modular forms and Ramanujan’s Mathematics Subject Classification. theta function identities that u (12 n + 4) = 6 a ( n ) , where a ( n ) = |E n | . Note that as a consequence of our work, we obtain a new proofof this equality.Recall that a partition λ of n is a non-increasing sequence of positive integerssumming to n . The integer n is then called the size of λ , also denoted by | λ | . Let P be the set of partitions of integers, and P n be the one of partitions of n . To any λ = ( λ , . . . , λ r ) ∈ P , we associate its Young diagram. This is a collection of boxesarranged in left- and top-aligned rows so that the number of boxes in its j th rowis λ j .The hook associated to a box b of the Young diagram of λ is the set of the boxesof the diagram to the right and below b . The number of boxes appearing in a hookis called its hooklength. We recall that a -core partition is a partition with nohooklength equal to . We denote by E the set of -core partitions. Furthermore,as already mentioned, we write E n for the set of λ ∈ E of size n . Recall that A.Granville and K. Ono showed in [9] that every natural number has a t -core partitionwhenever t ≥ . In the same paper, the authors proved [9, §3] that this is not thecase for t = 3 ; see also Remark 4.3.The construction in this paper is based on the characteristic vectors approachfor the -core partitions. In [8], Garvan, Kim and Stanton proved that the set E isin bijection with the set C = { ( c , c , c ) ∈ Z | c + c + c = 0 } . (2)Recently, the two authors proposed a new approach based on the Frobeniussymbol of a partition [7] to interpret this bijection using abaci theory. Note thatabaci methods were previously used in [15] by K. Ono and L. Sze to connect -corepartitions and class group structure of the integer rings of the imaginary quadraticfields Q ( √− n − . Such links were also established in many cases; see for exam-ple [1], [3] and [14].Now, we present more precisely our strategy. For any nonnegative integer n , wedefine a crank c : U n +4 → ( Z / Z ) . To prove that c is a crank, we will describea natural action of Z / Z over U n +4 and show that it permutes the fibers of c ;see Theorem 2.1. On the other hand, we will construct an injective map between E n and U n +4 (see Lemma 3.1), and we show that its image is a fiber of c . Inparticular, this is a set of representatives for the Z / Z -action. We then obtain away to parametrize the elements of U n +4 . See Theorem 3.2. Conversely, we canrecover E n from U n +4 . See Remark 3.4.In §4, we give the main result of the paper. Let K = Q ( √− . In 2011, G. Hanand K. Ono connect -core partitions and ideals of the ring of Eisenstein integers [11,Theorem 1.4] by proving an identity between modular forms. Ono has asked for adirect explanation for this phenomenon. We will see that our construction allowsto describe E n using this ring. In particular, we obtain in Theorem 4.1 an explicitbijection between the set of -core partitions and the ideals of the ring of Eisensteinintegers of norm n + 1 . This explains the result of Han and Ono.In the last section, we give some consequence of our approach. First, we connectthe numbers of -core partitions of n and of kn + k − for any integer k not divisibleby p and such that the -valuation at all prime with residue modulo is even.See Formula (13). This generalizes results of D. Hirschhorn, and J. A. Sellers [12], also proved by another method by Baruah and Nath in [2]. Then we will prove thatthe new equality a (3 n + (3 k +1 + 2) n + 3 k ) = a ( n ) a ( n + 3 k ) (3)holds for any integers n ≥ and k ≥ . Furthermore, using the crank c of §2,we construct an explicit bijection that explains it. This example illustrates theadvantage of the methods developed in the present work. Finally, we discuss aquestion of Han [10, 5.2].2. Construction of a crank
For any symmetric matrix A ∈ M ( R ) , we denote its corresponding orthogonalgroups by O ( A ) , that is O ( A ) = { Q ∈ GL ( R ) | Q T AQ = A } , where Q T denotes the transposed matrix of Q . When A = I is the identity matrix,we simply write O ( R ) for O ( I ) .Let k be a nonnegative integer. In this section, we consider the sets A k = { ( x, y ) ∈ R | x + y = k } and B k = { ( x, y ) ∈ R | x + 3 y = k } . Note that we can interpret the sets A k and B k using matrices. For that, we write D ∈ M ( R ) for the diagonal matrix with entries and . By abuse of notation, weidentify the elements of R with column vectors. Then we have X ∈ A k if and onlyif X T X = k . Similarly, Y ∈ B k if and only if Y T DY = k . Moreover, we remarkthat O ( R ) and O ( D ) act respectively on A k and B k by left multiplication.Now, we denote by P the diagonal matrix with entries and √ . In particular, P T P = D , and, by elementary bilinear algebra results, conjugation and multiplica-tion by P − induce respectively a group isomorphism between O ( R ) and O ( D ) and a bijection between A k and B k . We remark that U k is the subset of B k con-sisting of points with integral coordinates. Theorem 2.1.
