Partitions with Parts in a Finite Set and with PartsOutsideaFiniteSet

Transcription

1 Partitions with Parts in a Finite Set with PartsOutsideaFiniteSet Gert lmkvist CONTENTS Introduction Partitions into Parts in a Finite Set 3 Partitions with Parts Outside a Finite Set References Exact formulas for the number of partitions with parts in a finite set are proved without the use of partial fractions Corresponding formulas with parts outside of a finite set are found Several numerical examples are given INTRODUCTION In a recent paper, M Nathanson [Nathanson] finds the highest order term of the asymptotic formula for the number of partitions with parts in a finite set This is done without referring to the partial fraction expansion of the generating function In this paper, we essentially find the entire asymptotic formula without using partial fractions This is done by the metaphysical method described in [lmkvist 98] There the asymptotic formula was found using Fourier transformations of distributions in a rather obscure way ut when the generating function is rational, it is shown here that the method is perfectly sound This is based on the following result: Lemma e nz n + k Res( ( e z,z =0= k k In the second part of the paper, we consider partitions with parts outside a finite set This is a complement of the situation in the first part If = {a,,a k } 000 MS Subject Classification: Primary P8; Secondary 057, 34 Keywords: Partitions, asymptotics of partitions, additive number theory is the finite set p(, n = number of partitions of n with parts in, then we have the generating function f (x = ( x a = p(, nx n c KPeters,Ltd /00 $ 050 per page Experimental Mathematics :4, page 449

2 450 Experimental Mathematics, Vol (00, No 4 Let p(, n = the number of partitions of n with parts outside, thenwehave f (x = ( x j = F (x f (x = p(, nx n, j/ F (x = ( x j = p(nx n is the generating function of ordinary partitions p(n Since the expansions of F (x f (x nearx =are known, we also know the expansion of f (x nearx = This gives the first term in the asymptotic formula for p(, n We also do the same thing for x near x = to get the second term Since f (x is not rational, we do not have a proof, but numerically the approximation is excellent if n is not too close to the largest number in PRTITIONS INTO PRTS IN FINITE SET We start by proving that the metaphysical method is correct when the generating function is rational Theorem Let f(x be a rational function with expansion f(x = c n x n Let x = α be a pole of f(x of order k ssumethat f(αe t = k a j t j + b j t j when t 0+ the pole α gives rise to a term α n k a j n j (j! in the asymptotic expansion of c n as n Proof: It is enough to prove the theorem for Now f(x = (x α k = ( k α k ( k α k ( x α k n + k k α n x n f(αe t = ( k α k ( e t k ssume that k ( e t k = d j t j + O( We have to show that k n j n + k d j (j! = k Consider e nt ( e t k = i=0 n i t i i! k d j t j + O( The coefficient of t of the right-h side is k d j n j (j! so we are done if we can prove the following lemma: Lemma We have e nz n + k Res( ( e z,z =0= k k Proof of Lemma : Let so Clearly ut e nz c(k, n =Res( ( e z,z =0 k e nz ( e z k = e nz e(n z ( e z + k ( e z k, c(k, n =c(k,n+c(k, n c(,n=foralln ( e z k ( e z k e z = ( e z k = d k dz ( e z k has residue 0 Hence c(k, 0 = c(k, 0 = for all k Now we do induction on both k n n + k n + k c(k,n = + k k

3 lmqvist: Partitions with Parts in a Finite Set with Parts Outside a Finite Set 45 In order to state our next result, we introduce the following notation Let k x j t/ sinh(x j t/ = σ m (x,, x k t m m=0 σ m is a certain symmetric function (polynomial of x,, x k similar to those introduced by Hirzebruch in topology Clearly σ 0 =σ m =0ifm is odd Let = {a,, a k } be a set of positive integers without a common factor f (x = ( x a = p(, nx n Theorem 3 Let ξ = n + k a j the first approximation of p(, n, coming from the pole x =of f(x is Φ (n = k ξ k j a σ j (a,, a k (k j! Proof: We have f(e t = ( e at = exp( t a sinh(at/ = exp( t at k at/ a sinh(at/ = exp( t a σ j (a,, a k t (k j a The exponential exp( t a means just a translation (Taylor s theorem: n n + a = ξ Thus the pole x = gives the contribution to p(, n Φ (n = k ξ k j σ j (a,, a k a (k j! j Example 4 (Partitions into primes f(x = = {, 3, 5, 7,, 3, 7} a=,3,,7 ( x a = Let c n x n at/ = sinh(at/ 4 t t t gives with k =7ξ = n + a = n +9 ξ 6 Φ (n = ξ ξ If n = 0000, then while Even better is with c 0000 = Φ (0000 = c 999 = Φ (999 = This example is rather special since the a j :s are pairwise coprime In general, if f(x = x p ( x b (p, b = for all b, we will get a contribution Φ p (n = α n p ( αb α thesum runs overthenontrivial p-th roots of unity Inourexample,wewillget If p =3,then so if n = 0000, then Φ 3 = 3 α = 3 Φ (n = ( n 8 ( α( α = (α + α =3 α α 0000 ( α ( α 5 ( α 7 ( α ( α 3 ( α 7 α ( α ( α 4 = 3 5 α ( α = 3 5 ( +α + α = 3 5 ( 4 = 8

