SOME ARITHMETIC PROPERTIES OF SHANKS'S ... - Semantic Scholar

SOME ARITHMETIC PROPERTIES OF SHANKS'S GENERALIZED EULER AND CLASS NUMBERS E. TESKE AND H. C. WILLIAMS Abstract. Over three decades ago Daniel Shanks discovered some integers c which generalize both the Euler numbers and the class numbers of imaginary quadratic elds. Little work appears to have been done on these numbers since their discovery apart from a few scattered results of Shanks that appear in some of his papers and in his correspondence with Mohan Lal. In this paper we investigate some arithmetic properties of these numbers and generalize Shanks's results. In particular, we establish some periodic properties of c modulo m and we partially answer the question of what primes can be divisors of c for a given value of a. a;n

a;n

a;n

1. Introduction Let D be a fundamental discriminant of an imaginary quadratic eld; then D 1 (mod 4) or D 8; 12 (mod 16). If we de ne the character by (n) = D (n) = (D=n), where (D=n) is the Kronecker symbol, then the analytic class number formula gives p (1.1) L(1; ) = 2h(?a)=(w jDj) where j a = (2 ? jjD(2) (1.2) j)2 ; p h(?a) is the class number of the quadratic eld ( ?a), and w is the number of p roots of unity in ( ?a) (w = 2 if jDj > 4; w = 6 if jDj = 3; w = 4 if jDj = 4). Here Q

Q

(1.3) If we put

L(s; ) =

1 X

n=1

(n)n?s :

1 ?a X ?s La (s) = 2n + 1 (2n + 1) ; n=0

Date : October 13, 1998. 1991 Mathematics Subject Classi cation. 11B68, 11R11. Secondary 11R29, 11Y40. Research of the second author supported by NSERC of Canada grant #A7649. 1

2

E. TESKE AND H. C. WILLIAMS

where (?a=2n+1) is the Jacobi symbol, then by using the Euler product representation of L(s; ), we get (1.4) La (s) = (1 ? (2)2?s)L(s; ) : In [15] Shanks showed that if we de ne ca;n by 2n+1 p ca;n a (2n)! = La (2n + 1) (a > 1) (1.5) 2a and 1 2n+1 c1;n = L (2n + 1) (a = 1) ; 2 2 (2n)! 1

then ca;n is an integer for a 1. He proved this interesting result by rst noting that if a = m2 b, where b is squarefree, then Y ?b ?s 1 ? p p Lb (s) : La (s) = pjm Thus, if cb;n is an integer, then so is ca;n; indeed, cb;n j ca;n. This reduced the problem to showing that ca;n is an integer if a is squarefree. In the sequel we will assume that a always represents a squarefree positive integer. Under this condition Shanks derived the recurrence n X (1.6) ca;n?i(?a2 )i 2n 2i = Ca;n i=0 where 8 (aX ?1)=2 > k (a ? 4k)2n n > when a 3 (mod 4) n X ( ? 1) > : 2k +1 2k+1 1, then ( ca;0 = 3h(?a) if a 3 (mod 4) and a 6= 3 (1.7) h(?a) otherwise ; hence ca;0 2 . Furthermore, c1;n = En, the n?th Euler number. Because the values p of ca;n are integers and generalize the class number of ( ?a) in one direction and the Euler numbers in the other, Shanks called them \generalized Euler and class numbers". Z

Q

His interest in the ca;n numbers developed from his investigation of certain number theoretic constants. (See Shanks [14], p. 136.) In particular, he was concerned about how to obtain very precise values for ha , a number needed for determining the asymptotic density (under a conjecture of Hardy and Littlewood) of primes of the form x2 + a. In [13] he produced a technique for evaluating ha which required accurate values of La (s) for s = 1; 2; 3; : ::. In the case of La (2n + 1), we see from (1.5) that as soon as the value of ca;n is known, La (2n + 1) can be readily determined as precisely as needed. On the other hand, since ca;n is an integer, only