Assume k = 12 n + 4 for a nonnegative integer n . Then there is acyclic subgroup G of O ( D ) of order that acts freely on U k . Moreover, G acts onthe fibers of the map c : U k −→ ( Z / Z ) , ( x, y ) (2 x, x + 2 y ) , where m denotes the residue class of m in Z / Z .Proof. The orthogonal group is well-known in dimension . It consists of rotationsand orthogonal symmetry. It follows that O ( D ) = P − O ( R ) P = (cid:26)(cid:18) √ (cid:19) (cid:18) cos θ − ε sin θ sin θ cos θ (cid:19) (cid:18) √ (cid:19) | θ ∈ R , ε ∈ {− , } (cid:27) = (cid:26)(cid:18) cos θ − ε √ θ √ sin θ cos θ (cid:19) | θ ∈ R , ε ∈ {− , } (cid:27) , where P is defined as above. Now, consider the element R of O ( D ) associatedwith the parameters θ = π/ and ε = 1 in the above description, that is R = 12 (cid:18) −
31 1 (cid:19) . OLIVIER BRUNAT AND RISHI NATH
This is the conjugate by P − of the rotation matrix R π/ of angle π/ of O ( R ) .In particular, R has order .We set G = h R i . Then G acts on B k by left multiplication. Let ( x, y ) ∈ B k .Write X = ( x y ) T . Denote by H the G -stabilizer of ( x, y ) . It is a subgroup of G oforder , , or . The order of H is not , because ( x, y ) = (0 , . We now excludethe orders and . Suppose H has order . Since G is cyclic, it only has onesubgroup of order , hence H = h R i , and R X = − X . It follows that X = − X ,that is X = 0 which is not possible. Suppose H has order . Then H = h R i . Thecondition R X = X gives the system (cid:26) − x − y = 0 x − y = 0 . Here again, X = 0 . It follows that H is trivial and the G -action is free. Each G -orbit then has size .As a subgroup of O ( D ) , G acts naturally on B k by left multiplication. We nowcheck that this action stabilizes U k . Let ( x, y ) ∈ U k . Then x and y are integers and x + 3 y = k . Furthermore, since k is even by assumption, we deduce that x ≡ y mod 2 , and x and y have the same parity. Thus, the coordinates of R (cid:18) xy (cid:19) = 12 (cid:18) x − yx + y (cid:19) (4)are integers, as required.Now, we study the residue modulo of the coordinates of a vector of U k underthe action R . First, we observe that if x = 2 x ′ is even, then x ′ ≡ x mod 3 , andif x = 2 x ′ + 1 is an odd number, then x ′ ≡ x −
1) mod 3 .Let X T = ( x, y ) ∈ U k . Write ( a, b ) for the coordinates of RX . As remarkedabove, x and y have the same parity. Assume first that x = 2 x ′ and y = 2 y ′ areboth even. Then (4) gives a = x ′ − y ′ ≡ x mod 3 and b = x ′ + y ′ ≡ x + y ) mod 3 . Assume now that x = 2 x ′ + 1 and y = 2 y ′ + 1 are odd. Then a = x ′ − y ′ − ≡ x − − ≡ x mod 3 and b = x ′ + y ′ + 1 ≡ x −
1) + 2( y −
1) + 1 ≡ x + y ) mod 3 . In any cases, we obtain a ≡ x mod 3 and b ≡ x + y ) mod 3 , (5)which does not depend on the parity of x and y . Note that c ( x, y ) = ( a, b ) . Since ( x, y ) ∈ U k , we have x ≡ because k ≡ by assumption.Hence, x ≡ ± and there is no condition on the residue modulo of y . Thepossible residue classes modulo for the coordinates of ( x, y ) are elements of { (1 , , (2 , , (1 , , (2 , , (1 ,
2) (2 , } . (6)Now, using (5), we compute the residue modulo of the coordinates of RX fora vector X whose coordinates has residue in (6). We summarize the result in thefollowing graph. The vertices are labeled by the residues modulo of the coordinatesof the vector and an arrow between two vertices represents a multiplication by R . (1 ,
1) (2 , , , , , We already remark that the Z / Z -residue vector of ( x, y ) lies in (6). Then by thegraph, the elements c ( R i ( x, y )) for ≤ i ≤ are exactly the ones of the set (6). Inparticular, the image of c is the set (6) and multiplication by R permutes cyclicallythe fibers of c . (cid:3) Remark 2.2.