4 45 Experimental Mathematics, Vol (00, No 4 Similarly one finds One checks that Φ 5 = 5 ; Φ 7 = 5 49 ; Φ = ; Φ 3 = 0; Φ 7 = 7 Φ + Φ + Φ 3 + Φ 5 + Φ 7 + Φ + Φ 3 + Φ 7 = c 0000 Example 5 (Partitions into Fibonacci numbers F =,F =, F n = F n + F n be the Fibonacci numbers Consider 0 f(x = ( x Fj = c n x n we get the first approximation as before Φ (n = ξ 9 r ξ ξ5 If n = 000, then while ξ ξ r = ξ = n + 0 F j = n + 3 c 000 = Φ (000 = Let so the error is We compute the second term coming from x = Thus f( e t = 6963 t t t + giving Φ (n = ( n n n In particular Φ (000 = = which is much smaller than the error So, what is going on? We compute the other terms as well Let Φ k denote the contributions from the k :th roots of unity = that have not been treated before We get with α =exp(πi/3 f(exp(πi/3 t = 503 t +( i t + Φ 3 (000 = α α (α α i (α 000 α Now let α =exp(πi/5 4 α 000j f(α j e t = 5 4 (t +99t + 99 Φ 5 (000 = ( = Further computations using Maple give Φ 4 = 56 ; Φ 8 = 6 ; Φ 3 = 3 3 ; Φ = ; Φ 34 = 7 7 ; Φ 55 = 38 ; Φ 89 = Hence Φ 89 is the largest term! This depends on the fact that ( α( α ( α 3 is very small when α = exp(πi/89 When n becomes large, Φ (n will of course dominate

5 lmqvist: Partitions with Parts in a Finite Set with Parts Outside a Finite Set 453 Remark 6 Partions into all Fibonacci numbers is a much more difficult problem One needs to study the Fibonacci ζ function ζ Fib (s = F s n= n In particular one needs the values ζ Fib (0 ζ Fib (0, etc Example 7 (Partitions into at most r parts Let p(r, n = the number of partitions of n into parts r = the number of partitions of n into at most r partswe have the generating function f r (x = r jt/ sinh(jt/ Hence f r (e t = t r r! r ( x j = p(r, nx n = exp( = exp( = exp( r r k= k= r(r + exp t 4 k= log( sinh(jt/ jt/ k j k t k k(k! k k+ (r + t k k(k +! k k+ (r + t k k(k +! we get the first approximation of p(r, n withξ = n + r(r +/4 Φ (n = r! exp If r =, we get Φ (,n= k=! 0! k k+ (r + D k k(k +! D = d dξ ξ r (r! ξ ξ8 + 90ξ ξ We compute a few of the other terms: Φ (,n= ( n 5! 4! ξ ξ 4 539ξ ; Φ 3 (,n= 3 8 (ξ 77 πn cos ξ cos πn 3 + 4ξ sin πn 3 3 ; ; Φ 4 (,n= sinπn 4 Φ 5 (,n= sin((n + π 5 sin((n + π 5 4 sin( π sin( π 5 ξ π cos((n+ 5 + π (5 cot( π sin( 5 5 cot(π ; π cos((n+ + 5 sin( π 5 ( cot( π 5 +5cot(π 5 cos( nπ Φ 6 (,n= 3 08 Numerically we consider n =000 We find It follows that p(, 000 = Φ = ; Φ = ; Φ 3 = 40497; Φ 4 =009; Φ 5 =657; Φ 6 = 0005 Φ + Φ + Φ 3 + Φ 4 + Φ 5 + Φ 6 = Remark 8 Since r ( e jt =( e rt we have (using Taylor s formula, r ( e jt, Φ (r,n=φ (r, n Φ (r, n r 3 PRTITIONS WITH PRTS OUTSIDE FINITE SET Let = {a,, a k } be a finite set of positive integers Let p(, n be the number of partitions of n into parts not contained in The generating function is f (x = ( x j p(, nx n = j/