GENERALIZED EULER AND CLASS NUMBERS

3

an approximate value of La (2n + 1) is needed in order to x exactly what ca;n is. In collaboration with Mohan Lal, Shanks used his method of computing La (s) by means of Epstein zeta-functions to tabulate values of La (s) to 45 decimals. By using (1.5), Stirling's formula and the fact that La (2n + 1; ) is asymptotic to 1, it is easy to see that 4n 2n r 4n 2n+1=2 ; ca;n e a hence, Shanks deduced that c100;9 6 1048 : Thus, given the level of accuracy for which they could compute La (s), they were able to produce a table of values for ca;n for a = 1(1)100 and n = 0(1)9 (see [16], p. 279). Of course, having only 45 decimals of La (2n+1), even for the range of n and a values above, is not enough to evaluate ca;n when ca;n is large, but Shanks discovered some congruences such as ca;n+2 ca;n (mod 60) (n 1) which are useful for determining the low order digits of ca;n . In [16] and in his 1968 correspondence with Lal, currently in the possession of the second author, he mentioned several such results, all without proof. He also mentioned in [16] another of his reasons for wanting to investigate the ca;n numbers. By results in x4 of [14], it is evident that for negative D, L(s; ) is rational for all s = 0; ?1; ?2; : : :. (This had actually been proved earlier by Leopoldt [11].) Shanks believed that these numbers, which are easily expressed in terms of his ca;n , would play a very signi cant role in a proof of the Extended Riemann Hypothesis that he felt might be produced from a proof of the Riemann Hypothesis. Whether this should prove to be the case or not, however, his ca;n numbers are of intrinsic interest and certainly merit further investigation. In this paper we shall derive a simple formula for evaluating ca;n, which is quite ecient as long as a is small. We use this to produce tables of values of ca;n. We then provide proofs of generalizations of all of Shanks's congruence results concerning ca;n. The methods that we will develop will work for all squarefree a, but because c1;n is de ned slightly dierently from ca;n (a > 1), we will assume for convenience that a > 1 in what follows. 2. Preliminary Results In order to compute values of ca;n and to derive some of their arithmetic properties, it is helpful to develop a formula for ca;n. A simple (and well known) approach to this is to make use of (1.3), together with the classical result on Gauss sums D p X jDj D e2ijm=jDj : D = (2.1) m j =1 j

4


From (1.3) we have

p

DL(2n + 1; ) =

hence, by (2.1)

p

DL(2n + 1; ) =

1 D pD X m=1

m m2n+1 ;

jDj X X D 1 e2ijm=jDj

j m=1 m2n+1 : Since L(2n + 1; ) is real, we observe on equating real and imaginary parts that jDX j?1 X p D 1 sin(2jm=jDj) : jDjL(2n + 1; ) = (2.2) m2n+1 j =1 j m=1 Since 0 < j=jDj < 1, we know by classical results (see, for example, Erdelyi et al [9], p. 38) that j 1 sin(2jm=jDj) (2)2n+1 (?1)n+1 X (2.3) = B 2n+1 jDj ; m2n+1 2(2n + 1)! m=1 where B2n+1(x) is the Bernoulli polynomial 2X n+1 2n + 1 (2.4) x2n+1? B B2n+1 (x) = =0 and B denotes the ? th Bernoulli number. Hence1 jDX j?1 D B2n+1(j=jDj) ; (2.5) ca;n = (?1)n+1 Mn ()D2n 2n + 1 j =1 j where ( 2n 2n+1 ? (2)) when a ?1 (mod 4) Mn () = 2 (2 1 otherwise : Now it is well known that D D (2.6) jDj ? j = ? j j =1

and (2.7) B2n+1(1 ? x) = ?B2n+1 (x) ([9], p. 37). It follows that j bjDXj=2c D j jDX j?1 D (2.8) B2n+1 jDj = 2 j B2n+1 jDj ; j =1 j j =1 and by (2.5) we get j n+1 n ()D2n bjD Xj=2c D (2.9) B ca;n = (?1) 2n2M +1 j 2n+1 jDj : j =1