Let k = 12 n + 4 be a positive integer.(i) In the proof the theorem, we only use the fact that k is even and has residue modulo . By Chinese Remainder Theorem, this condition is equivalent to k ≡ . Suppose that k = 6 q + 4 with q an odd integer. Then thereis m ∈ Z such that q = 2 m + 1 , and k = 12 m + 10 . Assume ( x, y ) ∈ U k .Then ( x, y ) satisfies equation (1), and by reducing the equality modulo , wefind that the equation x + 3 y = 10 has a solution over Z / Z . However, anexhaustive computation in Z / Z shows that this equation has no solution.Hence, U k = ∅ , and we only have to consider integers satisfying the assumptionof the theorem.(ii) Note that the group G constructed in Theorem 2.1 does not lie in the orthog-onal group over Z of D . This is a subgroup of matrices over R that stabilizesthe integral vectors of B k .(iii) Since G permutes cyclically the fibers of c , each fiber has the same cardinality.Hence, c is a crank for U k for any k a positive even integer. In particular, thisproves that | U k | ≡ . (iv) Each elements of (6) appear only once in all G -orbits. Hence, each G -orbitscontain only one vector whose coordinates have Z / Z -residue (1 , .3. Connection with -core partitions First, we roughly recall how to connect the set C of characteristic vectors givenin (2) and the Frobenius symbol of -core partitions. For more details, we referto [4, §3.2]. To any partition λ , we attach bijectively two sets of nonnegativeintegers A λ and L λ of the same size, respectively called the set of arms and of legsof λ . Geometrically, these sets can be interpreted on the Young diagram of λ : anarm of λ attached to a diagonal box d of the diagram is the number of boxes inthe same row and to the right of d . Similarly, the leg corresponding to d is the onein the same column and below d . This information is encoded into the Frobeniussymbol of λ , denoted by F ( λ ) = ( L λ | A λ ) .There is another geometric interpretation of the Frobenius symbol of λ in termsof a pointed t -abacus, where t is a positive integer. We recall it only for t = 3 . OLIVIER BRUNAT AND RISHI NATH
A pointed -abacus is an abacus with three runners labeled , and equippedwith a fence. On each runner, slots both over and under the fence are labeled bythe set of nonnegative integers. In each slot, a white or black bead is drawn sothat the number of black beads over the fence is finite and is equal to the one ofwhite beads under the fence. The pointed -abacus of λ is then obtained as follows.Let ≤ i ≤ and q be a positive integer. Then q + r ∈ A λ if and only if thepointed -abacus of λ has a black bead over the fence at position q on the runner i . Similarly, it has a white bead under the fence at position q on the runner − r if and only if q + r is a leg of λ .Let λ be a -core partition, and T its pointed -abacus. For ≤ i ≤ , if the i th-runner of T has no white bead under the fence, we write c i for the number ofblack beads over the fence. Otherwise, we write − c i for the number of white beadsunder the fence of the i th-runner of T . Then we set c λ = ( c , c , c ) . This tuple iscalled the characteristic vector of λ . In [4, §3.3], it is proved that the map ϕ : E −→ C , λ c λ is a well-defined bijection. In the following, we identify Z with C using the map Z → C , ( x, y ) ( − x − y, x, y ) . By this abuse of notation, we also will write c λ = ( x, y ) for the characteristic vector of λ .For positive integer n , we denote by C n for the elements of C corresponding to -core partitions of size n . Lemma 3.1.