6 454 Experimental Mathematics, Vol (00, No 4 Let F (x = ( x j = p(nx n be the generating function of ordinary partitions Let f (x = j ( x j f (x = F (x f (x Now by Hardy-Ramanujan s expansion, we know that t F (e t π exp ζ(t t, 4 we have f (e t = exp( t a t k a It follows that f (e t a exp( t a π sinh(at/ exp at/ Now t k+/ exp t( 4 + a at/ sinh(at/ ζ(t t 4 is just a translation t k means D k Formally we have t / exp(ζ(t ζ( j = t / j ζ( j n j 3/ j! j!γ(j / = πζ( D sinh 4ζ(n D = d dn This is essentially the first term in the Hardy- Ramanujan-Rademacher formula for p(n So let ξ = n 4 a p(, n is approximated by a Φ (, n = π sinh(ad/ D k+ sinh 4ζ(ξ ζ( ad/ D = d dξ Example 3 (Prime numbers Let = {, 3, 5, 7,, 3, 7} as before a=,3,,7 sinh(at/ at/ Thus we get with that =+ t t ξ = n 4 9 t t Φ (n = 5050 π ( + ζ( D D4 Let n = D 6 + D 9 sinh 4ζ(ξ 945 D = d dξ p(, 000 = while 3 terms give Φ (000 = so we get 3 correct digits out of 7 We will now give a similar formula for the second term coming from the singularity at x = ssume that = C consists of the even numbers in C of the odd f ( e t = ( e bt ( + e ct C = exp( t a b t Furthermore, we have for F (x = the functional equation ( x j F ( x = F (x 3 F (xf (x 4 bt/ sinh(bt/ C cosh(ct/

7 lmqvist: Partitions with Parts in a Finite Set with Parts Outside a Finite Set 455 which implies hence F ( e t t ζ( π exp 4 t t 4 f ( e t = F ( e t f ( e t = t ζ( = π exp 4 t t 4 b exp a t ( cosh(ct/ C sinh(bt/ bt/ We get Φ (n = b π sinh(bd/ ζ( bd/ ( cosh(cd/d + sinh ζ(ξ C ξ = n 4 a D = d dξ We apply this formula when = {, 3, 5, 7,, 3, 7} we have sinh(t/ t/ we get 3,5,,7 Φ (n = π ζ( ( cosh(ct/ = t t t D D D 6 + D 3 sinh ζ(ξ 63 ξ = n 4 9 If n = 000, we get taking 7 terms Φ (000 = Φ (000 + Φ (000 = , so we get 9 correct digits out of 7 Example 3 (Partitions with parts r Let = {,,, r } p(, n =p(r, n =thenumberof partitions of n into parts r Wehavethegenerating function f r (x = p(r, nx n = ( x j = F (x f r (x j=r Using the computations made in Example 7, we get f r (e t (r! = t r / π exp ζ(t (rt j j+ (r + 4 j(j +! tj (r! Φ (r, n = π j j+ (r exp j(j +! Dj Dr+ sinh( 4ζ(ξ ξ = n (r 4 D = d dξ s a numerical example, we take r =n = 00 p(, 00 = Φ (, 00 = We also compute giving Φ (, 00 = Φ + Φ = so we get 8 correct digits out of 6 In a forthcoming paper, the author will show how to compute higher-order terms hence compute p(, 00 to the last digit

8 456 Experimental Mathematics, Vol (00, No 4 s a further example, we take r =50n = 000 p(50, 000 = p(, n = j c j p(n j Φ (50, 000 = so we get digits out of 7 Here we stopped after 8 terms of Φ, sincetakingmoretermsdestroystheresult It is still surprising that we can use the formula when r is that big ( r> n ut when (r >n,thenξ is negative the formula is not valid p(r, n can be computed in the following way: Take r =00 n =000 0 p(00, 000 = p(j, j = = Remark 33 Maple computes p(n immediately with numbpart This makes it possible to compute p(, n very fast Let ( x a = c j x j j REFERENCES [lmkvist 98] G lmkvist symptotic Formulas Generalized Dedekind Sums Exp Math 7:4 (998, [Canfield 97] E R Canfield From Recursions to symptotics: On Szekeres Formula for the Number of Partitions ElectrJofComb4 (997, #R6 [Dixmier Nicolas 90] J Dixmier J L Nicolas Partitions sans petit somms In Tribute to Paul Erdös, editedbyaker,ollobas,hajnal, pp 5 Cambridge: Cambridge Univ Press, 990 [Nathanson] M Nathanson Partitions with Parts in a Finite Set ProcmerMathSoc8 (000, [Nicolas Sarközy 97] J L Nicolas Sarközy On Two Partition Problems cta Math Hungar 77 (997, 95 [Szekeres 5] G Szekeres n symptotic Formula in the Theory of Partitions Quart J of Math (95, [Szekeres 53] G Szekeres Some symptotic Formulae in the Theory of Partitions (II Quart J of Math 4 (953, 96 Gert lmkvist, Institute of lgebraic Meditation, PL 500, 4333 Höör, Sweden (gert@mathslthse Received December 5, 00; accepted in revised form May 3, 00