Pascal s Triangle (mod 8)

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1 Euop. J. Combinatoics (998) 9, Pascal s Tiangle (mod 8) JAMES G. HUARD, BLAIR K. SPEARMAN AND KENNETH S. WILLIAMS Lucas theoem gives a conguence fo a binomial coefficient modulo a pime. Davis and Webb (Euop. J. Combinatoics, (990), ) extended Lucas theoem to a pime powe modulus. Making use of thei esult, we count the numbe of times each esidue class occus in the nth ow of Pascal s tiangle (mod 8). Ou esults coect and extend those of Ganville (Ame. Math. Monthly, 99 (992), 38 33). c 998 Academic Pess Limited. INTRODUCTION Let n denote a nonnegative intege. The nth ow of Pascal s tiangle consists of the n + binomial coefficients ( n ) ( 0,,...,n). Fo integes t and m with 0 t < m, we denote by N n (t, m) the numbe of integes in the nth ow of Pascal s tiangle which ae conguent to t modulo m. Clealy if d is a positive intege dividing m then (m/d) j0 N n ( jd + t, m) N n (t, d), 0 t < d. (.) When d the ight-hand side of (.) is N n (0, ) n +. In 899 Glaishe [3] showed that N n (, 2) is always a powe of 2, see Theoem A. In 99 Davis and Webb [2] detemined N n (t, 4) fo t 0,, 2, 3, see Theoem B. Thei esults show that N n (, 4) is always a powe of 2 and that N n (3, 4) is eithe 0 o a powe of 2. These facts wee also obseved by Ganville [4] in 992. In addition Ganville found fo odd t that N n (t, 8) is eithe 0 o a powe of 2. Unfotunately some of Ganville s esults on the distibution of the odd binomial coefficients modulo 8 in the nth ow of Pascal s tiangle ae incoect. We label Ganville s five assetions following Figue 2 on p. 324 of [4] as (α), (β), (γ ), (δ), and (ε). (Note that in the woding descibing Figue 2 the assetion each u j 2 is not coect as the block of 0 s in (n) 2 futhemost to the ight may contain just one zeo, fo example, n 78 has (n) ) We comment on each of (α), (γ ), (δ), and (ε). (α) This assetion is false. Take n 3 so that (n) 2. Thus t 2 and thee ae no othe t j s. Hence n 3 falls unde (α). Howeve, the thid ow of Pascal s tiangle is, 3, 3, contadicting the assetion of (α). (γ ) This assetion is false. Take n 9 so that (n) Thus t, u 2, and t 2 2. Hence n 9 falls unde (γ ). Howeve, the fist half of the 9th ow of Pascal s tiangle modulo 8 is, 3, 3,, 4, 4, 4, 4, 6, 2 so that N 9 (, 8) 4, N 9 (7, 8) 0, contadicting Ganville s claim that N 9 (, 8) N 9 (7, 8). (δ) This assetion does not tell the full stoy. Take n 39 so that (n) Thus t, u 2, t 2 3, and n 39 falls unde (δ). Hee N 39 (t, 8) 4 (t, 3, 5, 7). On the othe hand if n 56 then (n) so that t, u 2, t 2 3, u 2 2, and n 56 falls unde (δ). Hee N 56 (, 8) N 56 (3, 8) 8, N 56 (5, 8) N 56 (7, 8) 0. (ε) This assetion is false. If n 3699 then (n) so that t 3, u 2, t 2 3, u 2 2, t 3 2, and n 3699 falls unde (ε). Howeve N 3699 (, 8) N 3699 (3, 8) 28, N 3699 (5, 8) N 3699 (7, 8) 0, contadicting Ganville s claim that N 3699 (, 8) N 3699 (3, 8) N 3699 (5, 8) N 3699 (7, 8) /98/ $25.00/0 ej97046 c 998 Academic Pess Limited