1 We will often write formulas involving both D and a. In such formulas D and a are always related by (1.2).


5

Using (2.9) in conjunction with (2.4), we were easily able to compute tables of ca;n for all squarefree values of a between 3 and 1000 and all values of n such that 0 n 120 when a < 20 and 0 n 40 otherwise. To obtain some insights concerning possible arithmetic properties of the ca;n values, we attempted to factor them. For all computations, we used the computer algebra system LiDIA [12]. Excerpts of some representative tables for a = 7; 11; 13 are presented below. The symbol Cn (Pn) denotes a composite (prime) cofactor of n digits with no prime factor below 106.

6


n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

ca;n 26 29 31 210 73 127 214 73 8831 214 11 23 73 89 701 217 173 8191 266447 218 31 151 145764975331 223 52181 131071 14859029 222 29 53 1459 23039 45599 524287 225 11 127 139 313 337 2389 23810336513 226 23 47 178481 27702422544842164391 230 31 47 601 1801 P22 230 73 17189 69151 262657 2074051907495337899 233 29 233 1103 2089 13866977 83889121 4897938881587 234 11 191 2147483647 P30 240 23 59 89 9281 9923 599479 P26 238 31 71 127 181 8191 122921 332273565017 P22 241 37 223 2399 30403 59359 616318177 P28 242 79 149 8191 121369 7059090067 227637367699153 48821335008117761009 246 11 13367 C56 246 43 53 431 9719 C55 249 23 31 73 137 151 631 23311 C53 250 29 691 2351 4513 C62 255 127 353 383 C69 254 11 103 1129 2143 11119 131071 587053 C57 257 53 83 6361 63703 69431 C68 258 23 31 89 881 3191 10711 201961 C69 262 29 541 3449 32377 524287 C75 262 193 179951 C91 265 11 1453 2203 C94 266 73 127 337 92737 649657 P89 273 31 83 113 503 8191 32083 C93 270 23 67 1097 3299 370213 C100 273 47 178481 C112 274 11 47 53 71 2063 228479 C109 278 37 439 C124 278 292 31 151 601 1801 100801 C115 281 23 89 127 C132 282 73 79 2687 3023 C131 287 11 73 2593 71119 75217 129953 262657 C119 Table 1. a = 7


n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

7

ca;n 23 33 25 32 52 17 27 33 17 43 71 29 34 53 19 4999 211 32 43 269 683 14923 213 33 52 787 2731 1183579 215 33 331 428708869630871 217 32 52 43691 1146770634498019 219 34 8297 174763 7168193 50233429 221 33 53 17 43 5419 96167 153246640574641 223 32 17 23 2659 2796203 707401237 60295211117 225 33 52 251 487 4051 3958601 44660027 127805701711 227 35 19 73 1297 87211 P28 229 32 55 59 39847 441101 1016263 3033169 6204475921023883 231 33 31 42677 1019251 35929499 715827883 6353757847606928069 233 33 52 67 179 683 20857 1788911 4902713 40607227 1761795619636066741 235 32 43 281 46919 86171 P43 237 34 52 17 37 1777 28657 25781083 4575315629 P31 239 33 17 2731 C60 241 32 53 83 C65 243 33 173 399577 935537 C59 245 34 52 19 23 331 1171 1627 49871 C59 247 32 47 283 C78 249 33 53 43 C83 251 33 307 2857 6529 43691 101411 C71 253 32 52 17 43 53 107 C88 255 35 17 79 107 683 1201 2971 C84 257 33 52 571 174763 C95 259 32 59 313 2833 37171 531793 C91 261 33 53 31 C110 263 34 19 43 313 5419 30841 C104 265 32 52 131 2731 28081 409891 C106 267 33 23 67 1823 C121 269 33 53 17 139 C127 271 32 17 71 61667 C130 273 34 52 37 1753 C135 275 33 251 331 4051 6389 C135 277 32 52 43 71 617 683 78233 C137 279 33 C157 281 36 53 19 89 163 19603 87211 135433 786211 C132 Table 2. a = 11

