Sums of Products of Bernoulli Numbers*

Transcription

1 journal of number theory 60, 234 (996) article no. 00 Sums of Products of Bernoulli Numbers* Karl Dilcher Department of Mathematics, Statistics, and Computing Science, Dalhousie University, Halifax, Nova Scotia, Canada, B3H 3J5 Communicated by W. Sinnott Received September, 994; revised November 6, 995 Closed expressions are obtained for sums of products of Bernoulli numbers of the 2n form ( 2j,..., 2j N ) B 2j }}}B 2jN, where the summation is extended over all nonnegative integers j,..., j N with j + j 2 +}}}+j N =n. Corresponding results are derived for Bernoulli polynomials, and for Euler numbers and polynomials. As easy corollaries we obtain formulas for sums of products of the Riemann zeta function at even integers and of other related infinite series. 996 Academic Press, Inc.. INTRODUCTION The Bernoulli numbers B n are defined by the generating function t e t & = n=0 t n B n, t <2?. (.) n! A well-nown relation among the Bernoulli numbers is (for n2) n& j=\ 2n 2j+ B 2jB 2n&2j =&(2n+) B 2n. (.2) This was found by many authors, including Euler; for references, see, e.g., [3]. Sitaramachandrarao and Davis [3] generalized (.2) to sums of products of 3 and 4 Bernoulli numbers \ 2n 2a, 2b,2c+ B 2aB 2b B 2c =(n+)(2n+) B 2n +n \n& 2+ B 2n&2, (.3) * Research supported in part by the Natural Sciences and Engineering Research Council of Canada X Copyright 996 by Academic Press, Inc. All rights of reproduction in any form reserved.

2 24 KARL DILCHER \ 2n 2a, 2b,2c,2d+ B 2aB 2b B 2c B 2d =& \2n B 2n& 4 3 n2 (2n&) B 2n&2, (.4) 2n where ( ) and ( 2n 2a, 2b,2c 2a, 2b,2c,2d ) are multinomial coefficients. The sum in (.3) ranges over all positive integers a, b, c with a+b+c=n (n3), and the sum in (.4) ranges over all positive integers a,..., d with a+ }}}+d=n (n4). The identities (.2)(.4) can be written in terms of the Riemann zeta function, via Euler's formula Thus, (.2) can be written as `(2n)=(&) n& (2?)2n B 2n. (.5) 2(2n)! n& j= `(2j) `(2n&2j)=(n+ 2 ) `(2n), n2, (.6) and similarly for (.3) and (.4). Sitaramachandrarao and Davis remared in [3] that it may be of interest to find formulas of the type (.2)(.4), (.6) for sums of products of N5 Bernoulli number, resp. zeta function factors. This was achieved by Sanarayanan [2] for N=5 and by Zhang [6] for N7. Before [6] appeared, Ramachandra and Sanarayanan [] proved that with we have Y(z) = n= Y N = (2z) 2n (&) n& B 2n (2n)! A j (z) z j d j Y dz j, (.7) where A j (z) are polynomials in z of degree at most () N, with rational coefficients. (In fact, a somewhat more general result was obtained). It is easy to see that (.7) will lead to expressions of the type (.2)(.4). It was remared in [] that the proof gives an algorithm for determining the A j (z); however, they are not explicitly given. It is one purpose of this paper to obtain explicit expressions, of the ind given in [2], [3], and [6], and valid for all N2 and all n (not

3 BERNOULLI NUMBERS 25 just for nn ). We will do this in Section 2. In Section 3 we prove corresponding results for the Bernoulli polynomials B n (x), and in Section 4 for the Euler polynomials E n (x). In both cases we obtain as easy consequences formulas for sums related to the Riemann zeta function. Section 5 contains some remars on related sums that occur in the literature. 2. BERNOULLI NUMBERS Before stating our first result, the generalization of (.2)(.4), we introduce some notation. We define the sequence b (N ) of rational numbers recursively by b () 0 = and b (N+) =& N b(n) + 4 b() &, (2.) with b (N ) =0 for <0 and for >[()2]. (Here and in what follows [x] denotes the greatest integer not exceceeding x). Furthermore, let f(t) = e t &, and f () (t) denote, as usual, the th derivative of f(t), with the convention that f (0) (t)=f(t). Lemma. For N we have \ 2 e t N N + & e &+ 2+ t \ [()2] = =0 b (N) f (N&2&) (t). (2.2) Proof. This is by induction on N. For N= we have e t + 2 e t & & 2 = e t & =f(t), which agrees with the right-hand side of (2.2). For N=2, the left-hand side of (2.2) becomes 4\ e2t +2e t + e 2t &2e t + & + = 4 4e t (e t &) 2=&f$(t), which again agrees with the right-hand side of (2.2).