2 46 J. G. Huad et al. This aticle was motivated by the desie to find the coect evaluation of N n (t, 8) when t is odd (see Theoem C (thid pat)). In addition ou method enables us to detemine the value of N n (t, 8) when t is even, a poblem not consideed by Ganville, see Theoem C (fist and second pats). Ou evaluation of N n (t, 8) involves the binay epesentation of n, namely n a 0 + a 2 + a a l 2 l, whee l 0, each a i 0o,anda l unless n 0 in which case l 0 and a 0 0. Fo bevity we wite a 0 a...a l fo the binay epesentation of n. Note that ou notation is the evese of Ganville s notation [4]. On occasion it is moe convenient to conside a 0 a...a l as a sting of 0 s and s. The context will make it clea which intepetation is being used. The length of the ith block of 0 s (espectively s) in a 0 a...a l is denoted by v i (espectively s i ). We conside n to begin with a block of 0 s and to finish with a block of s. Thus, the binay epesentation of n is and v 0, s 4, v 2, s 2 2, v 3 2, s 3, v 4 2, s 4 5, v 5, s 5. Thoughout this aticle denotes an abitay intege between 0 and n inclusive. The binay epesentation of is (with additional zeos at the ight-hand end if necessay) b 0 b...b l. The exact powe of 2 dividing the binomial coefficient ( n ) is given by a special case of Kumme s theoem [5]. PROPOSITION (KUMMER [5]). Let c(n, ) denote the numbe of caies when adding the binay epesentations of and n. Then 2 c(n,) ( ) n. Conside now the addition of the binay epesentation b 0 b...b l of to that of n to obtain the binay epesentation a 0 a...a l of n. If no cay occus in the (i )th position then thee is no cay in the ith position if b i a i, wheeas thee is a cay in the ith position if b i > a i. This simple obsevation enables us to say when c(n, ) 0,o2. PROPOSITION 2. (a) c(n, ) 0 b i a i (i 0,,...,l). (b) c(n, ) and the cay occus in the f th position (0 f l ) a f a f + 0, b f b f + 0, and b i a i (i f, f + ). (c) c(n, ) 2 and the caies occu in the f th and gth positions (0 f < g l ) a f a f + 0, b f b f + 0, a g a g+ 0, b g b g+ 0, if g f +, a f a f + a f +2 0, b f b f + b f +2 0, o a f a f + a f +2 00, b f b f + b f +2 0, if g f +, and ( denotes 0 o.) b i a i (i f, f +, g, g + ). If S denotes a nonempty sting of 0 s and s, we denote by n S the numbe of occuences of S in the sting a 0 a...a l. Fo example, if n then n 0 5, n 6, n 00 2, n 0 3, n 0 2, n 3, n 000, n 00.

3 Pascal s Tiangle (mod 8) 47 Popositions and 2(a) enable us to pove Glaishe s fomulae [3]. THEOREM A(GLAISHER [3]). N n (0, 2) n + 2 n, N n (, 2) 2 n. PROOF. We have N n (, 2) 0 ( n ) (mod 2) 0 c(n,)0 b 0,...,b l 0 (a 0 + ) (a l + ) 2 n. The fomula fo N n (0, 2) now follows fom (.) with d and m 2. Similaly we can pove Davis and Webb s fomulae [2]. THEOREM B(FIRST PART, DAVIS AND WEBB [2]). N n (0, 4) n + 2 n n 0 2 n, N n (2, 4) n 0 2 n. PROOF. Appealing to Popositions and 2(b), we have N n (2, 4) 0 ( n ) 2 (mod 4) f 0 a f a f + 0 l j0 j f, f + f 0 0 c(n,) cay in f th place (a j + ) f 0 a f a f + 0 f 0 a f a f n n 0 2 n. a 0,...,a f,a f +2,...,a l b 0,...,b f,b f +2,...,b l 0 b f b f + 0 Fom (.) with m 4, d 2, and t 0, we have N n (0, 4) + N n (2, 4) N n (0, 2), fom which the value of N n (0, 4) follows. Likewise we can use Popositions and 2(c) to detemine N n (0, 8) and N n (4, 8). THEOREM C(FIRST PART). N n (0, 8) n + (n 00 + )2 n n 0 2 n 2 n 0 (n 0 + 3)2 n 3, N n (4, 8) n 00 2 n + n 0 2 n 2 + n 0 (n 0 )2 n 3.

4 48 J. G. Huad et al. PROOF. We have N n (4, 8) 0 ( n ) 4 (mod 8) f 0 a f a f + a f l 3 f 0 a f a f + 0 f 0 a f a f + a f l 3 f 0 a f a f + 0 f 0 a f a f + a f c(n,)2 a 0,...,a f,a f +3,...,a l b 0,...,b f,b f +3,...,b l 0 b f b f + b f +2 0 g f +2 a g a g+ 0 l j0 j f, f +, f +2 g f +2 a g a g f 0 a f a f + a f a 0,...,a f,a f +2,...,a g,a g+2,...,a l b 0,...,b f,b f +2,...,b g,b g+2,...,b l 0 b f b f + b g b g+ 0 (a j + ) + 2 l j0 j f, f +,g,g+ f 0 a f a f + a f f 0 a f a f + a f (a j + ) 2 n + l 3 f 0 a f a f + 0 n 0 2 n 2 + n 00 2 n + n 0(n 0 ) 2 n 2. 2 The value of N n (0, 8) follows fom (.) with m 8, d 4, and t 0. a 0,...,a f,a f +3,...,a l b 0,...,b f,b f +3,...,b l 0 b f b f + b f +2 0 l j0 j f, f +, f +2 g f +2 a g a g+ 0 (a j + ) Although Kumme s esult (Poposition ) enabled us to detemine N n (, 2), N n (2, 4), and N n (4, 8), it is clea that we need a moe pecise conguence fo ( n ) to be able to detemine N n (t, 4) fo t, 3 and N n (t, 8) fo t, 2, 3, 5, 6, 7. The equied conguences fo ( n) modulo 4 and modulo 8 ae povided by the Davis Webb conguence, which is the subject of the next section. It is undestood thoughout that an empty sum has the value 0, an empty poduct the value, and 0 n, if n 0, 0, if n. 2. THE DAVIS WEBB CONGRUENCE In ode to state the Davis Webb conguence fo ( n ) modulo 2 h, we need the binay vesion of the symbol c d defined by Davis and Webb [] fo abitay nonnegative integes c c0 c...c s and d d 0 d...d s (whee additional zeos have been included at the ight-hand of eithe c o d, if necessay, to make thei binay epesentations the same length). If c 0 c...c i < d 0 d...d i fo i 0,,...,s, weset c 2 s+. d Othewise we let u denote the lagest intege between 0 and s inclusive fo which c 0 c...c u d 0 d...d u, and set ( ) c 2 s u c0 c...c u. d d 0 d...d u