8


n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

ca;n

2 151 2 128657 2 7 13 1087 2683 2 61 263 63477859 2 11 17 43 1663 11119 42223 2 7 31 2609 24419 4105617731 2 1951 3881 103020283 907058147 2 17 19 31 P25 2 7 13 19 97 P28 2 11 110729 135805619 20602775603 1861737262019 2 37 36494716387 366229898594827 405993740912827 2 7 73 971 530027 289340068611211 99757182964606125701 2 17 677 402991 41952180440214788539 P22 2 29 971 65609 967787 P40 2 7 11 13 31 C56 2 17 134059 C59 2 19 C69 2 7 19 197 10463 C67 2 139 C78 2 11 1061 C81 2 72 13 17 312 89 401 244121 C73 2 61 163 443 751 C86 2 31 47 C97 2 7 17 2797 7877 C96 2 11 241 409 2473 22573 312217 C92 2 19 43 53 149 6521 C106 2 7 13 19 C118 2 29 1741 C122 2 17 37 59 40493 931003 C117 2 7 11 31 61 P132 2 71 757 5023 P135 2 17 C147 2 7 13 67 C150 2 61 202567 C152 2 11 19 71 737207 C155 2 7 19 31 281 P164 2 17 33563 C170 2 31 1489 809189 C171 2 7 132 83 283 P180 2 11 17 547 212167 C183 Table 3. a = 13


9

Notice that when a = 7; 11, we get a large power of 2 as a factor of ca;n, a power that increases with increasing n, but when a = 13 it appears that 2 j j c13;n for all n. Also, it seems that there are more factors of ca;n when a = 7; 11 than when a = 13 and that very often (but not always) a prime p divides ca;(p?1)=2. We will explain these and other phenomena concerning ca;n in the sequel. It will be convenient to consider the following exponential sums. We de ne Sm () = Tm () =

jDj X

(j)j m

j =1 bjDX j=2c j =1

(j)j m :

We can now write (2.5) as n+1 2n + 1 n+1 Mn () 2X ( ? 1) (2.10) B jDj ?1S2n+1? () ca;n = 2n + 1 =0 and (2.9) as n+1 2n + 1 n+1Mn () 2X 2( ? 1) (2.11) B jDj ?1T2n+1? () : ca;n = 2n + 1 =0 Also, it is well known that (2.12) S0 () = 0 and by classical results of Dirichlet we have T0() = w2 (2 ? (2))h(?a) ; (2.13) S1 () = w2 Dh(?a) :

(2.14) Furthermore, Sm () = hence, by (2.6) (2.15)

bjDX j=2c j =1

(j)j m +

Sm () = Tm () ?

Putting m = 1, we get

bjDX j=2c j =1

(jDj ? j j)(jDj ? j)m ;

m m X

(?1)i jDjm?i Ti () : i i=0

S1 () = 2T1 () ? jDjT0() ; therefore, by (2.13) and (2.14), we get 2T1 () = w2 (1 ? (2))jDjh(?a) : (2.16) In his correspondence with Lal, Shanks noted, but did not provide a proof, that if a 3 (mod 8) and a 6= 3, then 3 j ca;n for all n. In fact, he must have had a proof

10


of this because it is easy to modify the method of Ayoub, Chowla and Wallum [2] to prove that S3 () = D(ca;1 ? 4D2 ca;0 )=12 whenever D 5 (mod 8). (See Lehmer, Lehmer, Shanks [10], p. 442.) Suppose 3 a. Since 3 j ca;0, it is clear that2 3 j ca;1. Now by (1.6) ca;1 ? a2ca;0 = Ca;1 ; hence 3 j Ca;1. For any n 1, we get Ca;n (?1)n+1 Ca;1 0 (mod 3) : If 3 j a, then Ca;n (?1)n+1 T2 () (?1)n+1 T0() 0 (mod 3) : It follows by (1.6) that 2n n X 2 i ca;n?i(?a ) 2i 0 (mod 3) : i=0 By induction on n, it is easy to see that 3 j ca;n if a 6= 3 and a 3 (mod 8). We have now proved the following theorem. Theorem 2.1. If a 3 (mod 8), a 6= 3, then 3 j ca;n for all n 0. -