Let n be a positive integer. Then the map ϑ : C n −→ U n +4 , ( x, y ) (6 x + 3 y + 1 , y + 1) is injective.Proof. For any -core partition λ with characteristic vector c λ = ( x, y ) ∈ Z , recallthat gives | λ | = 3 x + 3 xy + 3 y + x + 2 y . See for example [4, Corollary 3.20].However, x + 3 xy + 3 y + x + 2 y = 112 (6 x + 3 y + 1) + 14 (3 y + 1) + 13 . Furthermore, λ ∈ E n if and only if | λ | = n if and only if (6 x + 3 y + 1) + 3(3 y + 1) = 12 n + 4 . In particular, (6 x + 3 y + 1 , y + 1) ∈ U n +4 , and the map ϑ is well-defined. It isclearly injective. (cid:3) The next result connects E n and U n +4 . This gives a way to parameterize thesolutions of Equation (1). Theorem 3.2.
Let n be a nonnegative integer. Then U n +4 = { (6 x + 3 y + 1 , y + 1) , (3 x − y − , x + 3 y + 1) , ( − x − y − , x ) , ( − x − y − , − y − , ( − x + 3 y + 1 , − x − y − , (3 x + 6 y + 2 , − x ) | ( x, y ) ∈ C n } . Proof.
Write R for the matrix of O ( D ) described in the proof of Theorem 2.1.Since n + 4 is even, the group G = h R i acts on U n +4 , and each G -orbits hassize . Let ϑ : C n → U n +4 be the injective map given in Lemma 3.1. First, wewill prove that Im( ϑ ) is the set F n = { ( u, v ) ∈ U n +4 | u ≡ and v ≡ } . It is clear that
Im( ϑ ) ⊆ F n by definition of ϑ . Conversely, let ( u, v ) ∈ F n . Write z ∈ Z such that u = z + v . We have n + 4 = u + 3 v = ( z + v ) + 3 v = z + 2 zv + 4 v . Considering this equality modulo , we obtain that z ≡ , hence z is even.On the other hand, z + v and v have the same residue modulo since ( z + v, v ) ∈ F n .This implies that z ≡ . Hence, z is divisible by , and there is x ∈ Z suchthat z = 6 x . Furthermore, v ≡ . Thus, there is y ∈ Z such that v = 3 y + 1 ,and ϑ ( x, y ) = (6 x + 3 y + 1 , y + 1) = ( z + v, v ) = ( u, v ) , proving that F n ⊆ Im( ϑ ) . Hence, Im( ϑ ) = F n . However, we proved in Theorem 2.1 that F n is a set of representatives of the G -orbits on U n +4 . Note that for any X T = ( u, v ) ∈ U n +4 , RX = 12 (cid:18) u − vu + v (cid:19) , R X = 12 (cid:18) − u − vu − v (cid:19) , R X = − X, (7) R X = 12 (cid:18) − u + 3 v − u − v (cid:19) , R X = 12 (cid:18) u + 3 v − u + v (cid:19) , Finally, we conclude by computing the orbit of ϑ ( x, y ) for ( x, y ) ∈ C n using theserelations. (cid:3) Remark 3.3.
For any ( u, v ) ∈ U n +4 , we write G · ( u, v ) for its G -orbit. ByTheorem 3.2, we have u (12 n + 4) = |U n +4 | = X ( x,y ) ∈C n | G · ϑ ( x, y ) | = 6 |C n | = 6 |E n | . In particular, this proves [2, Theorem 3.1].
Remark 3.4.
By Theorem 2.1 and Theorem 3.2 the fiber of (1 , with respect tothe crank c is the set C n . Then we obtain a way to recover E n from U n +4 . Example 3.5.
We consider the equation x + 3 y = 448 over Z . First, we remarkthat
448 = 12 ·
37 + 4 . Then we have to describe E . On the other hand,
37 =4 × . By [4, Corollary 5.3] , E and E are in bijection. Now, we remark bylooking at its character table that the symmetric group S has exactly two defectzero characters labeled by the partitions λ = (5 , , and µ = (3 , , ) . These are the -core partitions of . Furthermore, we have F ( λ ) = (2 , | , and F ( µ ) = F ( λ ∗ ) = (4 , | , , and the corresponding pointed -abacus are f f λ µ OLIVIER BRUNAT AND RISHI NATH
The characteristic vectors of λ and µ are then ( − , , − and (1 , − , . Moreover,using the bijection [4, Corollary 5.3] , we deduce that the -core partitions of havecharacteristic vectors (3 , − , and (1 , , − , and ϑ ( − ,
1) = ( − , and ϑ (4 , −
3) = (16 , − . We conclude with Theorem 3.2 that U = { ( − , , ( − , − , (4 , − , (20 , − , (16 , , ( − , , (16 , − , (20 , , (4 , , ( − , , ( − , − , ( − , − } . Remark 3.6.