5 Pascal s Tiangle (mod 8) 49 TABLE. Values of [ c 0 ] d. 0 c 0 d TABLE 2. Values of [ c 0 c ] d 0 d. c 0 c d 0 d Thus, fo example, if c and d 39 00, we have s 5 and u 4so that ( ) ( ) The symbol c d is an extension of the odinay binomial coefficient since fo 0 d c we have u s and so c ( d c d). The odd pat of c [ d is denoted by c [ d]. The values of c d] fo s 0,, and 2 ae given in Tables 3 espectively. Fo ou puposes it is also convenient to set fo s c0 c...c s [ ] c d 0 d...d s odd pat of. d c0 c...c s d 0 d...d s The values of [ c d] fo s and 2 ae given in Tables 4 and 5 espectively. Fom Tables 5 we obtain the assetions of Lemma. LEMMA. [ ] c0 (a). d [ 0 ] c0 c (b) d 0 d [ ] c0 c (c) d 0 d [ ] c0 c (d) d 0 d (e) [ c0 c 0 d 0 d d 2 ]. (mod 4), if c 0 c, ( ) d 0+d (mod 4), if c 0 c. (mod 8), if c 0 c, ( ) d 0+d 5 d 0+d (mod 8), if c 0 c. (mod 4), if c 0 c, ( ) d 0+d (mod 4), if c 0 c.

6 50 J. G. Huad et al. TABLE 3. Values of [ c 0 c c 2 ] d 0 d d. 2 c 0 c c 2 d 0 d d TABLE 4. Values of [ c 0 c ] d 0 d. c 0 c d 0 d [ ] c0 0 (f) ( ) d 0 0 +c 0 (mod 4), ifc 0 d 0. [ ] 0 c2 (g) ( ) 0 d c 2 (mod 4). [ 2 ] c0 (h) ( ) d 0 d d d +d 2 (mod 4), if c 0 d 0. [ 2 ] c0 c (i) c 2 (mod 4), if c d 0 d d c 2 and c d. [ 2 ] 0 (j) 5 d 0 0 d d 0+d 2 (mod 8). [ 2 ] 0 (k) ( ) 0 d d d +d 2 (mod 8). [ 2 ] (l) ( ) d 0 d d d +d 2 5 d 0+d 2 (mod 8). [ 2 ] c0 c (m) c 2 (mod 8), if c d 0 d d 0 c c 0, 0,, and c i d i (i 0,, 2). 2 Let h be an intege with h 2. When l h, Davis and Webb [] have given a conguence fo ( n ) (mod p h ) fo any pime p. (Ganville s Poposition 2 in [4] is the special case of Davis and Webb s conguence when p ( n ).) When p 2 thei conguence can be expessed using Poposition in the fom: DAVIS WEBB CONGRUENCE (mod 2 h ). Fo 2 h l + ( ) [ ] n 2 c(n,) a0 a...a l h+ h 2 [ ] ai a i+...a i+h b 0 b...b h 2 b i b i+...b i+h (mod 2 h ). (2.)

7 Pascal s Tiangle (mod 8) 5 TABLE 5. Values of [ c 0 c c 2 ] d 0 d d. 2 c 0 c c 2 d 0 d d / /3 Ou next task is to make the Davis Webb conguence (mod 2 h ) explicit in cetain cases when h 2 and h 3 by means of Lemma. It is convenient to set E a i a i+ (b i + b i+ ), E 2 Fo an intege f with 0 f l we also set H f i f, f, f + a i a i+ a i a i+2 (b i + b i+ ). (b i + b i+2 ). DAVIS WEBB CONGRUENCE (mod 4). Fo l and c(n, ) 0, we have ( ) E (mod 4). PROOF. Taking h 2 and c(n, ) 0 in (2.), we obtain fo l ( ) [ ] n l a0 [ ] ai a i+ (mod 4). b i b i+ b 0 Appealing to Lemma (a)(d), we obtain l a i a i+ ( ) b i +b i+ ( ) l a i a i+ (b i + b i+ ) ( ) E (mod 4). DAVIS WEBB CONGRUENCE (mod 8). (a) Fo l 2 and c(n, ) 0, we have ( ) E 5 E 2 (mod 8), if a 0 a, ( ) E 5 b 0+b +E 2 (mod 8), if a 0 a.

