Factors of sums of powers of binomial coefficients


 Abner Goodwin
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1 ACTA ARITHMETICA LXXXVI.1 (1998) Factors of sums of powers of biomial coefficiets by Neil J. Cali (Clemso, S.C.) Dedicated to the memory of Paul Erdős 1. Itroductio. It is well ow that if ( ) a f,a = the f,0 = + 1, f,1 = 2, f,2 = ( ), ad it is possible to show (Wilf, persoal commuicatio, usig techiues i [8]) that for 3 a 9, there is o closed form for f,a as a sum of a fixed umber of hypergeometric terms. Similarly, usig asymptotic techiues, de Bruij has show [1] that if a 4, the h,a has o closed form, where ( ) a h,a = ( 1) (clearly, h +1,a = 0). I this paper we will prove that while o closed form may exist, there are iterestig divisibility properties of f,2a ad h,a. We will illustrate some of the techiues which may be applied to prove these sorts of results. Our mai results are: Theorem 1. For all positive ad a, 2 ( ( 1) ) a is divisible by ( ). Theorem 2. For all positive itegers a, m, j, [ ] a ( 1) j 1991 Mathematics Subject Classificatio: 05A10, 05A30, 11B65. [17]
2 18 N. J. Cali is divisible by t(, ), where the [ ] t(, ) is a aalogue of the odd part of ( ). are the biomial coefficiets, ad 2. Bacgroud. I attemptig to exted the results of previous wor [2], we were led to cosider factorizatios of sums of powers of biomial coefficiets. It uicly became clear that for eve expoets, small primes occurred as divisors i a regular fashio (Propositio 3), ad that this result could be exteded (Propositio 7) to odd expoets ad alteratig sums. Further ivestigatio revealed (Propositio 8) that for all alteratig sums, the primes dividig h,a coicided with those dividig ( ). This led us to cojecture, ad subseuetly to prove, Theorem 1; as part of our proof we obtai (Theorem 2) a correspodig result for biomial coefficiets. 3. Noalteratig sums Propositio 3. For every iteger m 1, if p is a prime i the iterval 2a( + 1) 1 < p < = + 1 m 2ma 1 m m m(2ma 1) the p f,2a. I particular, f,2a is divisible by all primes p for which 2a( + 1) 1 < p < = a 1 2a 1. The followig lemma will eable us to covert iformatio about divisors of f,a which are greater tha ito iformatio about divisors less tha. Lemma 4. Let = s s be the expasio of i base p (ad similarly for = s s ). The f,a (mod p). i=0 f i,a P r o o f. By Lucas Theorem (see for example Graville [6]), ( ) ( ) i (mod p) i=0 i where as usual, ( i ) i 0 (mod p) if i > i. Hece all the terms i the sum over for which i > i for some i disappear, givig as claimed. f,a = ( i i=0 i =0 ) a ( i i s s 1 s =0 s 1 =0... ) a (mod p) 0 0 =0 i=0 ( i i f i,a (mod p) i=0 ) a (mod p)