3. Properties of ca;n modulo 2k In a letter to Lal in February of 1968, Shanks mentioned that he noted \but did not formally prove" that if a 3 (mod 8) [a 6= 3], then 3 22n+1 j ca;n : This conjecture allowed him to determine the value of c11;9 (32 digits) from only a 25 digit approximation of L11 (9). We have already seen that 3 j ca;n whenever a 3 (mod 8) (a 6= 3). In this section we will prove that 22n+1 j ca;n whenever a 3 (mod 8), and we will establish several other divisibility results concerning ca;n with respect to powers of 2. By using the method of x2, we get (3.1)

p

jDjLa (2n + 1) =

jDj X X D 1 sin2(2m + 1)j=jDj j =1

(2m + 1)2n+1

j m=0

:

For any j 1, let x = j=jDj. By (2.3) 1 sin 4mx X 1 sin2(2m + 1)x (?1)n+1 (2)2n+1 B X 2n+1(x) ; + 2n+1 2n+1 = (2m) (2m + 1) 2(2n + 1)! m=1 m=0 where x < 1, and 1 sin 2m(2x) (?1)n+1 2n+1 B 1 sin 4mx X 1 X 2n+1(2x) ; = 2n+1 22n+1 2n+1 = (2m) m 2(2n + 1)! m=1 m=1 2 Indeed, if one examines values of c for n = 0; 1, one might begin to believe that c 0 c 1 for all a; but, in fact, as Shanks noted in a letter to Lal, c47 0 c47 1 . (c47 0 = 5, c47 1 = 27 32 7.) a;n

a;

;

-

;

;

;

j

a;


where 2x < 1. Since we have

sin 2m(2x ? 1) = sin 4mx ;

1 sin 4mx (?1)n+1 2n+1B X 2n+1 (2x ? 1) =

m=1 (2m)

2n+1

11

2(2n + 1)!

(1 < 2x < 2) :

It follows that 1 sin 2(2m + 1)x (?1)n+1 (2)2n+1B X 2n+1 (x) 2n+1 = (2m + 1) 2(2n + 1)! m=0 ( n+1 2n+1 B2n+1 (2x) when x < 1=2 ( ? 1) ? 2(2n + 1)! B (2x ? 1) when 1=2 < x < 1 : 2n+1 From this, (3.1) and (1.5), we get 0 jDj n +1 2 n 2 n a 2 @X D 22n+1B2n+1 j ca;n = (2 (??j1)(2)j)(2n + 1) j =1 j jDj

?

bjDX j=2c

D

2j

jDj X

D

2j

1 ?1 A :

j B2n+1 jDj ? j =bjDj=2c+1 j B2n+1 jDj By (2.8), (2.6) and (2.7), this becomes (3.2) 0bjDj=2c 1 n +1 2 n 2 n +1 X j 2j A : D 2 2n+1 @ ca;n = (2(??1)j(2)aj)(2n + 1) j =1 j 2 B2n+1 jDj ? B2n+1 jDj j =1

When a ?1 (mod 4) and n 1, we can write (3.2) as 2n 2n + 1 n+1 22n+1 X ? ( ? 1) a (2B ) 22n ? 22n? T2n+1? () ; (3.3) ca;n = a(2n + 1) =0 ? ? on using (2.4) and the fact that B2n+1 = 0. Since 2n+1 =(2n + 1) = 2?n1 =, we also have 2n 2n a X ?22n+1 ? 22n? T () : B (3.4) ca;n = (?1)n+1 22n+1 +1 2n? =0 + 1 Now it is well known that 2B 1 (mod 2); hence, if n > 1, then (3.5) 2n 2n + 1 X (2B ) ?22n ? 22n? T2n+1? () ?(2n + 1)a2n2B2nT1 () (mod 8) : a =0 If n = 1, then 2n 2n + 1 X ? (3.6) a (2B ) 22n ? 22n? T2n+1? () = ?6aT2 () + 3a2 T1() : =0 If a 3 (mod 8), then (2) = ?1 and T1 () h(?a) (mod 2) by (2.16); hence, by (3.3), we see that ca;n 0 (mod 22n+1) and 22n+1 j j ca;n if h(?a) is odd.

