Counting number of factorizations of a natural number
In this note we describe a new method of counting the number of unordered factorizations of a natural number by means of a generating function and a recurrence relation arising from it, which improves an earlier result in this direction.
Authors: Shamik Ghosh
Coun ting n um b er of factori zations of a natural n um b er Shamik Ghosh Department of Mathematics, Jada vpu r Univers ity , Kolk ata - 700 032, India. E-mail addr ess: sghosh@mat h.jdvu.ac.in Abstract In this note we describ e a new metho d o f countin g the num b er of unorder e d factorizations of a natural n umber by means of a generating function and a recurr ence relation arising from it, whic h improv es an earlier result in this direction. 1 In tro duction Consider the n a tur al num b er 18. It has 4 distin ct “factorizations”, namely 18 = 2 . 3 . 3 = 2 . 9 = 3 . 6 = 18 . Similarly , there are 9 wa ys of factorizing 36: 36 = 2 . 2 . 3 . 3 = 2 . 2 . 9 = 2 . 3 . 6 = 3 . 3 . 4 = 2 . 18 = 3 . 12 = 4 . 9 = 6 . 6 = 36. Our problem is to find this num b er for an y n a tur a l num b er n . Since we are not distinguishing b et w een 2 · 9 and 9 · 2, suc h a factorization is called unor der e d . A p artition of a natural n umb er n is a represen tation of n as the sum of an y num b er of p ositiv e in tegral parts, where the ord e r of the parts is irrelev ant . The n umb er of such partitions of n is known as the p ar tition function and is denoted b y p ( n ). Lik ewise the fu nction p ∗ ( n ) den otes the num b er of w a ys of expressing n as a pr o duct of p ositiv e integ ers greater than 1, the order of the factors in the pro du ct b eing irr elev an t. F or conv enience, p ∗ (1) is assumed to b e 1. Clearly p ∗ ( n ) is the n umb er of un ordered f actorizations of n . In 1983, Hughes and Sh alit [6] obtained a b ound for p ∗ ( n ), namely , p ∗ ( n ) 6 2 n √ 2 whic h w as then impro ved to p ∗ ( n ) 6 n by Mattics and Do d d [7] in 1986. By this time Canfi eld, Erd¨ os and Po merance [2] mo dified a result of Opp enheim regardin g the m aximal order of p ∗ ( n ). Th ey obtained another b ound for p ∗ ( n ) and describ ed an algorithm f or it. An a ve rage estimate for p ∗ ( n ) was giv en by Opp enh eim [8] which was also pr o v ed indep end en tly b y Szekeres and T uran [9]. Finally , in 1991, Harris and Su bbarao [5] came w ith a generating function and a recursion formula for p ∗ ( n ). One ma y consider [1] and [3] for some problems 2000 AMS Mathematics Subje ct Classific ation: Primary 11N25 Key wor ds and phr ases: Unordered factorization, partition function. 