Consider the map conj : U n +4 −→ U n +4 , ( u, v ) (cid:18)
12 ( − u + 3 v ) ,
12 ( u + v ) (cid:19) . This map is well-defined, because for ( u, v ) ∈ U n +4 , we have ( − u + 3 v ) ∈ Z and ( u + v ) ∈ Z since u ≡ v mod 2 , and (cid:18)
12 ( − u + 3 v ) (cid:19) + (cid:18)
12 ( u + v ) (cid:19) = u + 3 v = 12 n + 4 . Now, for a partition λ of n , we write λ ∗ for its conjugate partition. By [4, Propo-sition 3.12], if c λ = ( x, y ) , then c λ ∗ = ( − x, x + y ) . Furthermore, for any -corepartition λ , we have conj ◦ ϑ ( c λ ) = ϑ ( c λ ∗ ) . Hence, the image under ϑ of the set of self-conjugate -cores of n consisting of thevectors ( u, v ) ∈ U n +4 such that ( u, v ) lies in the -eigenspace of S = 12 (cid:18) − (cid:19) . This is the set of ( u, u ) ∈ Z such that u ≡ and u = 3 n + 1 . Inparticular, there is at most one self-conjugate -core partition of n . Let ( u, u ) besuch an element and k be any integer not divisible by . Let ε ∈ {− , } be suchthat εk ≡ . Then εku ≡ and ( εku ) = 3 (cid:18) k n + k − (cid:19) + 1 . It follows that the map ϑ − ( u, u ) ϑ − ( εku, εku ) from the set of self-conjugate -core partitions of n into the ones of -core partitions of k n + k − is well-definedand bijective. Hence, asc ( n ) = asc (cid:18) k n + k − (cid:19) , where asc ( n ) denotes the number of self-conjugate -core partitions of n . Thisgeneralizes [2, Theorem 3.6].Note that the group generated by R and S is a dihedral group of order . Connection with the ring of Eisenstein integers
Theorem 3.2 shows that we can recover E n from U n +4 . See also Remark 3.4. Inthis section, we give another way to determine U n +4 using the ring of Eisensteinintegers R = Z [ j ] , where j = e iπ/ . We write N : R → N , z zz for the normof R , where z denotes the complex conjugation. Many results here are standard,but we recall them for the convenience of the reader. Let u, v ∈ Z . We have u + v √ i = u + v + v j , thus u + v √ i ∈ R if and only if u and v have the sameparity. Note also that, in this case, we have N u v √ i ! = 14 ( u + 3 v ) . However, as remarked in the proof of Theorem 2.1, each element of U n +4 has thisproperty, hence the map ρ : U n +4 −→ R , ( x, y ) x y √ i (8)is well-defined. Furthermore, ( u, v ) ∈ U n +4 if and only if N ( ρ ( u, v )) = 3 n + 1 . Write Π and Π for the sets of prime numbers with residue and modulo ,respectively. For any integer k , we denote by ν p ( k ) the p -adic valuation of k .By [13, Proposition 9.1.4], each p ∈ Π is irreducible in R , and for p ∈ Π , there isan irreducible element x p ∈ R such that p = N ( x p ) . Furthermore, we define V n = Y p ∈ Π { ≤ j p ≤ ν p (3 n + 1) } , (9)and for any j = ( j p | p ∈ Π ) ∈ V n , we set x j = Y p ∈ Π x j p p x pν p (3 n +1) − j p . (10) Theorem 4.1.
