Extension of Stanleys Theorem for Partitions

EXTENSION OF ST ANLEY’S THEOR EM F OR P AR TITIONS Manosij Ghosh Dastidar South Point High Sch o ol, Kolkata, India gdmanosij@g mail.com Soura v Sen Gupta Indian Statistic al Institute, Kolka ta, India sg.sourav@g mail.com Abstract In this pap er w e presen t an extension of Stanley’s theorem relat ed to partit io ns of p ositive in tegers. Stanley’s theorem states a r elat io n b et we en “ the sum of the num b ers of distinct mem b ers in the partitions of a p ositiv e inte ger n ” and “the total n umber of 1’s that o ccur in the part itions of n ”. Our generalization states a similar relation b et w een “the sum of the n umbers of distinct mem b ers in the partitio ns of n ” and the tota l n um b er o f 2’s or 3’s or any general k that o ccur in t he partitions of n and the subsequen t in tegers. W e a lso apply this result to obta in an array of in teresting corollaries, including alternate pro ofs and analogues of some of the ve ry w ell-know n results in the theory of partitions. W e extend Ramanujan’s results on congruence b ehavior of the ‘num b er of pa rtition’ function p ( n ) to get a nalogous results for the ‘n um b er of o ccurrences of an elemen t k in partitions of n ’. Moreo ve r, w e presen t a n alternate pro of of Ramanujan’s results in this pap er. 1. In t ro duction P artitio ning a p ositiv e inte ger n as sum of certain p ositiv e in tegers is a w ell know n problem in the domain of n um b er theory [4]. One of the very w ell referred results in this area is the one presen ted b y Sta nley [6] whic h states the following. “The total numb er of 1’s that o c cur among al l unor der e d p artitions of a p ositive inte ger is e qual to the sum o f the n umb ers of d istinct memb ers of those p artitions.” One direction of generalizing Stanley’s t heorem is the Elder’s theorem [5] whic h states that “total numb er of o c curr enc es of an in te ger k among al l unor der e d p artitions of n i s e qual to the numb er of o c c asions that a p art o c curs k or mor e times in a p artition.” In case of Elder’s statemen t, a part it ion whic h con ta ins r pa r ts that eac h o ccur k or more times con t r ibutes r to the sum in question. 1.1. Our Con t ribution In this pap er w e generalize Stanley’s theorem in a differen t direction. Le t S ( n ) be the num b er of distinct mem b ers presen t in the uno r dered partitions o f n and Q k ( n ) b e the num b er of o ccurrences of k in all the unordered par t it ions of n . Then Stanley’s theorem sa ys t hat giv en an y p ositiv e in teger n , t he set of unordered partitions of n satisfy S ( n ) = Q 1 ( n ). Let us presen t a n example for n = 4 corresp o nding to Stanley’s theorem in T able 1. Stanley’s theorem for n = 4 P artitio ns Num b er of Distinct Mem b ers Num b er of 1’s 4 1 0 3+1 2 1 2+2 1 0 2+1+1 2 2 1+1+1+1 1 4 T otal 7 7 T able 1: Example of Stanley’s theorem W e exten d this to pro v e tha t giv en a n y t wo p ositiv e in tegers n and k , S ( n ) = P k − 1 i =0 Q k ( n + i ). 