3 Sums of powers of biomial coefficiets 19 Corollary 5. If l < p ad p f l,a the p f l+jp,a for all positive itegers j. We are ow i a positio to prove Propositio 3. We proceed i two stages: first, the case whe < p. Lemma 6. Let p be a prime i the iterval < p < (2a(+1) 1)/(2a 1). The p f,2a. P r o o f. Let p = + r where r > 0. The we have ( ) 2a p r ( ) 2a p r f,2a = (mod p) p r ( ) 2a r + 1 ( 1) 2a (mod p) p r ( ) 2a r + 1 (mod p) p r ( ) 2a ( + 1)( + 2)... ( + r 1) (mod p). (r 1)! If we write x (0) = 1 ad x (r) for the polyomial x(x + 1)... (x + r 1), the this last sum becomes p r ( ) ( + 2a 1)(r 1). (r 1)! We ow observe that the polyomials x (0), x (1),..., x (d) form a iteger basis for the space of all iteger polyomials of degree at most d. Hece there exist itegers c 0, c 1,..., c (r 1)(2a 1) so that Thus f,2a (( + 1) (r 1) ) 2a 1 = 1 p r (r 1)! 2a (r 1)(2a 1) j=0 (r 1)(2a 1) j=0 (r 1)(2a 1) 1 (r 1)! 2a j=0 c j ( + r) (j). c j ( + 1) (r 1) ( + r) (j) p r c j ( + 1) (r+j 1) (r 1)(2a 1) 1 (p r + 1) (r+j) (r 1)! 2a c j. r + j j=0
4 20 N. J. Cali Now, if r + (r 1)(2a 1) < p, the each of the terms i the sum is divisible by p, ad (r 1)! is ot divisible by p; hece f,2a is divisible by p. But ad r + (r 1)(2a 1) = 2ra 2a + 1 = 2pa a 2a + 1 2pa a 2a + 1 < p if ad oly if 2a( + 1) 1 p < 2a 1 completig the proof of the lemma. Now, suppose that = (m 1)p + l with l > 0 ad 2a(l + 1) 1 l < p <. 2ma 1 The, by Lemma 6, p divides f l,2a ad hece by Corollary 5, p divides f,2a. But l < p if ad oly if < mp, ad 2a(l + 1) 1 p < 2a 1 if ad oly if 2a( (m 1)p) 1 p <, 2a 1 that is, if 2a( + 1) 1 p <. 2ma 1 Thus, if 2a( + 1) 1 < p < m 2ma 1 the p divides f,2a, completig the proof of Propositio Alteratig sums. We ote that o similar result holds for the case of odd powers of biomial coefficiets (with the trivial exceptio of a = 1). Ideed, except for the power of 2 dividig f,2a+1 (which we discuss i Lemma 12), the factorizatios of sums of odd powers seem to exhibit o structure; for example, f 28,3 = However, for alteratig sums of odd powers, we have Propositio 7. p divides h,2a+1 for primes i the itervals (2a + 1)( + 1) 1 < p < = + 1 m m(2a + 1) 1 m m m(2a + 1) 1.
5 Sums of powers of biomial coefficiets 21 P r o o f. Ideed, by examiig the proof of Propositio 3, we see that if we defie ( ( )) a g,a = ( 1) so that g,2a = f,2a ad g,2a+1 = h,2a+1, the g,a is divisible by all primes i each of the itervals ( + 1)a 1 < p < m ma 1 so Propositios 3 ad 7 are really the same result. For all alteratig sums we have Propositio 8. If p ( ) the p h,a. P r o o f. Clearly 2 divides h,a if ad oly if 2 divides the middle term, ( ) a, ( as all of the other terms cacel (mod 2). Sice 2 divides ), 2 divides h,a. Now let p be a odd prime dividig ( ) ; we will show that p divides h,a. By Kummer s theorem, at least oe of the digits of writte i base p is odd (sice if all are eve, the there are o carries i computig + = i base p). Let the digits of i base p be () s, () s 1,..., () 1, () 0. The as i Lemma 4, 2 ( ) a ( () i ( ) a ) ( 1) ( 1) ()i i i=0 i =0 (sice p is odd, ( 1) = ( 1) s ). Now, sice p ( ), at least oe of the digits of i base p is odd; but the the correspodig term i the product is zero, ad so p h,a, completig the proof of Propositio 8. After computig some examples, it is atural to cojecture (ad the, of course, to prove!) Theorem The mai theorems. We will prove Theorem 1 by cosiderig  biomial coefficiets. Defiitios. Let be a positive iteger. Throughout we will deote the umber of 1 s i the biary expasio of by l() (so that 2 l() ( ) ). We further defie the followig polyomials i a idetermiate : θ () = 1 1 = (the aalogue of ), φ () = d (1 d ) µ(/d) i
6 22 N. J. Cali (the th cyclotomic polyomial i ),! = θ i () (the aalogue of!), ad (the aalogue of ( ) ). Further, defie ad r(x, ) = j x [ ] = i=1!! ( )! (1 j ) = (1 )!, s(x, ) = t(, ) = 2j+1 x s(, ) s(/2, )s(/4, )s(/8, ).... (1 2j+1 ) Note that the apparetly ifiite product i the deomiator is i fact fiite, sice s(x, ) = 1 if x < 1. We ow mae some useful observatios about t(, ). First, so t(, ) = = = s(, ) = r(, ) r(/2, 2 ), s(, ) s(/2, ) 2 s(/2, ) s(/4, ) 2 s(/4, ) s(/8, ) 2 r(, ) r(/2, ) 2 r(/2, ) r(/4, ) 2 r(/4, ) r(/8, ) 2 r(/2, 2 ) r(/4, 2 ) 2 r(/4, 2 ) r(/8, 2 ) 2 r(, ) r(/2, ) 2 r(/2, ) r(/4, ) 2 r(/2, 2 ) r(/4, 2 ) 2 r(/8, 2 ) r(/16, 2 ) 2 r(/4, ) r(/8, ) 2 r(/4, 2 ) r(/8, 2 ) 2 s(/8, ) s(/16, ) r(/8, ) r(/16, ) 2 r(/8, 2 ) r(/16, 2 ) 2 where agai, the apparetly ifiite product is i fact fiite. Now, sice r(x, ) r(x/2, ) 2 r(x, 2 ) r(x/2, 2 ) 2 { 1 if x is eve, 1 2 if x is odd,...
7 as 1, we see that Sums of powers of biomial coefficiets 23 lim t(, ) = 1 Further, t( + 1, ) has a factor 1, so ( ) 2 l. lim t( + 1, ) = 0. 1 I other words, sice 2 l ( ), we may regard t(, ) as the aalogue of the largest odd factor of ( ). Lemma 9. We have t(, ) = m φ m() where the product is over those odd m for which /m is odd. P r o o f. Clearly, if m is eve the φ m () does ot divide t(, ). Suppose m is odd; the φ m () divides s(, ) exactly /m /2 times, ad hece φ m () divides t(, ) m /2 2m /2 /2... 4m 2 j /2... m times. Now, by cosiderig the biary expasio of /m, it is immediate that this is 0 if /m is eve, ad 1 if /m is odd. Lemma 10. Let m,, be oegative itegers ad write = m +, = m +, = ( ) m + ( ), where, are the least oegative residues of, (mod m). The [ ] [ ] ( ) (mod φ m ()) where [ ] is tae to be 0 if <. P r o o f. See [3], [4] or [7]. Proof of Theorem 2. It is eough to show that if m ad = /m are odd, the [ ] a φ m () ( 1) j. But, from Lemma 10, [ ] a [ ] ( 1) j a ( ) a =0 =0 ( 1) + j (mod φ m ()) ( [ ] a )( ( ) a ) = ( 1) j ( 1) mj =0 =0 ad sice m ad are odd, the secod sum is zero, ad we are doe.