1 asso ciated with p ∗ ( n ). F or a list of v alues and compu ter programming one m ay consider the website: h ttp://www.researc h.att.com/cgi-bin/ac c ess.cgi/as/njas/sequences/ (sequence no. A00105 5). In this note, we describ e a new metho d for counting p ∗ ( n ) and obtain a generating fu nction for it whic h is f ollo wed by a recurrence relatio n that generalizes the one giv en b y Harris and Sub barao [5] as wel l as the one giv en by Euler for p ( n ). The final r ecursion form u la imp ro ve s the one g iven in [5] as it con tains less n umber of terms . It is imp ortan t to n ote that we w ish to devel op an algebraic approac h to the pr oblem which migh t b e helpf ul for other sim ilar situations in fu ture. Also we note some errors in describin g an equiv alen t form of the recurrence relation in [5]. Through ou t the note we denote set of all natural n umbers , non-negativ e in tegers, integ ers and rational num b ers b y N , Z + 0 , Z , Q resp ectiv ely . 2 Represen tation of n um b ers b y p olynomials Consider the monoid ( N , · ) of natur al num b er s under u sual m u ltiplication. F or an y natural n umber n , let S ( n ) b e the subm on oid of ( N , · ), generated b y the s et of prime factors of n , i.e., if the p rime factorizat ion of n is n = p n 1 1 p n 2 2 . . . p n k k , (2.1) where p i are distinct primes, n i ∈ N for all i = 1 , 2 , . . . , k , ( k ∈ N ), then S ( n ) = p r 1 1 p r 2 2 . . . p r k k ∈ N | r i ∈ Z + 0 , i = 1 , 2 , . . . , k . W e sho w that S ( n ) has an int eresting algebraic structure. Defin e the partial ordering ≤ · on S ( n ) b y a ≤ · b ⇐ ⇒ a divides b. This orderin g on S ( n ) is, in fact, a lattice ordering where a ∨ b = lcm ( a, b ) and a ∧ b = gcd( a, b ) for all a, b ∈ S ( n ). Moreo ve r this lattice is d istr ibutiv e and b ound ed b elo w b y 1. A monoid S is called a lattic e-or der e d semigr oup if it has a lattice ordering that satisfies a ( b ∨ c ) = ab ∨ ac and ( b ∨ c ) a = ba ∨ ca , for all a, b, c ∈ S . No w for all a, b, c ∈ S ( n ), a { lcm( b, c ) } = lcm( ab, ac ). S o w e ha v e the follo wing prop osition: Prop osition 2.1. F or any natur al numb er n , ( S ( n ) , · , ≤ · ) is a lattic e-or der e d semigr oup. Definition 2.2. No w corresp ond ing to eac h n atural n umb er n w e asso ciate a p olynomial in the p olynomial semiring Z + 0 [ x ] as f ( x ; n ) = n 1 + n 2 x + n 3 x 2 + · · · + n k x k − 1 , where n h as th e prime factoriza tion (2.1). 2 Next w e defin e a b inary relation ≦ on Z + 0 [ x ] as follo ws: Let f ( x ) = a 0 + a 1 x + a 2 x 2 + · · · + a m x m and g ( x ) = b 0 + b 1 x + b 2 x 2 + · · · + b n x n . Th en f ( x ) ≦ g ( x ) ⇐ ⇒ m 6 n and a i 6 b i for all i = 0 , 1 , 2 , . . . , m. Clearly ≦ is a partial ord ering on Z + 0 [ x ]. Finally , let us den ote the set of all p olynomials in Z + 0 [ x ] of degree less th an k b y P k [ x ]. Theorem 2.3. ( P k [ x ] , + , ≦ ) is a lattic e- or der e d semigr oup which is isomorphic t o ( S ( n ) , · , ≤ · ) , wher e n has the prime factorization (2.1). Pr o of. Let f ( x ) = a 1 + a 2 x + a 3 x 2 + · · · + a k x k − 1 , g ( x ) = b 1 + b 2 x + b 3 x 2 + · · · + b k