12


Theorem 3.1. If a 3 (mod 8) and n 1, then 22n+1 j ca;n. If 2 h(?a), then 22n+1 j j ca;n; otherwise, 22n+2 j ca;n. -

If a 7 (mod 8), then (2) = 1 and T1 () = 0 by (2.16). Thus if n > 1, we nd from (3.3) and (3.5) that 22n+4 j ca;n. When n = 1 we get (3.7) ca;1 ?16a3T2 () (mod 64) from (3.5). To proceed further we need a simple lemma. Lemma 3.2. If D 6 5 (mod 8), then S3 () S1 () (mod 4). Proof. We have

jDj X D

j(j 2 ? 1) j j =1 D bjDXj=2c D D 2 2 i i = 2 2 T1 () (mod 4) : i=1

S3 () ? S1 () =

The result now follows easily from (2.16). Corollary 3.2.1. If D 1 (mod 8), then 4 j T2(). Proof. From (2.15) with m = 3, we get

S3 () = 2T3() ? 3jDjT2 () + 3D2T1 () ? jDj3T0 () ; hence, by the lemma, we get S1 () = 2T1 () ? jDjT0 () 2T3 () ? 3jDjT2() ? T1 () ? jDjT0() (mod 4) : Since T3 () T1 () 0 (mod 2), we get jDjT2() 3T1 () 0 (mod 4).

By Corollary 3.2.1 and (3.7), we get 64 j ca;1. Theorem 3.3. If a 7 (mod 8) and n 1, then 22n+4 j ca;n. We next consider the case of a 6 ?1 (mod 4). In this case we can write (3.2) as 2n 2n + 1 1)n+1 X ?1 ca;n = (? (3.8) 2n + 1 =1 B (2 ? 1)(2a) T2n+1? () : It is very easy to see that (3.9) T2n+1 () T1 () = 2ah(?a) (mod 8) ; and if a 6= 2, we get 2 j h(?a); thus 4 j T2n+1() (n 0). It follows from (3.8) that 1)n+1 ? 2n + 1 T () + 2n + 1 aT () (mod 16) : (3.10) ca;n (? 2n?1 2n + 1 2 2n 2 We note further that T () when 2 n T2n () T2 () when 2 j n (mod 16) : 0 -


Thus, if 2 j n we get and by (3.10),

13

T2n()=2 h(?a) (mod 8) ; ca;n h(?a) (mod 8) :

If 2 n, we must investigate T2 () (mod 16). It is a simple matter to verify that for any odd j, we have j 2 5 ? 4 2j (mod 16) ; hence, jDX j=2 2D (mod 16) : T2 () 5T0() ? 4 j =1 j If 8 D, then a 1 (mod 4) and 2D = ?8a is a fundamental discriminant. By (2.13) we get jDj X 2D = 2h(?2a) ; j =1 j and since (2D=(jDj ? j)) = (2D=j), we can deduce that jDX j=2 2D = h(?2a) : j =1 j If 8 j D, then D=8 must be odd and either D=8 or 4D=8 is a fundamental discriminant. We get jDX j=2 jDX j=2 2D = j=2 4(D=8) = jDX D=8 = 0 j j j =1 j j =1 j =1 by (2.12). By putting all these results in (3.10) we get the theorem below. Theorem 3.4. If a 6 ?1 (mod 4), then ca;n (?1)nh(?a) (mod 8) when 2 j a or a 1 (mod 4) and 2 j n. If a 1 (mod 4) and 2 n, then ca;n h(?a) + 2h(?2a) (mod 8) : -