The set U n +4 is nonempty if and only if ν p (3 n + 1) is even forall p ∈ Π . In this case, we set q n = Y p ∈ Π p ν p (3 n +1) and y j = x j q n for j ∈ V n . (11) Then, for any j ∈ V n , there are integers u j and v j with residue modulo suchthat y j = 12 (cid:16) u j + v j i √ (cid:17) , and C n = { ϑ − ( u j , v j ) | j ∈ V n } , where ϑ : C n → U n +4 is the map defined in Lemma 3.1.Proof. First, we have n + 1 = Y p ∈ Π p ν p (3 n +1) · Y p ∈ Π p ν p (3 n +1) = Y p ∈ Π x ν p (3 n +1) p x ν p (3 n +1) p · Y p ∈ Π p ν p (3 n +1) , that gives a factorization into irreducible elements in R . Now, we note that thereis z = a + b i √ ∈ R such that n + 1 = N ( z ) = zz if and only if a + 3 b = 12 n + 4 if and only if ( a, b ) ∈ U n +4 . Assume there is z ∈ R such that n + 1 = N ( z ) .By [13, Proposition 1.4.2], R is a Euclidean domain. Hence, for any p ∈ Π suchthat ν p (3 n + 1) = 0 , p divides z or z in R . We can assume without loss of generalitythat p divides z . Then p = p divides z , proving that p divides n +1 in Z . It followsthat ν p (3 n + 1) is even. Furthermore, by the uniqueness (up to a unit element)of the decomposition into irreducible elements in a Euclidean ring, we deduce firstthat q n divides z , and then that there are ξ ∈ R × and integers ≤ j p ≤ ν p (3 n + 1) and ≤ j ′ p ≤ ν p (3 n + 1) for any p ∈ Π , such that z = ξq n Y p ∈ Π x j p p x pj ′ p . Hence, n + 1 = zz = q n Y p ∈ Π x j p + j ′ p p x pj p + j ′ p . By the uniqueness of the decomposition, we deduce that j p + j ′ p = ν p (3 n + 1) . Itfollows that z = ξq n Y p ∈ Π x j p p x pν p (3 n +1) − j p = ξq n x j = ξy j , (12)with j = ( j p | p ∈ Π ) . Conversely, any factorization in R of n + 1 of the form zz is as in (12) for some ξ ∈ R × and j ∈ V n .Now, by [13, Proposition 9.1.1], R × has order and R × = h j + 1 i . Note that R × acts on ρ ( U n +4 ) by multiplication since the elements of R × have norm . Onthe other hand, if z = ( a + bi √ for a, b ∈ Z , then ( j + 1) z = 12 (cid:18)
12 ( a − b ) + 12 ( a + b ) i (cid:19) . Comparing with (4), we see that h R i acts on C n like R × on ρ ( U n +4 ) . In particular,by Theorem 3.2, for any j ∈ V n , there is ξ ∈ R × such that ξq n x j = ( u j + v j i ) satisfies u j ≡ v j ≡ . Then ϑ ( C n ) = { ( u j , v j ) | j ∈ V n } by Theorem 3.2, asrequired. (cid:3) Remark 4.2.
By Theorem 3.2 and Theorem 4.1, the description of the set of -corepartitions of n is reduced to(i) The determination of the prime decomposition in Z of n + 1 .(ii) The decomposition of the prime p ∈ Π dividing n + 1 into irreducible ele-ments in Z [ j ] . This factorization is of the form zz for irreducible elements z and z of Z [ j ] . Remark 4.3.
By Theorem 4.1, we have |E n | = |C n | = |{ ( u j , v j ) | j ∈ V n }| = |V n | = Y p ∈ Π ( ν p (3 n + 1) + 1) , and we recover a well-known result first proven by Granville and Ono [9], and alsofund in [12, Theorem 6] and [2, Corollary 3.3].Theorem 3.2 can also be useful to find the set of -core partitions of an integer n ; a hard problem in general. Example 4.4.
We determine the -core partitions of n = 100 . First, considerthe equation x + 3 y = 4 · . Then describe U using Theorem 4.1. Weremark that
301 = 7 × , and and lie in Π . However, N (2 + i √ and
43 = N (4 + 3 i √ . Set x = 2 + i √ and x = 4 + 3 i √ , and consider α = x x = 12 ( − − i √ , α = x x = 12 (34 − i √ ,α = x x = 12 (34 + 4 i √ and α = x x = 12 ( − i √ , Thus, y , = α and y , = α , and y , = (1 + j ) α = 12 ( −
11 + 19 i √ and y , = (1 + j ) α = 12 ( − − i √ . Thus, F = { ( − , − , ( − , , ( − , − , (34 , } . Now, as in the proof of Theorem 3.2, we obtain ϑ (3 , −
7) = ( − , − , ϑ ( − ,
6) = ( − , , ϑ ( − , −
4) = ( − , − and ϑ (5 ,
1) = (34 , . Hence, C = { (4 , , − , ( − , − , , (7 , − , − , ( − , , } . Let λ be the -core partition with characteristic vector (4 , , − . Using [4, Lemma3.19] , we obtain F ( λ ) = (18 , , , , , , | , , , , , , , and we deduce that λ = (10 , , , , , , , , , ) . Similarly, we find that the -core partitions of with characteristic vectors ( − , − , , (7 , − , − and ( − , , are respectively (18 , , , , , , , , , , ) , (19 , , , , , , , , , and (14 , , , , , , , , , , ) . Applications
Generalization of Hirschhorn-Sellers formula.