x k − 1 ∈ P k [ x ]. Then it is routine to v erify that f ∨ g = c 1 + c 2 x + c 3 x 2 + · · · + c k x k − 1 and f ∧ g = d 1 + d 2 x + d 3 x 2 + · · · + d k x k − 1 , where c i = max { a i , b i } and d i = min { a i , b i } . Th us ( P k [ x ] , ≦ ) is a lattice. Obviously , ( P k [ x ] , + ) is an ab elian m on oid where the id en tit y elemen t is the zero p olynomial. No w c ho ose f ( x ) = k P i =1 a i x i − 1 , g ( x ) = k P i =1 b i x i − 1 , h ( x ) = k P i =1 c i x i − 1 ∈ P k [ x ]. Then f + ( g ∨ h ) = k P i =1 a i + max { b i , c i } x i − 1 = k P i =1 max { a i + b i , a i + c i } x i − 1 = ( f + g ) ∨ ( f + h ). Therefore ( P k [ x ] , + , ≦ ) is a lattice-ordered semigroup. No w define a map ψ : S ( n ) − → P k [ x ] by ψ ( m ) = f ( x ; m ) for all m ∈ S ( n ). That ψ is bijectiv e follo ws from Definition 2.2. Let m 1 = k Q i =1 p r 1 i i , m 2 = k Q i =1 p r 2 i i ∈ S ( n ). Then m 1 ≤ · m 2 ⇐ ⇒ m 1 divides m 2 ⇐ ⇒ r 1 i 6 r 2 i for eac h i = 1 , 2 , . . . , k ⇐ ⇒ r 11 + r 12 x + r 13 x 2 + · · · + r 1 k x k − 1 ≦ r 21 + r 22 x + r 23 x 2 + · · · + r 2 k x k − 1 ⇐ ⇒ f ( x ; m 1 ) ≦ f ( x ; m 2 ) ⇐ ⇒ ψ ( m 1 ) ≦ ψ ( m 2 ) . Also m 1 m 2 = k Q i =1 p r 1 i + r 2 i i . Then f ( x ; m 1 m 2 ) = k P i =1 ( r 1 i + r 2 i ) x i − 1 = k P i =1 r 1 i x i − 1 + k P i =1 r 2 i x i − 1 = f ( x ; m 1 ) + f ( x ; m 2 ). T h us ψ ( m 1 m 2 ) = ψ ( m 1 ) + ψ ( m 2 ). T herefore ψ is an isomorphism, i.e., ( S ( n ) , · , ≤ · ) ∼ = ( P k [ x ] , + , ≦ ), as required. Definition 2.4. F or an y f ( x ) ∈ Z + 0 [ x ], let p ( f ) denote the num b er of partitions of the p olynomial f ( x ) in terms of addition of p olynomials (not all distinct) in Z + 0 [ x ], where the order of addition is irrelev ant . W e assume p (0) = 1. 3 F or example, the d istinct partitions of the p olynomial 2 + x in Z + 0 [ x ] are 2 + x = (1) + (1) + ( x ) = (1) + (1 + x ) = (2) + ( x ) = (2 + x ) Note that f ( x ; 12) = 2 + x and compare the ab ov e partitio ns w ith usual factorizati ons: 12 = 2 . 2 . 3 = 2 . (2 . 3) = 2 2 . 3 = 12 . Theorem 2.5. F or any natur al numb er n , p ∗ ( n ) = p ( f ( x ; n )) . Pr o of. Let n ∈ N and F ( n ) denote the set of all factors of n . Then ( F ( n ) , ≤ · ) is a su b lattice of S ( n ). On the other hand, defi ne 1 the set S ( f ( x )) = g ( x ) ∈ Z + 0 [ x ] | g ( x ) ≦ f ( x ) . Then S ( f ( x )) is a sublattice of P k [ x ], where f ( x ) = f ( x ; n ) and n has the prime fac torization (2.1). By Theorem 2.3, it follo ws that the restriction of the map ψ on F ( n ) is a latt ice isomorphism from ( F ( n ) , ≤ · ) on to S ( f ( x ; n )) , ≦ . In deed, let m = p r 1 1 p r 2 2 . . . p r k k ∈ F ( n ) for some r 1 , r 2 , . . . , r k ∈ Z + 0 . Since m is a factor of n , w e ha ve r i 6 n i for eac h i = 1 , 2 , . . . , k . Hence ψ ( m ) = f ( x ; m ) = r 1 + r 2 x + r 3 x 2 + · · · + r k x k − 1 ≦ n 1 + n 2 x + n 3 x 2 + · · · + n k