1.2. Idea for Extension The main idea b ehind the extension is the observ ation that Stanley’s theorem can b e ex- tended to a relation b et w een the distinct mem b ers of the partit ions of n and the num b er of o ccurrences of an y po sitiv e in teger k in the partition table of n and subseq uen t in t egers. Let us first presen t a few examples to illustrate our motiv ation. k = 1 : The sum of the num b ers of distinct members of all the unordered partitio ns of a p ositiv e integer n is equal to the tota l num b er of 1’s that o ccur among t he pa rtitions of n . [This is Stanley’s theorem] k = 2 : The sum of the num b ers of distinct members of all the unordered partitio ns of a p ositiv e integer n is equal to the tota l num b er of 2’s that o ccur among t he pa rtitions of n and n + 1. k = 3 : The sum of the num b ers of distinct members of all the unordered partitio ns of a p ositiv e integer n is equal to the tota l num b er of 3’s that o ccur among t he pa rtitions of n , n + 1 a nd n + 2. One ma y note that Sta nley’s theorem is the sp ecial case with k = 1, as men tioned ab ov e. The claim of t his pa p er is that the statements made ab ov e a re correct and can b e extended to a n y v alue of k . The g eneral result is f o rmally presen ted in the next section. 2 2. Extension of Stanley’s Theorem Before we discuss the results, let us set some notational con v en tions. W e men tio n S ( n ) and Q k ( n ) once again for the sak e of completion. • P ( n ): Num b er of unordered partitio ns o f a p ositiv e in teger n . • S ( n ): Num b er o f di stinct memb ers presen t in the unordered partitions of n . • R k ( n ): Num b er of partitio ns o f n con taining the p ositiv e in teger k . • Q k ( n ): Num b er of o ccurrences of k in all the unordered pa r titions of n . Based o n the notation stated ab o ve , Stanley’s theorem can b e expressed as follows. Theorem 1 (Stanley’s Theorem) Given any p ositive inte ger n , the set of unor der e d p ar- titions of n satisfy S ( n ) = Q 1 ( n ) . The extension to Stanley’s theorem is as stated in Theorem 2. But b efore that, w e will pro ve the f o llo wing t wo tec hnical results whic h are used in the pro of of Theorem 2. 2.1. Preliminary T ec hnical R esults Lemma 1 Given any p ositive inte ger n and any p ositive inte ger k , Q k ( n + k ) = Q k ( n ) + R k ( n + k ) . Pr o of. Let us denote the set of partitions of n containing k as X = { A 1 , A 2 , . . . , A r } , where r = R k ( n ) is the nu m b er of suc h partitio ns, a nd the partitions are written in a dditiv e form ( e.g., 5 = 2 + 2 + 1). The set defined ab ov e give s us n = A 1 = A 2 = · · · = A r , and th us n + k = A 1 + k = A 2 + k = · · · = A r + k . Now all the partitions in this collection Y = { A 1 + k , A 2 + k , . . . , A r + k } are partitions of n + k con taining k , but this is not a n exhaustiv e list. There are additional partitio ns of n + k containing k . One ma y no te that the list Y con tains r = R k ( n ) partitions, whereas the num b er o f partitions of n + k containing k is R k ( n + k ). Th us, there a re R k ( n + k ) − R k ( n ) additional partitions of n + k containing k . In this direction, we claim the follow ing. 3 Claim: In eac h of the additio na l R k ( n + k ) − R k ( n ) partitions o f n + k con taining k , the mem b er k o ccurs only once. Pro of: Supp ose the claim is false, that is, there exists a partition B 1 6∈ Y of n + k that con tains more than one cop y o f k . Then B 1 − k m ust con tain at least one cop y o f k . Note that n + k = B 1 , and th us n = B 1 − k , whic h implies that B 1 − k is a partit io n of n con taining k . If so, then one must hav e B 1 − k ∈ X and therefore B 1 ∈ Y . But this p oses a con tradiction, and hence the claim is true. In view of our previous discussion and the claim ab ov e, w e o btain the following • Each partition of n + k in Y con tains k , and there are r = R k ( n ) partitions in Y . • There exist R k ( n + k ) − R k ( n ) additional part it ions of n + k whic h contain k . • Each of these additiona l partitions con tain exactly one copy of k . Moreo ve r, one ma y note that the total nu m b er of k ’s in the partitions b elonging to X is Q k ( n ), as w e selected all partitions of n con taining k . So, total num b er of k ’s in the partitions b elonging to Y is Q k ( n ) + R k ( n ), as we hav e added a k to eac h partitio n from X . Th us, t he t o tal n umber of k ’s o ccurring in the partitions of n + k is g iv en b y Q k ( n + k ) = [ Q k ( n ) + R k ( n )] + [ R k ( n + k ) − R k ( n )] = Q k ( n ) + R k ( n + k ) . The analysis holds for all n, k > 0, and hence the result. ✷ Lemma 2 Given any p ositive inte ger n and any p ositive inte ger k , P ( n ) = R k ( n + k ) . Pr o of. Let us denote the set of all partitions of n a s X = { A 1 , A 2 , . . . , A p } , where p = P ( n ) is the num b er of all partitions, and the pa r t itions are written in additiv e form ( e.g., 5 = 2 + 2 + 1). The set defined ab ov e give s us n = A 1 = A 2 = · · · = A p , and th us n + k = A 1 + k = A 2 + k = · · · = A p + k . Now all t he pa rtitions in this collection Y = { A 1 + k , A 2 + k , . . . , A p + k } are partitions of n + k con taining k . In this direction, w e claim the following. 4 Claim: There are no other part itions of n + k , apart fro m those in Y , that contain k . Pro of: Supp ose the claim is false, that is, there exists a partition B 1 6∈ Y of n + k that con tains k . Note that n + k = B 1 , and th us n = B 1 − k , whic h implies that B 1 − k is a partition of n . If so, then one m ust ha ve B 1 − k ∈ X and therefore B 1 ∈ Y . But this p oses a contradiction, and hence the claim is true. In view of our previous discussion and the claim ab ov e, w e o btain the following • Each partition of n + k in Y con tains k , and there are p = P ( n ) part it ions in Y . • There exist no additional partitio ns of n + k whic h con tain k . Th us, w e hav e the n um b er of partitions of n + k con taining k as R k ( n + k ) = | Y | = P ( n ). The analysis holds for all n, k > 0, and hence the result. ✷ 2.2. The Main Result No w that w e hav e prov ed the t wo lemmas required for our result, w e can state and pr ov e the main theorem of this pap er, as fo llo ws. Theorem 2 (Extension of Stanley’s Theorem) Given any p ositive inte ger n and any p ositive inte ger k , S ( n ) = Q k ( n ) + Q k ( n + 1) + Q k ( n + 2) + · · · + Q k ( n + k − 1 ) = k − 1 X i =0 Q k ( n + i ) . Pr o of. W e shall prov e this result b y induction ov er k , as follow s. Base Case ( k = 1 ): S ( n ) = Q 1 ( n ) is true for all n b y Theorem 1 (Stanley’s theorem). Assumption fo r k : L et us assume that for some k ≥ 1 , and for all n > 0, S ( n ) = k − 1 X i =0 Q k ( n + i ) . Pro of f o r k + 1 : Here, we hav e to pro v e that S ( n ) = P k i =0 Q k +1 ( n + i ) for all n > 0. Let us pro ve this result b y induction on n , as follows. 5 n = 1 : F or this base case of induction, we hav e k X i =0 Q k +1 (1 + i ) = k − 1 X i =0 Q k +1 (1 + i ) + Q k +1 ( k + 1) = 0 + 1 = 1 = Q 1 (1) = S (1) . n : Let us assume that f or some n ≥ 1, S ( n ) = Q 1 ( n ) = P k i =0 Q k +1 ( n + i ) . n + 1 : F or this case with n + 1, we ha ve S ( n + 1) = Q 1 ( n + 1) = Q 1 ( n ) + R 1 ( n + 1) , from L emma 1 = Q 1 ( n ) + P ( n ) , from Lemma 2 = k X i =0 Q k +1 ( n + i ) + R k +1 ( n + k + 1) , from assumption and Lemma 2 = k X i =0 Q k +1 ( n + i ) + [ Q k +1 ( n + k + 1) − Q k +1 ( n )] , from Lemma 1 = k +1 X i =1 Q