Binary Partitions Revisited

Transcription

1 Binary Partitions Revisited Øystein J. Rødseth and James A. Sellers Department of Mathematics University of Bergen Johs. Brunsgt. N 5008 Bergen, Norway rodseth@mi.uib.no Department of Science and Mathematics Cedarville University 5 N. Main St. Cedarville, Ohio 4534 sellersj@cedarville.edu April 3, 00 Abstract The restricted binary partition function b k (n) enumerates the number of ways to represent n as n a 0 + a + + a j with 0 a 0 a... a j < k. We study the question of how large a power of divides the difference b k ( r+ n) b k ( r n) for fixed k 3, r, and all n.

2 Introduction Let b(n) denote the number of partitions of the positive integer n into powers of. That is, b(n) is the number of ways to represent n as n a 0 + a + with a i Z and 0 a 0 a... We call b(n) the binary partition function. Churchhouse [] conjectured that () b( r+ n) b( r n) 0 (mod 3r/+ ) for r. Moreover, he conjectured that this result is exact; i.e. no higher power of divides the left hand side if n is odd. Churchhouse s conjecture was first proven in [6]. Subsequently, others produced proofs, including Gupta [3], [4], [5], and Andrews []. In [7] we proved a number of congruences for the restricted m-ary partition function with similar consequences for the ordinary (unrestricted) m-ary partition function. However, in the binary case m, these congruences reduce to mere trivialities. The object of this paper is to establish some alternative results for the restricted binary partition function b k (n), which is the number of ways to represent n as n a 0 + a + + a j with 0 a 0 a... a j < k. We also have that b k (n) equals the number of representations of n of the form n c 0 + c + c + with 0 c i < k. We now present the two theorems which we prove below. Theorem For r k we have () b k ( r+ n) b k ( r n) 0 (mod 3r/+ ).

3 Notice that for a given n, b(n) b k (n) for sufficiently large k, so that Theorem implies (). Although not exact, Theorem is best possible in the following sense: For r k 3, no higher power of divides the left hand side of () if n (mod k r ). Furthermore, for r k, we have (3) b k ( k n) b k ( k 3k/ n(n + ) n) (mod 3k/ ), k 3. If we replace n by n in (3), we get an exact result, which is the case t of the next theorem. Theorem For k 3 and t, we have b k ( k+t n) b k ( k+t n) 0 (mod 3k/+t ). Moreover, this result is exact. We prove Theorems and by considering various aspects of the generating function for b k (n). Theorem follows from Lemma below, while Theorem follows from Lemma 3. Indeed, Lemmata and 3 give somewhat stronger results than those stated in Theorems and, but the stronger results are also more complicated. Auxiliaries In the following we write π(a) for the largest integer π such that π divides the nonzero integer a. Notice that π(a) < π(c) implies π(±a ± c) π(a), π(a) π(c) implies π(±a ± c) > π(a). We regard π(0) > c for any integer c as valid. All power series in this paper will be elements of Z[[q]], the ring of formal power series in q with coefficients in Z. We define a Z-linear operator U : Z[[q]] Z[[q]] 3

4 via U n a(n)q n n a(n)q n. Notice that if f(q), g(q) Z[[q]], then (4) U(f(q)g(q )) (Uf(q))g(q). Moreover, if f(q) n a(n)qn Z[[q]], g(q) n c(n)qn Z[[q]], and M is a positive integer, then we have if and only if, for all n, In the work below we shall use the following identity for binomial coefficients: (5) f(q) g(q) (mod M) (in Z[[q]]) a(n) c(n) (mod M) (in Z). ( ) n + r r i r/ ( )( ) i n + i ( ) r i i r r i i The truth of this relation follows by expanding both sides of the identity ( q) n ( q( q)) n, and comparing the coefficient of q r on each side of the equation. We now begin to develop the machinery needed to prove our two theorems. First, let Then (6) so that h i h i (q) h i Uh r n q, i 0. ( q) i+ ( ) n + i q n, i ( ) n + r q n. r n 4