-

-

4. Periodic Properties of ca;n modulo m As mentioned earlier, Shanks had observed that (4.1) ca;n+2 ca;n (mod 60) when n 1. He also mentioned in a letter to Lal that if n = ca;n+2 ? ca;n, then (4.2) n+2 n (mod 100) (n 1) ;

14


furthermore, in a subsequent letter he noted that 4ca;3 3ca;1 + ca;9 (mod 100) ; ca;3 ca;9 (mod 819) ; ca;1 ca;9 (mod 17) : The rst of these last three follows on repeated application of (4.2), but all of the others suggest that there must be some general underlying periodic properties of the ca;n modulo m. In this section we will derive some of these properties. If by Bn we denote the generalized Bernoulli numbers of Leopoldt [11] (these numbers had in fact been mentioned as early as 1890 by Berger [3] and later in 1952 by Ankeny, Artin and Chowla[1]), then as noted by Carlitz [6] nX ?1 n

jDjj ?1Bj Sn?j () : j j =0 It follows from (2.10) that for n 1 B 2n+1 ca;n = (?1)n+1 Mn () 2n + 1 : (4.3) In 1959 Carlitz showed that B2n+1 is an integer as long as D 6= ?p, where p is a prime ?1 (mod 4). He also showed that in the exceptional case, B2n+1=(2n+1) would be an integer as long as p 1 + g2n+1 , where g is any primitive root of pk (k = 1; 2; 3; : : :). However, it is a very easy matter to show that if p j j g2n+1 + 1, then p j 22n+1 ? (2); hence, by (4.2) and Carlitz's reasoning in x3 of [6], we see that ca;n must always be an integer. Thus, an alternate proof of the integrality of ca;n can easily be derived from the work of Carlitz. Bn =

-

Carlitz also proved in [6] that Bn satis es the Kummer-type congruence r Bn+1+su r X r ? s (?1) 0 (mod (pn; per )) ; s n + 1 + su s=0

where p is any prime, pe?1(p ? 1) j u and either p D or jDj 6= p. Thus, by (4.3) we see that r X (?1)s(u=2?1) rs ca;n+su=2 0 (mod (p2n; per )) (4.4) s=0 -

when 2 j u, pe?1 (p ? 1) j u and a 6 ?1 (mod 4). In fact, it can be shown that (4.4) holds for all squarefree values of a. We rst note that if a ?1 (mod 4), then (4.4) holds when p = 2 by Theorems 3.1 and 3.3. We may therefore assume that p is odd. If, for a xed value of n, we de ne k = ak Bk+1 =(k + 1) ; k = 22n+1(22n+1 ? 2k )Tk () ;


then by (3.4) we get (4.5)

2n 2n X n +1 2n? (?1) ca;n = =0

15

:

By Kummer's congruence for the Bernoulli numbers we have for any positive integer m r B r X m+1+su 0 (mod (pm ; per )) : r ? s (?1) s m + 1 + su s=0 Also, r r X r ? s (?1) am+su = am (au ? 1)r 0 (mod pr ) (r m) : s s=0 Thus, by Theorem 10 of Carlitz [4], we get r r X r r ? s (4.6) (?1) s m+su 0 (mod p ) (r m) : s=0 Since r X (?1)r?s rs m+su s=0 = 22n+1

bjDX j=2c

k=1 0 (mod pr ) ;

? (k) 22n+1km (ku ? 1)r ? (2k)m ((2k)u ? 1)r

we see by (4.6) and results in Carlitz [5] that r r X r ? s

m+su 0 (mod per ) (er m) ; (?1) s s=0 where we de ne k k X k? :

k = =0 By (4.5) we get r r X r ? s (?1)n+1+su=2ca;n+su=2 0 (mod per ) (er 2n) : (?1) s s=0 Thus, we have proved the following theorem. Theorem 4.1. For all squarefree values of a, (4.4) must hold for any prime p. De nition 4.1. For a given positive integer m, put L = lcm[pe?1(p ? 1)=2 : pe j j m] ;

where the least common multiple is evaluated over all the distinct odd primes which divide m. We de ne ( when k 1 (m) = Lk?2 2 L when k 2 ;

16


where 2k j j m.