k +1 ( n + i ) = k X i =0 Q k +1 ( n + i + 1) = k X i =0 Q k +1 (( n + 1) + i ) Th us, b y induction on n , the result is pro v ed in case of k + 1 f o r all n > 0, and hence b y induction on k , the result is pro v ed f o r a ll k > 0 as w ell. ✷ 2.3. Illustrativ e Example In T able 2, w e presen t a mo dified version of the initial example ( n = 4) as an evidence t o this observ ation and the v alidity of Theorem 2. Recall from T able 1 that S ( 4) = 7. Eac h cell in the main segmen t of T able 2 r epresen ts the v alue of Q k ( n ) corresp onding to the ro w n and column k . The cells marked (-) are the v alues of Q k ( n ) that w e do not r equire in Theorem 2, and hence hav e b een omitted from the table. Theorem 2 for n = 4 n ↓ k → 1 2 3 4 4 Q 1 (4) = 7 Q 2 (4) = 3 Q 3 (4) = 1 Q 4 (4) = 1 5 - Q 2 (5) = 4 Q 3 (5) = 2 Q 4 (5) = 1 6 - - Q 3 (6) = 4 Q 4 (6) = 2 7 - - - Q 4 (7) = 3 T otal S (4) = 7 S (4) = 7 S (4) = 7 S (4) = 7 T able 2 : Example of our Extension of Stanley’s Theorem 6 3. Corollaries of the Extension In this section w e pro p ose a few corolla ries which follow directly from t he Lemmas 1 and 2, or from Theorem 2 itself. Our r esults can b e exploited to provide alternate pro ofs of existing facts and analogues to established results in t he field of in teger partitions. 3.1. Alternate Pro ofs of Exist ing Results One may find the following tw o results existing in the current literature [2, 3] on partitions of integers. W e shall provide a lternate pro of s to these results using Lemmas 1 and 2. Result 1 Given a ny p ositive in te ger n , abid ing b y the pr ev ious notation, one has Q 1 ( n ) = n − 1 X i =0 P ( i ) . Pr o of. F ro m Lemma 1, w e know that Q k ( n + k ) = Q k ( n ) + R k ( n + k ) for any pair of p ositiv e in tegers ( n, k ). Also, fr o m Lemma 2, w e get that R k ( n + k ) = P ( n ) for a ll pairs ( n, k ) . Cho osing k = 1 and using these tw o lemmas appropriately , w e obtain the following. Q 1 ( n ) = Q 1 (( n − 1) + 1) = Q 1 ( n − 1) + R 1 ( n ) = Q 1 ( n − 1) + P ( n − 1 ) F ollowing an iterativ e mo del based on this relation, one may easily obtain the result as Q 1 ( n ) = n X i =1 P ( i − 1) = n − 1 X i =0 P ( i ) . Hence the result. ✷ Result 2 Given a ny p ositive in te ger n , abid ing b y the pr ev ious notation, one has Q k ( n ) = n − 1 X i =0 i ≡ n mo d k P ( i ) . Pr o of. Once a gain, w e use the Lemmas 1 and 2 to get Q k ( n ) = Q k (( n − k ) + k ) = Q k ( n − k ) + R k ( n ) = Q k ( n − k ) + P ( n − k ) . An iterative pro cess applied on this relatio n pro duces Q k ( n ) = P j > 0 P ( n − j k ). Recalling the fact t ha t P ( n ) = 0 for n < 0, w e obtain Q k ( n ) = X j > 0 P ( n − j k ) = n − 1 X i =0 i = n − j k , j ∈ Z P ( i ) = n − 1 X i =0 i ≡ n mo d k P ( i ) . Hence the result. ✷ 7 3.2. Alternate Pro of of T heorem 2 Here w e pro vide an alternate pro of of Theorem 2, using the t wo results prov ed in the previous subsection, instead of the lemmas that we used in t he original pro of . Theorem 3 (Theorem 2 Restated) Given any p air of p ositive inte gers ( n, k ) , S ( n ) = Q k ( n ) + Q k ( n + 1) + Q k ( n + 2) + · · · + Q k ( n + k − 1 ) = k − 1 X i =0 Q k ( n + i ) . Pr o of. W e hav e S ( n ) = Q 1 ( n ) = P n − 1 i =0 P ( i ) by com bining Stanley’s t heorem and Result 1 as stated ab ov e. Let us define the f ollo wing set o f par t itions. P n = { P (0) , P (1) , P (2) , . . . , P ( n − 1) } . Then, S ( n ) is equal to the sum ov er all these partitions in P n . Giv en the p o sitive in