x k − 1 = f ( x ; n ) wh ic h imp lies ψ ( m ) ∈ S ( f ( x ; n )). Con ve rsely , let g ( x ) ∈ S ( f ( x ; n )). Then deg g ( x ) 6 deg f ( x ) = k − 1. Let g ( x ) = b 1 + b 2 x + b 3 x 2 + · · · + b k x k − 1 for some b 1 , b 2 , . . . , b k ∈ Z + 0 . T hen b i 6 n i for eac h i = 1 , 2 , . . . , k and g ( x ) = g ( x ; p b 1 1 p b 2 2 . . . p b k k ) = ψ ( p b 1 1 p b 2 2 . . . p b k k ), which implies ψ ( F ( n )) = S ( f ( x ; n )). Th us w e h a v e ( F ( n ) , ≤ · ) ∼ = S ( f ( x ; n )) , ≦ . Al so since f ( x ; m 1 m 2 ) = f ( x ; m 1 ) + f ( x ; m 2 ) for all m 1 , m 2 ∈ F ( n ), there exists a one-to-one corresp ondence b et we en factoriza tions n = m 1 m 2 . . . m r of n with partitions f ( x ; m 1 ) + f ( x ; m 2 ) + · · · + f ( x ; m r ) = f ( x ; m 1 m 2 . . . m r ) = f ( x ; n ) of f ( x ; n ). Th u s we ha ve p ∗ ( n ) = p ( f ( x ; n )). Corollary 2.6. L et p b e a prime numb er and n ∈ N . Then p ∗ ( p n ) = p ( n ) . R emark 2.7 . It is clear that the v alue of p ∗ ( n ) is indep en den t of the particular p rimes in vol ved in the prime factorizat ion expression of n , i.e., if n has the prime factorizatio n (2.1) and m = q n 1 1 q n 2 2 . . . q n k k , where q i are d istinct pr im es, then p ∗ ( m ) = p ∗ ( n ). Thus the p olynomial f ( x ; n ) in Definition 2.2 may b e d ifferen t for different arrangemen t of primes in the prime factorization 1 S ( f ( x ) ) is call ed the s e cti on of f ( x ) in Z + 0 [ x ]. 4 expression of n . But th e v alue of p ( f ( x ; n )) r emains same for eac h su c h arrangemen ts. In particular, p (2 + x ) = p ∗ (2 2 . 3) = 4 = p ∗ (2 . 3 2 ) = p (1 + 2 x ). Ind eed, the distinct partitions of the p olynomial 1 + 2 x in Z + 0 [ x ] are 1 + 2 x = (1) + ( x ) + ( x ) = (1 + x ) + ( x ) = (1) + (2 x ) = (1 + 2 x ) More generally , p ∗ ( p 2 q ) = 4 for an y p air of d istinct primes p, q . 3 Generating function and recurrence relations Let n b e a natur al num b er. W e kn o w that the n u mb er of partitions, p ( n ) of n is give n [4] by the follo wing classical generating function found by Euler: F ( x ) = 1 (1 − x )(1 − x 2 )(1 − x 3 ) . . . (3.1) = ∞ Y n =1 1 1 − x n (3.2) = ∞ Y i =1 1 + ∞ X n =1 x ni (3.3) = 1 + ∞ X n =1 p ( n ) x n . (3.4) In the ab ov e equ alities, since (3.3) provides all p ossible p ositiv e in tegral p ow ers of x less than n (with all p ossib le m u ltiplicities), pr o duct of these te rm s pro duce the term x n as man y times as n can b e expr essed as a sum of p ositiv e intege rs wh ic h is exactly the n umb er of partitions of n , i.e., the term x n o ccurs p ( n ) times. So w e get the coefficien t p ( n ) of x n in (3.4). Similarly , if w e wish to find the n umber of p artitions of the p olynomial f ( x ) in the p olynomial semiring Z + 0 [ x ], we ha v e to consider the pr o duct of summations which p ro vides a ll p ossible polynomials in Z + 0 [ x ] less