8 52 J. G. Huad et al. (b) Fo l 2 and c(n, ), we have 2( ) +a f +a f +2 +H f (mod 8), whee f (0 f l ) is the position of the cay when adding the binay epesentations of and n, and a, a l+ 0. PROOF. (a) Taking h 3 and c(n, ) 0 in (2.), we obtain fo l 2 ( ) [ ] n l 2 a0 a [ ] ai a i+ a i+2 (mod 8). b 0 b b i b i+ b i+2 By Poposition 2(a) we have b i a i fo i 0,...,l. Appealing to Lemma (j)(k)(l)(m), we obtain mod 8: ( ) [ n a0 a ] l 2 b 0 b l 2 a i a i+ a i+2 [ a0 a ] l 2 b 0 b i a i a i+ a i b l 2 i +b i+2 ( ) b i++b i+2 5 b i +b i+2 a i a i+2 [ ] l 2 a0 a b 0 b [ ] a0 a 5 b 0 b a i a i+2 l 2 5 b l 2 i +b i+2 a i a i+ a i+2 0 a i+ a i+2 5 b l i +b i+2 (b i +b i+2 ) a i a i+2 i a i a i+ ( ) ( ) b i++b i+2 ( ) b i +b i+ l (b i +b i+ ) i a i a i+ ( ) b i++b i+2 If a 0 a, then, by Lemma (c), we have ( n ) ( ) E 5 E 2 (mod 8). If a 0 a then, by Lemma (c), we have ( ) b 0+b 5 b 0+b 5 E 2 ( ) E (b 0 +b ) ( ) E 5 b 0+b +E 2 (mod 8). (b) Taking h 3 and c(n, ) in (2.), we obtain fo l 2 ( ) [ ] n l 2 a0 a 2 [ ] ai a i+ a i+2 (mod 8). b 0 b b i b i+ b i+2 We let f (0 f l ) be the position of the cay so that by Poposition 2(b).

9 Pascal s Tiangle (mod 8) 53 a f a f + 0, b f b f + 0, and a k b k fo k f, f +. We have, by Lemma (h)(i)(l), [ ] l 2 a0 a b 0 b [ a0 a i f 2, f, f ] l b 0 b [ ] a0 a b 0 b ( ) l i i f, f, f + ( ) i f, f, f + a i a i+ [ ] ai a i+ a i+2 b i b i+ b i+2 l i i f, f, f + a i a i+ (b i +b i+ ) [ ] ai a i a i+ b i b i b i+ (b i +b i+ ) ( ) H f (mod 4). We now conside fou cases: (i) f 0; (ii) f ; (iii) 2 f l 2; and (iv) f l. In each case we must detemine P l 2 i f 2, f, f [ ai a i+ a i+2 b i b i+ b i+2 ] (mod 4). Case (i). f 0. In this case we have P [ 0 a2 0 b 2 ] ( ) a 2 (mod 4), by Lemma (g), so that 2( ) H f ( ) a 2 2( ) +a f +a f +2 +H f (mod 8). Case (ii). f. Hee P [ a0 0 b 0 0 ] [ 0 a3 0 b 3 ] ( ) +a 0 ( ) a 3 ( ) +a 0+a 3 (mod 4), by Lemma (f)(g), so that 2( ) H f ( ) +a 0+a 3 2( ) +a f +a f +2 +H f (mod 8). Case (iii). 2 f l 2. Hee [ ] [ a P f 2 a f 0 a f 0 b f 2 b f b f 0 ] [ 0 a f +2 0 b f +2 ] ( ) +a f +a f +2 (mod 4), by Lemma (e)(f)(g), so that 2( ) H f ( ) +a f +a f +2 2( ) +a f +a f +2 +H f (mod 8).