t eger k , w e can distribute P n o ver some disjoint copies of congruence classes as f ollo ws. P n = { P ( i ) , P ( i + 1 ) , . . . , P ( i + k − 1) | i ≡ n mo d k } = k − 1 [ j =0 { P ( i + j ) | 0 ≤ i + j < n, i ≡ n mo d k } F rom the ab o ve -men tioned distribution o f P n , one ma y easily deduce that S ( n ) = Q 1 ( n ) = n − 1 X i =0 P ( i ) = k − 1 X j =0 n − 1 X i + j =0 i ≡ n mod k P ( i + j ) = k − 1 X j =0 n − 1 X i + j =0 i + j ≡ n + j mod k P ( i + j ) = k − 1 X j =0 Q k ( n + j ) , by Result 2 . Hence the result, whic h holds true for any p ositive integral v alues of n and k . ✷ Though the results pro v ed in the prev ious s ubsection ex isted in the literature, there ex ists no attempt to deriv e this generalization of Stanley’s theorem using these. This is t o the b est of o ur kno wledge ab out the literature o n in teger partitions. 8 3.3. Analogue of Raman ujan’s Results In the theory of integer pa rtitions, an arra y of elegant congr uence relations fo r partition function P ( n ) w ere prop osed by Ra man ujan. He pro v ed tha t for ev ery non-negativ e n ∈ Z , p (5 n + 4) ≡ 0 (mo d 5) , p (7 n + 5) ≡ 0 (mo d 7) , p (11 n + 6) ≡ 0 (mo d 11) , and he conjectured that there exist suc h congruence mo dulo arbitrary p ow ers of 5, 7, 1 1 . A lot of eminen t mathematicians hav e work ed on similar results for a long time, and the b est result till date is: “there exist suc h congruence relations for all non-negativ e in tegers whic h are co-prime t o 6”. This result w as prov ed b y Ahlg r en and Ono [1]. In this light, w e prop ose a simple analogue to the Ramanujan results tha t fo llo w directly from Lemmas 1 and 2. In simple w ords, our first result can b e stated as “for any n on-ne gative inte ger n , the numb er of times 5 o c curs in the unor der e d p artitions of 5 n + 4 is divisible by 5” . The ot her t wo r esults fall in the same line with divisibilit y b y 7 and 11 resp ectiv ely . Sta ted formally , the analogues of Ramanujan results that we prop ose a r e for the function Q k ( n ), as follows . Theorem 4 Given any non -ne gative in te ge r n , fol lowing the notation as b efo r e, one has Q 5 (5 n + 4) ≡ 0 (mo d 5) , Q 7 (7 n + 5) ≡ 0 (mo d 7) , Q 11 (11 n + 6) ≡ 0 (mo d 11) . Pr o of. Let us prov e the case for 5, and the other results will follow from similar idea. W e will pro ve the result for 5 using an induction on n , as follows. n = 0 : In this case, Q 5 (4) = 0 ≡ 0 (mo d 5) is trivially true. n = 1 : One can verify that Q 5 (9) = 5 ≡ 0 (mo d 5). n : Let us assume that Q 5 (5 n + 4) ≡ 0 (mo d 5) for some integer n > 1. n + 1 : F rom L emma 1 and 2, using k = 5, we obtain the following. Q 5 (5( n + 1) + 4) = Q 5 ((5 n + 4) + 5) = Q 5 (5 n + 4) + R 5 ((5 n + 4) + 5) = Q 5 (5 n + 4) + P (5 n + 4) ≡ 0 (mo d 5) The final congruence ho lds from our a ssumption and the first Ramanujan congruence. Hence, b y induction on n , the first r esult for mo dulus 5 is prov ed. One can similarly prov e the results for mo duli 7 and 11 using Lemmas 1 and 2 appropriately . ✷ One can a lso deriv e analogous results f o r higher order Ramanujan congruence, suc h a s Q 5 (25 n + 24) ≡ 0 (mo d 5 2 ), Q 5 (125 n + 99) ≡ 0 (mo d 5 3 ) etc., in a similar fashion. 