5 It follows from (5) and (6) that (7) for r 0. (8) Uh r i r/ ( ) i ( ) r i i r r i Next, we recursively define K r K r (q) by ( ) K 3 h and K i+ U q K i for i. We have the following lemma regarding K r. Lemma For i r, there exist γ r (i) Z such that h i (9) r K r γ r (i)h i+. i Moreover, 3r + i π(γ r (i)), where equality holds if and only if i or r + i is odd. Note. In the following we set γ r (i) 0 if i r. Proof. We use induction on r. The lemma is true for r thanks to (8). Suppose that the lemma is true for r replaced by r for some r 3. Then we have (0) and () r K r γ r (i)h i+, i 3(r ) + i π(γ r (i)), 5

6 where equality holds if and only if i or r + i is even (and i r ). By (0), (8), and (7), we find K r r γ r (j)uh j+ j r γ r (j) j i r min(r,i ) jmax(,i ) min(r,i) i jmax(,i ) j+ i j/ + ( ) ( ) i+j i j i j + i h i ( ) ( ) i+j i j i γ r (j)h i j + i ( ) i + ( ) i+j+ i j γ r (j)h i+. j + i Thus (9) holds with () γ r (i) min(r,i) jmax(,i ) ( ) i + ( ) i+j+ i j γ r (j), j + i so that all values γ r (i) are integers. Now we have γ r () γ r () + γ r (), where 3(r ) + 3r + π( γ r ()) +. If r is odd, then 3r + π( γ r ()), while 3(r ) + 3r + π(γ r ()) >, so that 3r + (3) π(γ r ()). If r is even, then 3r + π( γ r ()) +, 6

7 while 3(r ) + 3r + π(γ r ()), and (3) holds in this case also. Next, let i r. By (), we then have (4) γ r (i) i+ γ r (i ) i (i + )γ r (i) +, where, by (), 3(r ) + j 3r + i (5) π( ) min (i j + ) +. j i+ Now consider i. We have 3(r ) + 3r + π( 3 γ r ()) 3 +. If r is odd, then 3(r ) + π( 3r + 3γ r ()) > + >, so that, by (4) and (5), 3r + π(γ r ()). If r is even, then 3(r ) + π( 3r + 3γ r ()) +, and it follows that 3r + π(γ r ()) >. Finally, if 3 i r, then 3(r ) + i π( i 3r + i (i + )γ r (i)) i + >, so that, by (), (4), and (5), 3r + i π(γ r (i)) with equality if and only if r + i is odd. This implies the result stated in Lemma. 7

8 3 Proof of Theorem With b k (0), the generating function for b k (n) is B k (q) b k (n)q n n0 k i0 q i, k 0, where, in particular, B 0 (q). Notice that, for k, (6) Thanks to (4), we have for k, Furthermore, so that, for k 3, By (7), we now have B k (q) q B k (q ). UB k (q) (U q )B k (q) q B k (q) ( q) B k (q ) from (6) (n + )q n B k (q ). n0 U B k (q) (n + )q n B k (q), n0 U B k (q) B k (q) (n + )q n B k (q) n (h + h 0 )B k (q) (h + h )B k 3 (q ). U 3 B k (q) UB k (q) (Uh + Uh )B k 3 (q) 3 h B k 3 (q) K B k 3 (q). 8

9 Moreover, since U(K i B k i (q)) U( q K ib k i (q )) K i+ B k i (q), induction on r gives U r+ B k (q) U r B k (q) K r+ B k r (q) for r k. Thus we have (7) (b k ( r+ n) b k ( r n))q n K r+ B k r (q) n for r k. Theorem now follows from Lemma. Next we turn to the remarks following the statement of Theorem. For r, we have, by Lemma, (8) K r+ (q) 3r/+ h (q) (mod 3r/+3 ). If we now put r k in (7), (3) follows by (8) and (6). (9) Let Since B k (q) d r (n)q n h (q)b k r (q). n k i0 ( q) i we then have, for r k 3, d r (n + )q n so that n0 d r (n + )q n n0 ( q) k ( q) k r + ( q ) k r 3 + ( q) k r 3 + (mod ), (mod ), q (mod ). q k r 3 9