Let n be a positive inte-ger. Let k be a positive integer not divisible by . Write m be for the product ofprime factors of k with residue modulo . Assume m is a square. Then a (cid:18) kn + k − (cid:19) = a ( n ) Y p ∈ Π ν p ( k ) + ν p (3 n + 1) + 1 ν p (3 n + 1) + 1 , (13)where a ( n ) = |E n | .Indeed, we first observe that m ≡ since m is a square, and it followsthat k ≡ . Set N = kn + ( k − . We have N + 1 = k (3 n + 1) = Y p ∈ Π p ν p ( k )+ ν p (3 n +1) m ′ m, where m ′ is the product (with multiplicity) of the prime factors with residue modulo of n + 1 . By assumption, mm ′ is a square if and only if m ′ is. Hence, Theorem 4.1 implies that a ( n ) = 0 if and only if a ( N ) = 0 . Assume a ( n ) = 0 .Now, again by Theorem 4.1, we obtain a ( N ) = Y p ∈ Π ( ν p ( k ) + ν p (3 n + 1) + 1) = Y p ∈ Π ν p ( k ) + ν p (3 n + 1) + 1 ν p (3 n + 1) + 1 a ( n ) , as required. Remark 5.1. (i) When k is a square not divisible by and with no prime factors with residue modulo , then a (cid:18) kn + k − (cid:19) = a ( n ) . This generalizes [12, Corollary 8] and [2, Theorem 5].(ii) Assume k is not divisible by and has no prime factors congruent to modulo . In [4, Theorem 5.1], we prove the injectivity of the map C n −→ C k n + ( k − , ( x, y ) ( εkx, ε ( ky + q )) , where ε ∈ {− , } and q ∈ N are such that k = 3 q + ε . By (i), this map isbijective.5.2. An amazing equality.
In this part, we derive (3) from Theorem 4.1. Thenusing the crank c of §2 and the map ϑ of Lemma 3.1, we construct an explicitbijection that explains this equality.Let n and m be two positive integers such that n + 1 and m + 1 are coprime.Then the prime decomposition of (3 n + 1)(3 m + 1) is the “concatenation” of the oneof n + 1 with the one of m + 1 . Since (3 n + 1)(3 m + 1) = 3(3 mn + m + n ) + 1 ,Theorem 4.1 gives a (3 mn + m + n ) = a ( n ) a ( m ) . Assume m = n + 3 k for some k ∈ N . If d divides n + 1 and m + 1 , then n + 1 ≡ ≡ m + 1 mod d . Hence, n ≡ m mod d since d and are coprime, and k ≡ d . Using again that d and are coprime, we deduce that d = 1 . Theintegers n + 1 and m + 1 are then coprime, and Equality 3 holds.Now, we assume a ( n ) and a ( m ) are non-zero. For any positive integer t , andany j ∈ V t , we write x ( t ) j and y ( t ) j for the elements defined in (10) and in (11). Set N = 3 n + (3 k +1 + 2) n + 3 k . Since n + 1 and m + 1 are coprime, we obtain V N = V n × V m , x ( N ) j = x ( n ) j y ( m ) j and q N = q n q m , where j = ( j , j ) . Hence y ( N ) j = y ( n ) j y ( m ) j , and any factorizations of N + 1 = zz satisfies z = z n z m z n and z m are equal up to a unit of Z [ j ] to y ( n ) j and y ( m ) j ,respectively. However, by Theorem 4.1, these elements are equal (up to a unitelement) to ρ ◦ ϑ ( x, y ) and ρ ◦ ϑ ( x ′ , y ′ ) for some ( x, y ) ∈ C n and ( x ′ , y ′ ) ∈ C m , where ρ is the map defined in (8). Furthermore, ρ ◦ ϑ ( x, y ) ρ ◦ ϑ ( x ′ , y ′ ) = ( X, Y ) , where X = ( − x − y + 3 x ′ − y ′ + 18 xx ′ + 9 xy ′ + 9 yx ′ − yy ′ ) and Y = (9 yy ′ + 1 + 9 yx ′ + 3 y ′ + 3 x + 9 xy ′ + 3 x ′ + 3 y ) . We remark that c (2 X, Y ) = (2 , . By the graph page 4, we deduce that the element of U N +4 which lies in the imageof ϑ and on the h R i -orbit of (2 X, Y ) is R − (2 X, Y ) = (6 a + 3 b + 1 , b + 1) , where a = x + x ′ + 3 xx ′ + 3 xy ′ + 3 yx ′ and b = y + y ′ − xx ′ + 3 yy ′ . This proves that the map α : C n × C n +3 k −→ C n +(3 k +1 +2) n +3 k ) is defined for any ( x, y ) ∈ C n and any ( x ′ , y ′ ) ∈ C n +3 k by α (( x, y ) , ( x ′ , y ′ )) = ( x + x ′ + 3 xx ′ + 3 xy ′ + 3 yx ′ , y + y ′ − xx ′ + 3 yy ′ ) is a bijection.5.3. Remarks on a question of Han.