10 54 J. G. Huad et al. Case (iv). f l. Hee [ ] [ al 3 a P l 2 0 al 2 0 b l 3 b l 2 b l 2 0 ] ( ) +a l 2 (mod 4), by Lemma (e)(f), so that 2( ) H f ( ) +a l 2 2( ) +a f +a f +2 +H f (mod 8). Ou final task in this section is to give the mechanism wheeby we can count the numbe of integes (0 n) fo which ( n ) is in a paticula esidue class (mod 4) o (mod 8). This mechanism is povided by the next lemma. LEMMA 2. Let c 0,...,c l be integes. Then b 0,...,b l 0 ( ) l c i b i PROOF. We have b 0,...,b l 0 ( ) l c i b i 2 n, if c i 0 (mod 2) fo each i 0,,...,l with a i, 0, if c i (mod 2) fo some i (0 i l) with a i. l ( ai ) ( ) c i b i b i 0 l ( a i ) ( ) c i b i b i 0 l ( + ( ) c i ) a i l 2, if c i 0 (mod 2) fo each i with a i, a i 0, if c i (mod 2) fo some i with a i. In applying Lemma 2 in the evaluation of N n (t, 4)(t, 3) and N n (t, 8)(t, 2, 3, 5, 6, 7) a numbe of finite sums involving E and E 2 aise. These sums ae evaluated in Lemmas 3 7. LEMMA 3. b 0,...,b l 0 ( ) E 0 n 2 n. PROOF. Set S a 0,...,a l b 0,...,b l 0 ( )E. Fo j 0,,...,l let c j denote the numbe of occuences of b j in E a i a i+ (b i + b i+ ), so that S a 0,...,a l b 0,...,b l 0 ( ) l c i b i. If n 0 then c j 0 (0 j l) so S 2 n by Lemma 2. If n > 0 let u be the least intege (0 u l ) such that a u a u+. Then c u and S 0 by Lemma 2. LEMMA 4. b 0,...,b l 0 ( ) E 2 2 n, if n 0 n 0, 0, if n 0 > 0 o n > 0.

11 Pascal s Tiangle (mod 8) 55 PROOF. Set S a 0,...,a l b 0,...,b l 0 ( )E 2. Fo j 0,,...,l let c j denote the numbe of occuences of b j in E 2 a i a i+2 (b i + b i+2 ). If n 0 n 0 then c j 0 (0 j l) so, by Lemma 2, we have S 2 n.ifn 0 > 0 o n > 0 let s be the least intege such that a s a s+2 (0 s l 2). Then c s and S 0 by Lemma 2. Befoe stating the next lemma, we emind the eade that the length of the ith block of 0 s in a 0 a...a l is denoted by v i and the length of the ith block of s by s i. We conside a 0 a...a l to stat with a block of 0 s and finish with a block of s. LEMMA 5. Fo n 0, if n 0 > 0 o n > 0, ( ) E +E 2 0, if n 0 n 0 and some s i 2, b 0,...,b l 0 2 n, if n 0 n 0 and each s i o 3. PROOF. The lemma is easily checked fo l 0,, 2 so we may suppose that l 3. Let S a 0,...,a l b 0,...,b l 0 ( )E +E 2. Fo j 0,,...,l let c j denote the numbe of occuences of b j in E + E 2 a i a i+ (b i + b i+ ) + a i a i+ a i+2 0 (b i + b i+2 ) + a i a i+ a i+2 (b i + b i+2 ). Suppose fist that n 0 n 0 and some s i 2, whee i. Hence thee exists an intege u (0 u l ) such that a u a u+, a u 0ifu, and a u+2 0if u l 2. Let u be the least such intege. Then c u and S 0 by Lemma 2. Suppose next that n 0 n 0 and each s i o3. Let j (0 j l) be an intege such that a j. If j 0 and a 0, then a 2 0 and c j 0. If j 0 and a, then a 2 and c j 2. If j l and a l 0, then a l 2 0 and c j 0. If j l and a l, then a l 2 and c j 2. Now suppose j l. If a j a j a j+ 00 then c j 0. If a j a j a j+ 0 then j 2 and a j 2 a j a j a j+ 0 so that c j 2. If a j a j a j+ 0 then j l 2 and a j a j a j+ a j+2 0 so that c j 2. If a j a j a j+ then c j 2. Hence c j is even fo evey j with a j. Thus, by Lemma 2, we have S 2 n. Now suppose that n 0 > 0. Let s be the least intege such that a s a s+ a s+2 0. If s 0 then c s. If s and a s 0 then c s. If s and a 0 then c 0. If s 2, a s, and a s 2 0 then c s. If s 2, a s, and a s 2 then c s 3. Hence, by Lemma 2, we have S 0. Finally suppose that n > 0. Let w be the least intege such that a w a w+ a w+2 a w+3. Then c w+ 3 and, by Lemma 2, we have S 0. LEMMA 6. If a 0 a then b 0,...,b l 0 ( ) b 0+b +E 2 0.