4 26 KARL DILCHER For the induction step we first note that we have the following identity (for N) which is easy to verify \ e t N+ N+ + & 2 e &+ 2+ t \ =& d N dt{\ 2 + 4{\ 2 e t N N + & e &+ 2+ t \ e t N N +) & e &+ 2+ = t \. Assuming (2.2) is true for all exponents up to N, we thus obtain \ 2 e t + & e &+ 2+ t \ =& d Ndt [()2] =& [N2] = =0 { =0 [()2] =0 N b(n) & N b(n) [((N+)&)2] = =0 b (N) f (N&2&) (t)+ 4 [N2] f (N&2) (t)+ = + 4 b() & = f (N&2) (t) b (N+) f (N+&2&) (t), [(N&2)2] =0 4 b() & = b () f (N&2&2) (t) f (N&2) (t) where the last step follows from (2.). This completes the proof. Let N be given. Our aim is to evaluate the sum S N (n) = \ 2n 2j,2j 2,..., 2j N+ B 2j B 2j2 }}}B 2jN, (2.3) where the sum is taen over all nonnegative integers j,..., j N such that j +}}}+j N =n, and where is the multinomial coefficient. 2n \ 2j,..., 2j N+ = (2n)! (2j )!}}}(2j N )! K Theorem. For 2n>N we have S N (n)= (2n)! (2n&N)! [()2] =0 b (N) B 2n&2 2n&2. (2.4)

5 BERNOULLI NUMBERS 27 Before proving the theorem, we use it to write down some special cases. First we tabulate, by way of the recursion (2.), a few values of b (N ) N & & 6 & & 20 & 2 & With this table and (2.4) we obtain for N= the trivial S (n)=b 2n, and further S 2 (n)=&(2n&) B 2n ; (2.5) S 3 (n)= 2 (2n&)(2n&2) B 2n+ 2 n(2n&) B 2n&2; (2.6) S 4 (n)=& 6 (2n&)(2n&2)(2n&3) B 2n& 3 2n(2n&)(2n&3) B 2n&2; (2.7) these are equivalent to (.2)(.4) if we tae into account the slightly different ranges of summation. The first new identities are obtained for N=5 and N=6 S 5 (n)= \2n& 4 + B 2n+ 5 2 n(2n&)(2n&3)(2n&4) B 2n&2 + 8 n(2n&)(2n&2)(2n&3) B 2n&4; (2.8) S 6 (n)=& \2n& 5 + B 2n& 6 n(2n&)(2n&3)(2n&4)(2n&5) B 2n&2 & n(2n&)(2n&2)(2n&3)(2n&5) B 2n&4. (2.9) Proof of Theorem. We rewrite (.) as t e t & =& 2 t+ j= B 2j (2j)! t2j ; (2.0) hence f(t)= t & 2 + j= B 2j 2j t 2j& (2j&)!, (2.)