9 3.4. Alternate Pro of of R aman ujan’s R esults In this section, w e prop ose a nov el alternate pro of of Ra ma nujan’s congruence results for the partition function P ( n ). T o the b est of o ur kno wledge, this approac h has not b een prop o sed in the literature till date. F o r this a lternate pro of , w e shall use the following relation, whic h w e obtained in t he previous section. P (5 n + 4) = Q k (5( n + 1) + 4) − Q k (5 n + 4) And in tur n, w e will require the study of Q k ( n ) in terms of generating functions to prov e Raman ujan’s results for P ( n ). Let us prov e the first relation (case with mo dulo 5), and the other pro of s will follow similarly . 3.4.1. Generating F unction of Q k ( n ) The generating function for the pa r tition function P ( n ) is giv en b y F ( x ) = ∞ X m =0 P ( m ) · x m = ∞ Y n =1 1 1 − x n where w e assume P (0 ) = 1. In this fo rm ula, w e count the co efficien t o f x m on b oth sides, where the coefficien t on t he right hand side is the r esult of coun t ing all p ossible w a ys that x m is generated by multiplyin g smaller or equal p ow ers of x . This ob viously give s the n umber of pa rtitions of m in to smaller or equal parts. What we require for Q k ( n ) is to coun t the num b er of k ’s o ccurring in each of these partitions. Th us, w e w an t to (i) add r to t he count if x r k is a mem b er inv olved fr o m the righ t hand side, and (ii) not coun t an y of the partitions where no p ow er of x k is inv olv ed. This in t uitio n giv es rise to the following generating function for Q k ( n ). G k ( x ) = ∞ X m =0 Q k ( m ) · x m = 1 · x k + 2 · x 2 k + 3 · x 3 k + · · · (1 − x ) · (1 − x 2 ) · · · (1 − x k − 1 ) · (1 − x k +1 ) · · · = (1 − x k ) · ( x k + 2 x 2 k + 3 x 3 k + · · · ) · ∞ Y n =1 1 1 − x n = ( x k + x 2 k + x 3 k + · · · ) · ∞ Y n =1 1 1 − x n = x k 1 − x k · ∞ Y n =1 1 1 − x n . Our next goal is to use the generating f unction G 5 ( x ) to pro v e Q 5 (5 n + 4) ≡ 0 (mo d 5). 10 3.4.2. Pro of of Q 5 (5 n + 4) ≡ 0 (mo d 5) W e prov e t his along the same line as Ramanujan’s pro of for P (5 n + 4) . Notice that we ha ve x · [(1 − x ) · (1 − x 2 ) · (1 − x 3 ) · · · ] 4 = x · (1 − 3 x + 5 x 2 − 7 x 3 + · · · ) · (1 − x − x 2 + x 3 + · · · ) = ∞ X µ =0 ∞ X ν = −∞ ( − 1) µ + ν · (2 µ + 1) · x 1+ 1 2 µ ( µ +1)+ 1 2 ν (3 ν +1) (1) In the ab ov e expression, let us consider the co efficien t of x 5 n . In this case, w e hav e 1 + 1 2 µ ( µ + 1) + 1 2 ν (3 ν + 1) ≡ 0 (mo d 5) ⇒ 8 + 4 µ ( µ + 1) + 4 ν (3 ν + 1) ≡ 0 (mo d 5) ⇒ (2 µ + 1) 2 + 2( ν + 1) 2 ≡ 0 (mo d 5) (2) Note that (2 µ + 1) 2 is congruen t to 0 , 1 or 4 mo dulo 5, whereas 2( ν + 1) 2 is congruen t to 0, 2 o r 3 mo dulo 5. Th us, to satisfy Equation (2), w e m ust hav e b oth (2 µ + 1) 2 and 2( ν + 1) 2 congruen t to 0 mo dulo 5 , i.e., (2 µ + 1) ≡ 0 (mo d 5) to b e sp ecific. Using t his in conjunction with Equation (1), w e get that the co efficien t of x 5 n in x · [(1 − x ) · (1 − x 2 ) · (1 − x 3 ) · · · ] 4 is divisible by 5. Again, considering the co efficien ts of 1 1 − x 5 and 1 (1 − x ) 5 mo dulo 5, w e get 1 (1 − x ) 5 ≡ 1 1 − x 5 (mo d 5) ⇒ 1 − x 5 (1 − x ) 5 ≡ 1 (mo d 5) . Th us, a ll the co efficien ts (except the first one) in t he expression 1 − x 5 (1 − x ) 5 · 1 − x 10 (1 − x 2 ) 5 · 1 − x 15 (1 − x 3 ) 5 · · · = [(1 − x 5 ) · (1 − x 10 ) · · · ] · ∞ Y n =1 1 (1 − x n ) 5 are divisible b y 5. Hence, t he co efficien t of x 5 n in the expansion of x · [(1 − x 5 ) · (1 − x 10 ) · · · ] · ∞ Y n =1 1 1 − x n = x · [(1 − x ) · (1 − x 2 ) · (1 − x 3 ) · · · ] 4 · [(1 − x 5 ) · (1 − x 10 ) · · · ] · ∞ Y n =1 1 (1 − x n ) 5 is a lso divisible by 5, by the virtue of the previous discussion. This in turn prov es t ha t the co efficien t of x 5 n is the expansion of x · x 5 1 − x 5 · ∞ Y n =1 1 1 − x