12 56 J. G. Huad et al. PROOF. Set S a 0,...,a l b 0,...,b l 0 ( )b 0+b +E 2. Let c j (0 j l) denote the numbe of occuences of b j in b 0 + b + E 2 b 0 + b + a i a i+2 (b i + b i+2 ). Let k( k l) be the lagest intege such that a 0 a...a k... Then c k. Hence, by Lemma 2, S 0. LEMMA 7. If a 0 a then 0, if n 0 > 0 o n > 0, 0, if n 0 n 0 and some s i 2 with i 2, ( ) b 0+b +E +E 2 0, if n 0 n 0 and each s i o 3, b 0,...,b l 0 2 n, if n 0 n 0, s 2, and each s i o 3 with i 2. PROOF. Set S a 0,...,a l b 0,...,b l 0 ( )b 0+b +E +E 2. Fo j 0,,...,l, let c j denote the numbe of occuences of b j in b 0 + b + a i a i+ (b i + b i+ ) + a i a i+2 (b i + b i+2 ). Suppose fist that n 0 > 0. Let s (0 s l 2) be the least intege such that a s a s+ a s+2 0. As a 0 a we have s. If s then a s a 0 and c 3. If s 2 and a s 0, then a s 2 0 and s 4, so that c s. If s 2 and a s 2 a s then c s 3. If s 2 and a s 2 0, a s, then s 3 and a s 3 0, so that c s. Hence, by Lemma 2, S 0. Suppose next that n > 0. Let s (0 s l 3) be the least intege such that a s a s+ a s+2 a s+3. If s 0 then c 0 3. If s then a s 0 and c s+ 3. Hence, by Lemma 2, S 0. Now suppose that n 0 n 0 and some s i 2 with i 2, say a s a s+. As a 0 a and n 0 0 we have s 4. Clealy a s 2 a s 00. Hence c s, and, by Lemma 2, we have S 0. Next suppose that n 0 n 0 and each s i o 3. Then, as a 0 a, we must have a 2. Thus c 0 3 and, by Lemma 2, we have S 0. Finally suppose that n 0 n 0, s 2, and each s i (i 2) o 3. Clealy c 0 c 2, c i 0ifa i a i a i+ 00 (4 i l ), c l 0ifa l a l 0, and c i c i c i+ 2ifa i a i a i+ (5 i l ). Thus c i is even fo all i with a i so that, by Lemma 2, we have S 2 n. In Section 3 we use the Davis Webb conguence (mod 4) to detemine N n (, 4) and N n (3, 4), theeby epoving the fomulae due to Davis and Webb [2] (see Theoem B (second pat)). In Sections 4 and 5 we employ the Davis Webb conguence (mod 8) to detemine N n (t, 8) (t 2, 6) (see Theoem C (second pat) in Section 4) and N n (t, 8) (t, 3, 5, 7) (see Theoem C (thid pat) in Section 5). 3. EVALUATION OF N n (, 4) AND N n (3, 4) In this section we illustate ou methods by e-establishing the fomulae fo N n (, 4) and N n (3, 4) due to Davis and Webb [2].

13 THEOREM B(SECOND PART, DAVIS AND WEBB [2]). 2 n, if n 0, N n (, 4) 2 n N n (3, 4), if n > 0, Pascal s Tiangle (mod 8) 57 0, if n 0, 2 n, if n > 0. PROOF. It is easily checked that the fomulae hold fo n 0, so that we may take n 2. Thus l. Fo t and 3, we have N n (t, 4) 0 ( n ) t(mod4) b 0,...,b l 0 ( ) E t(mod4) by Poposition, Poposition 2(a), and the Davis Webb conguence (mod 4). Hence N n (t, 4) b 0,...,b l 0 E 2 (t ) (mod 2) 2 b 0,...,b l 0 2 n + a 0 2 ( ),...,a l 2 (t ) b 0,...,b l 0, ( + ( ) 2 (t )+E ) ( ) E 2 n + ( ) 2 (t ) 2 n, if n 0, 2 n, if n > 0, by Lemma EVALUATION OF N n (2, 8) AND N n (6, 8) In this section we evaluate N n (2, 8) and N n (6, 8). THEOREM C(SECOND PART). n 0 2 n n 00 2 n, if n 0, N n (2, 8) n 0 2 n 2 n 0 2 n 2 + n 00 2 n, if n, n 0 2 n 2, if n 2. n 00 2 n, if n 0, N n (6, 8) n 0 2 n 2 + n 0 2 n 2 n 00 2 n, if n, n 0 2 n 2, if n 2. PROOF. It is easily checked that the theoem holds fo n 0,, 2, 3 (equivalently l 0, ). Hence we may assume that l 2. Fo t 2 and 6 we have, by Poposition, N n (t, 8) 0 ( n ) t (mod 8) 0 c(n,) ( n ) t (mod 8) f 0 0 c(n,) cay in f th place ( n ) t (mod 8) Befoe continuing it is convenient to intoduce some notation. Let S be a sting of 0 s and s of length k. Fo 0 i i + k l we set (( )) ai a i+...a i+k, if a i a i+...a i+k S, S 0, if a i a i+...a i+k S..