6 28 KARL DILCHER and therefore (N&2&)! f (N&2&) (t)=(&) t N&2 + j=[(n&2+)2] B 2j t 2j&&(N&2&) 2j (2j&N+2)!, (2.2) except when N&2&=0 in which case we have just (2.). Hence for N>2+ we have \ t 2 Now Lemma gives e t N N + = e &+ 2+ t \t t N f (N&2&) (t)=(&) (N&2&)! t 2 [()2] + =0 + b (N) j=[(n+)2] B 2j&2 2j&2 _ (N&2&)! t 2 + {(&) t 2j (2j&N)!. j=[(n+)2] B 2j&2 t 2j&2 (2j&N)!= & 2 a Nb (N) ()2 tn, (2.3) where a n =0 when N is even, and a N = when N is odd (from (2.)). On the other hand, with (2.0) we get t e t + 2 e t & = B 2j n=0 (2j)! t2j, and raising this to the Nth power gives, with (2.3), \ t 2 e t N + = e &+ t n=0 S N (n) t 2n (2n)!. (2.4) Finally, we compare coefficients of t 2n in (2.4) and (2.3), and thus obtain (2.4) for 2n>N, i.e., for n[(n+2)2]. K The proof of Theorem immediately gives expressions for S N (n) also in the case n[n2]. First we need a lemma about some special values of the coefficients b (N ).

7 BERNOULLI NUMBERS 29 Lemma 2. (a) (b) b (N) For N we have 0 =(&) ()!; b (N) ()2 =2&N+ for N odd. Proof. (a) By definition we have b () 0 = and b(n+) 0 =&(N ) b (N ) 0 ; the result now follows immediately by induction. (b) In this case (2.) gives b () 0 = and b(n+2) (N+)2 = 4 b(n) ()2 Theorem 2.. The result follows again by induction. K (a) (b) S 2n (n)= (2n)! n& (2n)! + 4 n 2n&2 b(2n) B 2n&2 ; =0 S N (n)=(&) (N&2n&)! (2n)! b (N) n, 0n _ 2 &, and in particular (c) S 2n+ (n)=(2n)! 4 &n. Proof. Parts (a) and (b) follow directly from (2.3) and (2.4), by comparing coefficients of t N. Note that for odd N, the two terms involving t N in (2.3) cancel each other, by Lemma 2(b). Part (c) follows from (b), with Lemma 2(b). K Although the recurrence relation (2.) provides a convenient way of determining the coefficients b (N ) occurring in Theorems and 2, it might be of interest of express the b (N ) in closed form in terms of nown functions. This can be done by first defining polynomials which have the b (N ) as coefficients. It is then easy to establish a three-term recurrence between these polynomials, which in turn gives rise to a generating function. Standard methods concerning generating functions are then used to obtain the explicit expression. Here and in the remainder of the paper we use the well-nown Stirling numbers of the first ind s(n, ). For definition an properties, see, e.g., [2, p. 22ff.]; a table can be found in [2, p. 30]. Lemma 3. b (N ) For N and 0[()2] we have 2 =(&) N i=0\ i + s(n&i, N&2) (N&i)! \ i. (2.5) 2+ Proof. While the procedure described above was used to obtain this expression, it now suffices to verify that the right-hand side of (2.5)

8 30 KARL DILCHER satisfies the recursion (2.). First we note that (2.5) gives b () 0 =s(, )=. Also from (2.5) we get, after changing the order of summation, 4 b() 2 & =(&) N i=2 2 (N&i)!\ i&2+ _s(n+&i, N+&2). (2.6) Using the ``triangular'' recurrence relation for the s(n, ) (see, e.g., [2, p. 24]), we obtain & 2 N b(n ) =(&) N i=0 Now we note that 2 (N&i)!\ i + [s(n+&i, N+&2) +(N&i) s(n&i, N+&2)] 2 =(&) { N 2 (N&i)!\ s(n+&i, N+&2) i + i= i= 2}2 (N&i)!\ i&+ s(n+&i, N+&2) =. (2.7) \ i + +2 \ i&+ + \ i&2+ = \ N i + + \ i&+ N = \N+ i + = N+&i\ N&i+ N. Hence with (2.6), (2.7) and (2.5) we obtain (2.), which was to be shown. (Note that the terms for i=0, in (2.6) and for i=0, 2+ in the last sum in (2.7) all vanish). K 3. BERNOULLI POLYNOMIALS The Bernoulli polynomials B n (x) can be defined by the generating function te xt e t & = n=0 B n (x) tn, t <2?. (3.) n!