8 24 N. J. Cali We observe ow that both sides of Theorem 2 are iteger polyomials; thus whe we evaluate them at = 1, the left had side (if ozero) will divide ( the right had side. But we have already observed that t(, 1) = ) /2 l, ad hece we have proved Corollary 11. ( ) 2 ( ) a 2 l() ( 1). To prove Theorem 1 it remais to show that 2 ( ) a 2 l() ( 1). We prove a stroger result by iductio. ad Lemma 12. For all positive itegers a ad, ( ) a 2 l() 2 l() ( ) a ( 1). P r o o f. The assertio is clearly true whe = 1. Assume ow that it holds for all values less tha. For each 1 i l(), let m = 2 i ad let,,,, ( ), ( ) be defied as i Lemma 10. Writig ( [ ] a )( ( ) a ) w i () = we have [ ] a w i () (mod φ 2 i()). By our iductio hypothesis, sice l() = l( )+l( ), 2 l() w i (1) for each i. We ow wish to combie these euivaleces modulo θ 2 l()() = φ 2 ()φ 4 ()φ 8 ()... φ 2 l()() ad evaluate them at = 1. To do this, defie π 1 = 1 2 l 1 ad π i = 1 (1 ) for i = 2, 3,..., l(). 2l i+1 The, settig u i () = φ 2 ()φ 4 ()... φ 2 i 1()π i φ 2 i+1()... φ 2 l()()
9 we have Sums of powers of biomial coefficiets 25 u 1 () = 1 2 l 1 (1 + 2 )(1 + 4 )... (1 + 2 l() 1 ) 1 (mod (1 + )) ad for i 2, u i () 1 2 l i+1 (1 2 )(1 + 2 )(1 + 4 )... (1 + 2 i 2 )(1 + 2i )... (1 + 2l() 1 ) 1 i 1 (1 2 )(1 + 2i )(1 + 2i+1 )... (1 + 2l() 1 ) 2l i+1 1 (mod (1 + 2i 1 )). Further, if i j, the u i () 0 (mod φ 2 j ()). Hece, that is, [ ] a l() w i ()u i () (mod θ 2 l()()), [ ] a i=1 l() = P ()θ 2 l()() + w i ()u i () where we wish to coclude that P () is a iteger polyomial. Observe that it is sufficiet to prove that each w i ()u i () is a iteger polyomial, sice θ 2 l() is moic. To do this, cosider w i (). First, observe that w i () is divisible by 2 l( ) by our iductive hypothesis, sice < ; further, if is odd, so is, ad hece the biomial sum i w i () is symmetric ad its coefficiets are eve; if is eve, the l( ) i 1, ad i each case, 2 l i w i () (that is, each coefficiet of w i () is divisible by 2 l i+1 ). Thus, for each i, w i ()u i () is a iteger polyomial. We have thus prove that [ ] a i=1 l() = P ()θ 2 l()() + w i ()u i () where P () has iteger coefficiets. Now, settig = 1 i both sides, we observe that u i (1) is a iteger for each i, 2 l() w i (1) for each i (ideed, u i (1) = 0 for i 2, ad u 1 (1) = 1), ad that θ 2 l()(1) = 2 l(). Hece each term o the right is divisible by 2 l(), provig that ( ) a 2 l(). i=1
10 26 N. J. Cali To prove that 2 l() we proceed similarly, settig ( [ v i () = ] a ( ) a ( 1) )( ( ) ( 1) a ), with the oly major differece beig i the proof that l() i=1 v i()u i () is a iteger polyomial: i this case, if is eve, thigs wor as above, ad if is odd, the we have is odd, ad v i () is idetically eual to 0. Note that we eed to have already prove the lemma for oalteratig sums to prove the alteratig case. This completes the proof of Lemma 12 ad thus of Theorem 1. We gratefully acowledge may iformative discussios with Professors Joatha M. Borwei, Ira Gessel, Adrew J. Graville ad Herbert S. Wilf. Refereces [1] N. G. de Bruij, Asymptotic Methods i Aalysis, Dover, New Yor, [2] N. J. Cali, A curious biomial idetity, Discrete Math. 131 (1994), [3] M.D. Choi, G. A. Elliott ad N. Yui, Gauss polyomials ad the rotatio algebra, Ivet. Math. 99 (1990), [4] J. Désarméie, U aalogue des cogrueces de Kummer pour les ombres d Euler, Europea J. Combi. 3 (1982), [5] A. J. Graville, Zaphod Beeblebrox s brai ad the fiftyith row of Pascal s triagle, Amer. Math. Mothly 99 (1992), [6], The arithmetic properties of biomial coefficiets, i: Proceedigs of the Orgaic Mathematics Worshop, 1996, idex.html (URL verified September 10, 1997). [7] G. Olive, Geeralized powers, Amer. Math. Mothly 72 (1965), [8] M. Petovše, H. S. Wilf ad D. Zeilberger, A = B, A. K. Peters, Wellesley, Mass., Departmet of Mathematical Scieces Clemso Uiversity Clemso, South Carolia U.S.A. Received o ad i revised form o (3203)
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