n = x 5 · ( 1 + x 5 + x 10 + x 15 + · · · ) · x · ∞ Y n =1 1 1 − x n 11 is divisible by 5. Now , recall the generating function of Q 5 ( n ): G 5 ( x ) = ∞ X m =0 Q 5 ( m ) · x m = x 5 1 − x 5 · ∞ Y n =1 1 1 − x n . The discussion so fa r reve als tha t the co efficien t of x 5 n in the expansion of x · G 5 ( x ) is divisible by 5, i.e., t he co efficien t of x 5 n − 1 or x 5 n +4 in the expansion o f G 5 ( x ) is divisible by 5. Therefore, we hav e the desired result: Q 5 (5 n + 4) ≡ 0 (mo d 5) for all non-negative n ∈ Z . 3.4.3. Pro of of P (5 n + 4) ≡ 0 (mo d 5) Recall the follow ing result from the pro of of the analogue of Raman ujan’s results. P (5 n + 4) = Q 5 (5( n + 1) + 4) − Q 5 (5 n + 4) for all non- negativ e n ∈ Z . This readily prov ides us with P (5 n + 4 ) ≡ 0 (mo d 5) for all non-negativ e n ∈ Z . This giv es a new alternate pro of of Ramanujan’s congruence results using the concept of Q k ( n ). The results for k = 7 , 11 ma y b e deriv ed similarly . 4. Conclusion In this pap er w e hav e presen ted an extension of Stanley’s theorem f o r pa r t itions of p o sitiv e in tegers. There exists an extension o f Sta nley’s theorem in the literature, whic h had b een prop osed b y Elder in 1984 . But our generalization op ens a new line of though t in this direc- tion. W e hav e also pro p osed some intere sting corollaries to illustrate p oten tial a pplicatio ns of t his g eneralization tow ards the existing results in integer partitio ns. In particular, w e pr ov e a couple of existing results using alternativ e tec hniques and pro- p ose analo gues t o Raman ujan’s results on congruence b eha vior of f unctions related to par- titions. W e hav e a lso prop osed an alterate pro of of Raman uja n’s results using the concept of Q k ( n ) and its generating function. A couple of in teresting r esearc h problems related to this topic ma y b e to study Elder’s theorem in the lig ht of the results prop osed in this pap er, and to prop ose analogues to more in volv ed congruence results for P ( n ) prov ed in the literature. Ac kno wledgemen t: The author s w ould lik e to thank their advisor Professor Subhamo y Maitra of Applied Statistics Unit, Indian Statistical Institute, for n umerous technic al dis- cussions and un tiring supp ort throughout the pro ject. The authors also lik e to express their gratitude to w ards Professor P a n telimon Sta nica for his preliminary f eedbac k on the tec hnical con tent o f this pap er. 12 References [1] S. Ahlgren and K. Ono. C ongruenc es and c o nje ctur es for the p artition function. Pro- ceedings of the Conference o n q-series with Applications to Com binatorics, Number Theory and Ph ysics, AMS Con t emp orary Mathematics, V ol. 291, pp. 1–1 0 , 20 01. [2] N. J. A. Sloane. Sum { k=0..n } p( k) w her e p(k) = numb er of p artitions of k. The on- line encyclopedia of in teger sequences, 20 10. http://www. research.a tt.com/ ∼ njas/sequences/A000070 . [3] N. J. A. Sloane. T riangle T(n,k), n > =1, 1 < =k < = n, giving numb er of k’s in al l p artitions of n . The on-line encyclop edia of integer sequences, 201 0. http://www. research.a tt.com/ ∼ njas/sequences/A066633 . [4] Wikip edia, the Online Encyclop edia. Partition (Numb er The ory). 2 010. http://en.w ikipedia.o rg/wiki/Partition (number theory) [5] W olfram Math w orld. Elder’s The or em. 2010. http://math world.wolf ram.com/EldersTheorem.html [6] W olfram Math w orld. Stanley’s The or em. 2010. http://math world.wolf ram.com/StanleysTheorem.html 13