9 BERNOULLI NUMBERS 3 These polynomials satisfy a large number of identities (see e.g., [6, Ch. 50] or [9]), including the sum n =0\ n + B (x) B n& (y) =n(x+y&) B n& (x+y)&(n&) B n (x+y). (3.2) In this section we will generalize (3.2) to sums of products of N2 Bernoulli polynomials, and derive some consequences. In analogy to (2.3) we denote S N (n; x, x 2,..., x N ) = \ n j,..., j N+ B j (x ) B j2 (x 2 )}}}B jn (x N ), (3.3) where the sum is taen over all nonnegative integers j,..., j N such that j +}}}+j N =n. It is possible to proceed just as in Section 2. However, here we can mae use of the well-developed theory of higher-order Bernoulli polynomials B (N ) n ( y) defined by the generating function see, e.g., [0, p. 45 ff.]. Lemma 4. t N e yt (e t &) N= n=0 B (N) n (y) tn, t =2?; (3.4) n! Let y =x +x 2 +}}}+x n.then for nn we have S N (n; x,..., x N )=(&) N N+ \n _B (N) j (&) j\ j + (y) B n&j(y). (3.5) n&j Proof. In (3.) we replace x by x, x 2,..., x n and multiply these N expressions together. With (3.3) and (3.4) we immediately obtain S N (n; x,..., x N )=B (N) n (y)(x +x 2 +}}}+x N ). Now the right-hand side of (3.5) comes directly from Equation (87) in [0, p. 48]. K

10 32 KARL DILCHER Theorem 3. Let y=x +}}}+x n.then for nn we have S N (n; x,..., x N )=(&) N N+ \n _ { j =0\ N&j&+ (&) j + s(n, N&j+) y= B n&j(y). n&j (3.6) Proof. From Equation (52.2.2) in [6, p. 350] we obtain, after appropriate substitutions, \ j + B(N) j j (y)= Now (3.6) follows directly from (3.5). =0\ N&j&+ + s(n, N&j+) y. (3.7) Examples. (i) N=2 With s(2, 2)=, s(2, )=&, (3.6) immediately gives (3.2). (ii) N=3 With s(3, )=2, s(3, 2)=&3, and s(3, 3)= we obtain for n3 n! i! j!! B i(x) B j (y) B (z) = n(n&) 2 2 _\ x+ y+z&3 & 2+ 4& B n&2(x+y+z)&n(n&2) K _ 2+ \x+y+z&3 B n&(x+y+z)+ (n&)(n&2) B n (x+y+z), 2 (3.8) where the sum is taen over all nonnegative integers, i, j, with i+ j+=n. Corollary. If x +}}}+x N =0 then we have for nn S N (n; x,..., x N )=N N+ \n (&) &j s(n, N&j) B n&j n&j. (3.9) This follows immediately from (3.6) with y=0; note that B j (0)=B j.in particular, with x =x 2 =}}}=x N =0 the right-hand side of (3.9) gives an expression for \ n j,..., j N+ B j }}}B jn.

11 BERNOULLI NUMBERS 33 Compare with (2.4); the difference lies in the fact that the above sum includes B =& 2. Remar. For the special case of Bernoulli numbers, i.e., when x =}}}=x N =y=0, the identity (3.9) (or (3.6)) was proved by Vandiver [5, Eq. (40)]. The equation (3.5) in this particular case occurs in [5] as identity (42). Apart from formulas for `(2n), such as (.6) and its generalizations, Sitaramachandrarao and Davis [3] also deal with formulas involving the alternating sums (in the notation of []) For instance, they show that n& j= '(n) = = (&) &. (3.0) n '(2j) '(2n&2j)=(n& 2) `(2n)&'(2n), (3.) and a formula for a sum of products of 3 factors. Using Theorem 3, we can easily generalize this to any number of factors. Theorem 4. For 2nN we have (&)N+n& (2?) 2n '(2j )}}}'(2j N )= ()! (2n&N)! 2 N [()2] _ { 2j =0\ N&2j&+ + s(n, N&2j+) \ N 2+ = _ B 2n&2j(N2), (3.2) 2n&2j with the convention '(0)= 2, and the sum on the left-hand side taen over all nonnegative integers j,..., j N with j +}}}+j N =n. Proof. From the definition (3.0) it is clear that for n> we have '(n)=(&2 &n ) `(n), (3.3) and with Euler's formula (.5) and the well-nown identity B n ( 2)=&(&2 &n ) B n (see, e.g., [, p. 805]) we obtain '(2n)=(&) n (2?)2n 2(2n)! B 2n\ 2+. (3.4)