10 Repeated application of (4) now gives d r ( k r n + )q n n0 ( q) q (mod ), so that (0) d r ( k r n + ) (mod ), for r k 3. From (7) (0) it now follows that, for r k 3, the left hand side of () is not divisible by 3r/+3 if n (mod k r ). 4 Proof of Theorem By putting r k in (7), we see that () (b k ( k+t n) b k ( k+t n)q n U t K k (q). n With the goal of proving Theorem, we prove the following two lemmas regarding U t K r (q). We first consider the t case, UK r (q), as a basis case. Lemma For r, there exist δ r, (i) Z such that () UK r δ r, (i)h i, i where 3r + (3) π(δ r, ()), and 3r + i + (4) π(δ r, (i)) for i,..., r. Moreover, (4) holds with equality if and only if i or r + i is even. 0

11 Proof. By (9) and (7), we have UK r r γ r (j)uh j+ j r γ r (j) j i j+ i j+ ( ) min(r,i ) jmax(,i ) ( ) j+ i i j i j + i h i ( ) ( ) i+j+ i j i γ r (j)h i. j + i Thus () holds with δ r, (i) min(r,i ) jmax(,i ) ( ) ( ) i+j+ i j i γ r (j), j + i and all values δ r, (i) are integers. Moreover, δ r, () γ r (), so by Lemma, (3) holds. For i r, we have (5) δ r, (i) i γ r (i ) i iγ r (i) +, where (6) Note that (7) 3r + j π( ) min (i j + ) j i+ 3r + i r + (i ) π( i 3r + i + γ r (i )) i + with equality if and only if i or r + i is even. Furthermore, (8) 3r + i π( i 3r + i + iγ r (i)) i + π(i) + >, where we look separately at the cases i and i 3. Combining (5) (8) completes the proof of Lemma.

12 Lemma 3 For r and t, there exist δ r,t (i) Z such that (9) U t K r δ r,t (i)h i, i where 3r + (30) π(δ r,t ()) + t, and 3r + i + (3) π(δ r,t (i)) + i(t ) for i,..., r. Moreover, (3) holds with equality if and only if i or r + i is even. Proof. We use induction on t. By Lemma, Lemma 3 is true for t. Next, suppose that Lemma 3 is true for t replaced by t for some t. Using (7), we get U t K r so that (9) holds with δ r,t (i) δ r,t (j)uh j j δ r,t (j) j i j/ min(r,i) i ji min(r,i) ji j ( ) i ( ) j i i j j i h i ( ) i ( ) i+j i j δ r,t (j)h i, j i ( ) i ( ) i+j i j δ r,t (j), j i and all values δ r,t (i) are integers. Now we have δ r,t () δ r,t () δ r,t (). By the induction assumption, we have 3r + π(δ r,t ()) + + t,

13 and 3r + + 3r + π(δ r,t ()) + (t ) > + t. Thus, (30) follows. For i r, we have (3) δ r,t (i) i δ r,t (i) + 3, where 3r + j π( 3 ) min (i j j(t )), j i+ so that 3r + i + (33) π( 3 ) > + i(t ). Moreover, 3r + i π( i + (34) δ r,t (i)) + i(t ), with equality if and only if i or r + i is even. Combining (3) (34) completes the proof of Lemma 3. We are now in a position to prove Theorem. For k 3 and t, we have by Lemma 3 and (6), U t K k δ k,t ()h δ k,t () In particular, by (30) and (), nq n (mod 3k/+t ). n b k ( k+t n) b k ( k+t n) 3k/+t n (mod 3k/+t ), and the proof of Theorem is complete. References [] G. E. Andrews, The Theory of Partitions, Encyclopedia of Mathematics and its Applications, Vol., Addison-Wesley, Reading, Mass

14 [] R. F. Churchhouse, Congruence properties of the binary partition function, Proc. Camb. Phil. Soc. 66 (969), [3] H. Gupta, Proof of the Churchhouse conjecture concerning binary partitions, Proc. Camb. Phil. Soc. 70 (97), [4] H. Gupta, A simple proof of the Churchhouse conjecture concerning binary partitions, Indian J. Pure Appl. Math. 3 (97), [5] H. Gupta, A direct proof of the Churchhouse conjecture concerning binary partitions, Indian J. Math. 8 (976), 5. [6] Ø. J. Rødseth, Some arithmetical properties of m-ary partitions, Proc. Camb. Phil. Soc. 68 (970), [7] Ø. J. Rødseth and J. A. Sellers, On m-ary partition function congruences: A fresh look at a past problem, J. Number Theory 87 (00),