In [10, 5.2], Han conjectured a criteriacharacterizing integers with no -cores. We now discuss this question.(i) First, we remark that if N = 4 m n + (10 · m − − for some integers n ≥ and m ≥ , then N + 1 = 4 m − (12 n + 10) = 2 m − (6 n + 5) . However, n + 5 is odd, hence ν (3 N + 1) = 2 m − is odd. Since ∈ Π , thefirst part of Theorem 4.1 gives a ( N ) = 0 . This proves the point (i) of [10,5.2].(ii) Now, we focus on the point (ii) of [10, 5.2], asserting that if there are integers n ≥ , m ≥ , k ≥ with m k − k − such that N = (6 k − n + (6 k − m + 4 k − , (14)then a ( N ) = 0 . Consider the case N = 58 . We have a (58) = 2 . We can seethat by noting that N + 1 = 5 · and by applying Theorem 4.1. However,
58 = (6 k − m + 4 k − for m = 1 and k = 6 . Since m k − k − , satisfies the previousassumptions, but a (58) = 0 , giving a counterexample of the statement.Note that N + 1 = 3(6 k − n + 3(6 k − m + 12 k −
2= (6 k − n (6 k −
1) + 3 m + 2) . If we additionally assume that ν p (6 k − is odd whenever the prime p has residue modulo , then, for such a p , we have k − ≡ p ,and k − ≡ − k mod p . Suppose p divides (3 n (6 k −
1) + 3 m + 2) . Then m + 2 ≡ p , and m ≡ − p . Multiplying by k , we obtain m ≡ − k ≡ k − p, which is a contradiction. Hence, ν p (3 N + 1) is odd and p ∈ Π . Theorem 4.1again gives a ( N ) = 0 .(iii) Let N ∈ N be such that a ( N ) = 0 . By Theorem 4.1, there is p ∈ Π suchthat ν p (3 N + 1) is odd. In fact, we can assume that p = 2 , because thereare at least two distinct prime numbers with this property since N + 1 ≡ . Thus, p ν p (3 N +1) ≡ − , and p ν p (3 N +1) ≥ . It follows thatthere exists k ≥ such that p ν p (3 N +1) = 6 k − . Let now q ∈ N be such that N + 1 = p ν p (3 N +1) q . Then q is not divisible by p . Note that we must have q ≡ since N + 1 ≡ . Assume q > . Then q ≥ , and there is m ≥ such that q = 3 m + 2 . Furthermore, q p . The samecomputation as above gives that m k − p , and N = (6 k − m + 2) −
13 = (6 k − m + 4 k − . Thus N is as in (ii) with n = 0 . Finally, assume that q = 2 . We have N + 1 = 2(6 k −
1) = 2(6( k −
1) + 5) = 4 · k −
1) + 10 . Hence, N = 4( k −
1) + 13 (10 · − , and N is as in (i) for n = k − and m = 1 .Hence, the point (iii) of [10, 5.2] holds if we use the new assumptionsintroduced in (ii). Acknowledgements.
The authors wish to thank the referee for their construc-tive remarks and suggestions that improved the organization and emphasis of thepaper.
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Université Paris-Diderot Paris 7, Institut de mathématiques de Jussieu – ParisRive Gauche, UFR de mathématiques, Case 7012, 75205 Paris Cedex 13, France.
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