than f ( x ) (with all p ossible multiplicit ies) as in dices. Thus we ha ve the follo wing formal generating fu nction for p ( f ( x )): F ( x ) = Y g ∈ Z + 0 [ x ] ⋆ 1 1 − e g ( x ) (3.5) = Y g ∈ Z + 0 [ x ] ⋆ 1 + ∞ X n =1 e ng ( x ) (3.6) = 1 + X f ∈ Z + 0 [ x ] ⋆ p ( f ) e f ( x ) , (3.7) 5 where Z + 0 [ x ] ⋆ = Z + 0 [ x ] r { 0 } . R emark 3 .1 . Th e expressions (3.5)-(3.7) are merely formal in the sen se that for any particular f ( x ) ∈ Z + 0 [ x ], the co efficien ts of e f ( x ) in either sid e are same. So the con v ergence pr ob lem do es not arise here. How ever, if one insists on it, one ma y r ep lace e b y e 1 = 1 e in whic h case (3.5)-(3.7) are absolutely and uniformly con verge nt for all p ositiv e in tegral v alues of x . F or example, consider F (1) = Q g ∈ Z + 0 [ x ] ⋆ 1 1 − e g (1) 1 . Th e pr o duct Q g ∈ Z + 0 [ x ] ⋆ (1 − e g (1) 1 ) is con v ergent if P g ∈ Z + 0 [ x ] ⋆ e g (1) 1 is conv ergent . No w P g ∈ Z + 0 [ x ] ⋆ e g (1) 1 = ∞ P n =1 p ( n ) e n 1 whic h is absolutely and uniformly conv ergent for e 1 = 1 e [4]. No w (3.5) can b e written in the form: F ( x ) = Y g ∈ Z + 0 [ x ] ⋆ g i s pri mitive 1 ∞ Q n =1 (1 − e ng ( x ) ) (3.8) whic h is again b y (3.4), F ( x ) = Y g ∈ Z + 0 [ x ] ⋆ g is primi tive 1 + ∞ X n =1 p ( n ) e ng ( x ) (3.9) So w e ha ve the follo wing generating fun ction for p ( f ( x )): 1 + X f ∈ Z + 0 [ x ] ⋆ p ( f ) e f ( x ) = Y g ∈ Z + 0 [ x ] ⋆ g is primi tive 1 + ∞ X n =1 p ( n ) e ng ( x ) (3.10) Using this we d escrib e a metho d of calculating p ( f ( x )): p ( f ( x )) = co efficien t of e f ( x ) in Y 0 f ( x ; 12), as these term s hav e n o f urther con tribution in forming the term e f ( x ;12) . (ii) By the pr o cess of calculating p ( f ( x )), w e are getting all th e v alues of p ( g ( x )) for all g ( x ) 6 f ( x ) in Z + 0 [ x ]. F or example, the ab o ve calculation giv es u s p (1) = 1 , p (2) = 2 , p ( x ) = 1 , p (1 + x ) = 2 . Next w e wish to obtain a recurr en ce relation for p ( f ( x )). F rom (3.5) and (3.7), we get that 1 + X f ∈ Z + 0 [ x ] ⋆ p ( f ) e f ( x ) · Y g ∈ Z + 0 [ x ] ⋆ 1 − e g ( x ) = 1 . (3.12) No w taking formal deriv ativ es 2 on b oth sides of (3.12) we get, X f ∈ Z + 0 [ x ] ⋆ p ( f ) e f ( x ) · f ′ ( x ) · Y g ∈ Z + 0 [ x ] ⋆ 1 − e g ( x ) + 1 + X f ∈ Z + 0 [ x ] ⋆ p ( f ) e f ( x ) · X g ∈ Z + 0 [ x ] ⋆ n − e g ( x ) g ′ ( x ) Y g 1 ∈ Z + 0 [ x ] ⋆ g 1 6 = g 1 − e g 1 ( x ) o = 0 whic h implies X f ∈ Z + 0 [ x ] ⋆ p ( f ) e f ( x ) · f ′ ( x ) = 1 + X f ∈ Z + 0 [ x ] ⋆ p ( f ) e f ( x ) · X g ∈ Z + 0 [ x ] ⋆ n e g ( x ) g ′ ( x ) · 1 + ∞ X r =1 e r g ( x ) o . (3.13) Then equating the co efficien t of e f ( x ) of b oth sides w e get p ( f ) f ′ ( x ) = X r c ( f ) 1 r · f ′ ( x ) + X 0 rg ( x ) } . Consid- ering p (0) = 1 and replacing r g b y g we finally ha v e p ( f ) f ′ ( x ) = X 0
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