Corollary 4.1.1. If pe j j m and re 2n for each prime power which divides m,

then

r X

(4.7)

s=0

(?1)s( (m)?1)

r

r s ca;n+s (m) 0 (mod m ) :

By Corollary 4.1.1 with r = 1 we get (4.8) ca;n+ (m) (?1) (m) ca;n (mod m) as long as e 2n for all e such that pe j j m. This is a generalization of (4.1) as (60) = 2. Note further that (819) = 6 and (17) = 8. If we put r = 2 in (4.7), we get ca;n+2 (m) ? 2(?1) (m) ca;n+ (m) + ca;n 0 (mod m2 ) as long as e n for all pe j j m. Since (10) = 2, we get ca;n+4 ? 2ca;n+2 + ca;n 0 (mod 100) for n 1; this is the same as (4.2). 5. Primes that divide ca;n . If w(pe) is the least positive integer n such that pe j ca;n, then by (4.8) we must have w(pe) < e=2 + pe?1(p ? 1)=2. If, for example, we examine Table 1, we note that neither 5, 13 nor 19 can ever be a divisor of c7;n because for each of these values of p, p c7;n for all 1 n (p ? 1)=2. The general question of just what primes can divide ca;n for some n for a prime value of a seems to be very dicult. In this section we will provide a partial answer to this question. -

We have already seen that 3 j ca;1 when a 6= 3 and a 3 (mod 8); also, Shanks mentioned to Lal that 5 j ca;2 whenever a 1 (mod 5). We will now investigate the more general question of when pe j ca;n, where n is a multiple of pe?1 (p ? 1)=2 and p is an odd prime. By (5.5) of [6] and (2.14) it is easy to see that B2n+1 2 e 2n + 1 ? w (1 ? (p))h(?a) (mod p ) when p a and pe?1(p ? 1) j 2n. Hence, by (4.3) we get ?a 2(2 ? (2)) m ( p ? 1) = 2 e (5.1) ca;mpe?1 (p?1)=2 (?1) (2 ? j(2)j)w (1 ? p )h(?a) (mod p ) for any integer m > 0. 5.1. If p a and m is a positive integer, then pe j ca;mpe?1 (p?1)=2 if Theorem ?a = 1 or if pe j h(?a). p -

-


17

Thus, if ?pa = 1 or p a and p j h(?a), we nd that p j ca;(p?1)=2. In the case of p = 5, e = 1, we see that 5 j ca;2 if (?a=5) = 1 or a 1 (mod 5). We now consider this problem when p j a. We know by a result that goes back to Legendre that if = (j ? )=(p ? 1), where is the sum of the digits of j (> 0) when written in base p, then p j j j!. Thus, since 1, we get j ? 1 (j 1). It follows that if pe?1 (p ? 1) j 2n and e 2i, then for i 1 2n e?2i 2i 0 (mod p ) ; therefore, for all e 1, 2n 2 i (?a ) 2i 0 (mod pe) : By (1.6) we get (5.2) ca;n Ca;n (mod pe) : Now if a ?1 (mod 4), we get

k

Since p

-

Ca;n = (?1)n

(aX ?1)=2

k 2n a (a ? 4k) :

k=1 = 0 when p j k and (a ? 4k)2n 1 (mod pe) when

Ca;n (?1)n By (2.13) and (5.2) we get

(aX ?1)=2 k=1

p k, we have -

?a = (?1)n T () (mod pe ) : k

0

ca;mpe?1 (p?1)=2 (?1)m(p?1)=2 2(2 ?w(2)) h(?a) (mod pe ) :

If a 6 ?1 (mod 4), then 4 j D and D D (5.3) jDj=2 + j = ? j : By this result and (2.6) we get D D jDj=2 ? j = j : Since D = ?4a, it is now easy to show that X ?a n = (?1)n 21 T0 () (mod pe ) : Ca;n (?1) 2b + 1 2b+1