14 58 J. G. Huad et al. Now, by the Davis Webb conguence (mod 8), we have t (mod 8) 4 (t + 2) + a f + a f +2 + H f 0 (mod 2). Next, let E f 2 a i a i+ (b i + b i+ ), E i f +2 a i a i+ (b i + b i+ ), so that E + E H f. Hence the Davis Webb conguence (mod 8) becomes t (mod 8) 4 (t + 2) + a f + a f +2 + E + E 0 (mod 2). Hence N n (t, 8) f 0 a f a f + 0 a 0,...,a f,a f +2,...,a l b 0,...,b f,b f +2,...,b l 0 b f b f + 0 n 0 2 n ( ) 4 (t+2) l a f +2,...,a l b f +2,...,b l 0 Now, by Lemma 3,we have ( ) E. f 0 a f a f ( + ( ) 4 (t+2)+a f +a f +2 +E +E ) ( ) a f +a f +2 a 0,...,a f b 0,...,b f 0 ( ) E a 0,...,a f b 0,...,b f 0 ( ) E 0 n 2 n, a f +2,...,a l b f +2,...,b l 0 ( ) E 0 n 2 n, whee n is the numbe of s in a 0a...a f, n is the numbe of s in a f +2...a l, n is the numbe of occuences of in a 0 a...a f, and n is the numbe of occuences of in a f +2...a l. Hence a 0,...,a f a f +2,...,a l ( ) E ( ) E 0 n +n 2 n 0 n 2 n, if a +n f +2 0, 0 n 2 n, if a f +2. b 0,...,b f 0 b f +2,...,b l 0 Thus fo n > we have N n (t, 8) n 0. Next fo n we have N n (t, 8) n ( ) 4 (t+2) n 0 ( ) 4 (t+2) f 0 a f a f + a f +2 0 f 0 a f a f + a f +2 0 ( ) a f +a f +2 2 n ( ) a f ( (( )) ) n 0 2 n 2 + ( ) 4 (t 2) a0 a a 2 + n 0 00 n 0 n 0 + ( ) 4 (t 2) (2n 00 n 0 ),

15 Pascal s Tiangle (mod 8) 59 as Hence (( a0 a a 2 0 )) n 0 n 00 n 0. N n (2, 8) n n n 00 n 0 and Finally fo n 0 we have N n (t, 8) n ( ) 4 (t+2) N n (6, 8) n 0 2 n n 00 + n 0. f 0 a f a f + a f (( n 0 2 n 2 + ( ) 4 (t+2) a0 a a (( )) al 2 a l a l + 0 f a f a f + a f ( ) a f +a f +2 2 n )) (( al 2 a + l a l 0 0 }. ( ) a f Now as n 0 we have (( )) (( )) a0 a a 2 a0 a n n 00 n 0, (( )) al 2 a l a l n n 000 n 00 n 00 n 000, (( )) al 2 a l a l n 0 0 n 00 n 0 n 0 n 00, f a f a f + a f ( ) a f f a f a f a f + a f f a f a f a f + a f )) n 000 n 00, so that (( a0 a a f a f a f + a f )) (( al 2 a + l a l 0 0 ( ) a f n 0 + 2n 00. )) (( al 2 a l a l 0 )) Hence N n (t, 8) n 0 2 n 2 + ( ) 4 (t+2) ( n 0 + 2n 00 ) n 0 2 n n 00 2 n, if t 2, n 00 2 n, if t 6.

16 60 J. G. Huad et al. 5. EVALUATION OF N n (t, 8), t, 3, 5, 7 In this section we cay out the evaluation of N n (t, 8) fo t, 3, 5, 7 using the Davis Webb conguence (mod 8). THEOREM C(THIRD PART). Case n n n 0 n v s s i (i 2) N n (, 8) N n (3, 8) N n (5, 8) N n (7, 8) No. (i) 0 (ii) n (iii) 0 2 n 0 2 n 0 (iv) 0 0 (v) n n (vi) 0 2 n n (vii) 2 some 2 (viii) 2 none 2 2 n 2 n 0 0 (ix) (x), 2 some 2 (xi), 2 none 2 2 n 2 n 0 0 (xii) 0 some 2 (xiii) none 2 2 n 2 n 0 0 PROOF. It is easily checked that the theoem holds fo n 0,, 2, 3 (equivalently l 0, ). Hence we may assume that l 2. Fo t, 3, 5, 7 we have N n (t, 8) 0 ( n ) t (mod 8) 0 c(n,)0 ( n ) t (mod 8) b 0 ((,...,b l 0 a0 ( ) E 5 (b 0 +b ) a )) +E 2 t(mod 8), by Popositions and 2(a), and pat (a) of the Davis Webb conguence (mod 8). Set so that t ( ) α(t) 5 β(t) (mod 8). Hence α(t) (t )/2. β(t) (t 2 )/8, ( ) E 5 (b 0+b ) ( ( a 0 a ) ) +E 2 t (mod 8) (( a0 a E α(t) (mod 2), (b 0 + b ) )) + E 2 β(t) (mod 2). Thus N n (t, 8) 4 b 0,...,b l 0 + ( )α(t) 4 + ( )β(t) 4 + ( )α(t)+β(t) 4 ( ( + ( ) α(t) ( ) E ) + ( ) β(t) ( ) (b 0+b ) ( ( a 0 a ) ) ) +E 2 a 0,...,a l b 0,...,b l 0 b 0,...,b l 0 a 0,...,a l ( ) E ( ) (b 0+b ) ( ( a 0 a ) ) +E 2 b 0,...,b l 0 ( ) (b 0+b ) ( ( a 0 a ) ) +E +E 2.

17 Pascal s Tiangle (mod 8) 6 We teat the two cases a 0 a and a 0 a sepaately. If a 0 a then (( a 0 a )) 0 and appealing to Lemmas 3, 4 and 5, we obtain N n (t, 8) 2 n 2 ( ) + α(t) 2 n 2 }, if n 0 ( ) + β(t) 2 n 2 }, if n 0 n 0 0, if n > 0 0, if n 0 o n > 0 ( ) α(t)+β(t) 2 n 2, if n 0 n 0 and each s i o3 + 0, if n 0 n 0 and some s i 2; o n 0 > 0; o n > 0. Appealing to the case definitions given in the statement of the theoem we obtain the value of N n (t, 8). Cases (i), (iv), (xii). N n (t, 8) Case (ii). Hee n 0 so that each s i. N n (t, 8) 2 n 2 + ( ) α(t) 2 n 2 + ( ) β(t) 2 n 2 + ( ) α(t)+β(t) 2 n, if t, 0, if t 3, 5, 7. Case (iii). N n (t, 8) + ( ) α(t) 2 n, if t, 5, 0, if t 3, 7. Cases (v), (vi). Hee n > 0, n 0 implies that some s i 2. N n (t, 8) 2 n ( ) β(t) 2 n 2 2 n, if t, 7, + 0 0, if t 3, 5. Cases (vii), (viii), (ix). Hee v 0 and s 2 o 3 so that a 0 a, contadicting a 0 a. These cases cannot occu. Case (x). Hee v 0 and s o2.asa 0 a we have s. N n (t, 8) Case (xi) (Hee v 0 and s o2:asa 0 a we have s.) and Case (xiii). N n (t, 8) 2 n ( ) α(t)+β(t) 2 n 2 2 n, if t, 3, 0, if t 5, 7. If a 0 a, appealing to Lemmas 3, 6, and 7, we obtain ( ) α(t)+β(t) 2 n 2, if n 0 n 0, s 2, N n (t, 8) 2 n 2 + each s i (i 2) o3, 0, othewise. Cases (i), (iv), (ix), (x). N n (t, 8) + 0. Cases (ii), (iii). Hee a 0 a implies n > 0 so these cases cannot occu.

18 62 J. G. Huad et al. Case (v). Hee a 0 a implies v 0 so this case cannot occu. Case (vi). Hee a 0 a implies v 0 and s 2 so this case cannot occu. Case (vii). N n (t, 8) + 0. Case (xi) (Hee a 0 a implies v 0 and s 2.) and Case (viii). N n (t, 8) 2 n 2 + ( ) α(t)+β(t) 2 n 2 2 n, if t, 3, 0, if t 5, 7. Cases (xii), (xiii). Hee v > 0 contadicting a 0 a. These cases cannot occu. REFERENCES. K. S. Davis and W. A. Webb, Lucas theoem fo pime powes, Euop. J. Combinatoics, (990), K. S. Davis and W. A. Webb, Pascal s tiangle modulo 4, Fibonacci Quat., 29 (99), J. W. L. Glaishe, On the esidue of a binomial-theoem coefficient with espect to a pime modulus, Quat. J. Math., 30 (899), A. Ganville, Zaphod Beeblebox s bain and the fifty-ninth ow of Pascal s tiangle, Ame. Math. Monthly, 99 (992), E. E. Kumme, Übe die Eganzungssätze zu den allgemeinen Recipocitätsgesetzen, J. Reine Angew. Math., 44 (852), E. Lucas, Su les conguences des nombes euléiens et des coefficients difféentiels des fonctions tigonométiques, suivant un module pemie, Bull. Soc. Math. Fance, 6 (877 8), Received 8 August 996 and accepted on 2 Mach 997 JAMES G. HUARD Depatment of Mathematics, Canisius College, Buffalo, NY 4208, USA huad@canisius.edu BLAIR K. SPEARMAN Depatment of Mathematics and Statistics, Okanagan Univesity College, Kelowna, B.C. VV V7, Canada bkspeam@okanagan.bc.ca KENNETH S. WILLIAMS Depatment of Mathematics and Statistics, Caleton Univesity, Ottawa, Ontaio KS 5B6, Canada williams@math.caleton.ca