12 34 KARL DILCHER Hence we use Theorem 3 (or Lemma 4) with x =}}}=x N = 2, so that y=n2. With (3.4) we have B (N) j \ N 2+ t j j! = tn e Nt2 (e t &) N= t N (e t2 &e &t2 ) N, and from this it is clear that the generating function is an even function; hence we have B (N ) j (N2)=0 for odd j. Now (3.2) follows directly from (3.5) and (3.6). K Remar. The right-hand side of (3.2) can be written again in terms of '(2n) or`(2n). Indeed, the well-nown difference equation B n (x+)&b n (x) =nx n& (see e.g., [, p. 804]) implies for N even, and B 2n&2j\ N (N2)& 2+ =(2n&2j) B 2n&2j\ N [N2]& 2+ =(2n&2j) =0 \ for N odd. Now use (.5), resp. (3.4) = 2n&2j& +B 2n&2j 2n&2j& + +B 2+ 2n&2j\ 2+ Example. (.5) give N=2. Note that s(2, 2)= and B 2n ()=B 2n. Then (3.2) and n '(2j) '(2n&2j)=(n& 2 ) `(2n). This is equivalent to (3.) (via '(0)= 2 ). 4. EULER NUMBERS AND POLYNOMIALS In complete analogy to the method of Section 3 one can obtain results concerning the Euler polynomials. They are defined by the generating function 2 e t + ext = n=0 E n (x) tn, t <? (4.) n!

13 BERNOULLI NUMBERS 35 (see, e.g., [, p. 804]). Again, many identities are nown; see, e.g., [9] or [6, Ch. 5]. For instance, we have in analogy to (3.2), n =0\ n + E (x) E n& (y)=2(&x&y) E n (x+y)+2e n+ (x+y). (4.2) In analogy to (3.3) and (2.3) we denote T N (n; x,..., x N )= \ n j,..., j N+ E j (x )}}}E jn (x N ), (4.3) where the sum is again taen over all nonnegative integers j,..., j N such that j +}}}+j N =n. We use the nth order Euler polynomials defined by \ 2 N e ++ t e yt = n=0 E (N) n (y) tn, t <?, (4.4) n! (see, e.g., [9, p. 43 ff.]). The following lemma and theorem are proved in complete analogy to Lemma 4 and Theorem 3. Here Equation (88) in [0, p. 48] is used. Lemma 5. Let y=x +}}}+x N.Then for nn we have T N (n; x,..., x N )= 2 ()! (&) j\ j + B(N) j (y) E n+&j (y). (4.5) Theorem 5. Let y=x +}}}+x N.Then for nn we have T N (n; x,..., x N )= 2 ()! (&) j _ { j =0\ N&j&+ s(n, N&j+) + y= _E n+&j (y). (4.6) The Euler numbers are defined by the generating function 2 e t +e &t= n=0 It follows from a comparison with (4.) that t n E n, t <?2. (4.7) n! E n ( 2 )=2&n E n. (4.8)

14 36 KARL DILCHER The left-hand side of (4.7) is an even function. Hence the odd-index Euler numbers are zero, and we have E 2n ( 2 )=2&2n E 2n, E 2n+ ( 2 )=0. If we now use (4.5) and (4.6) with x =}}}=x n = 2 and use the fact that (N2)=0 for odd j, we obtain the following result. B (N ) j Theorem 6. For 2nN we have \ 2n 2j,..., 2j N+ E 2j }}}E 2jN = 22n+ ()! [()2] { 2j =0\ N&2j&+ + s(n, N&2j+) \ N 2+ = _E 2n+&2j\ N 2+. (4.9) Remar. The case N=3 was recently considered by Zhang [7]. Another sum dealt with in [3] is (in the notation of [, p. 807]) It is nown that ;(n) = =0 (&) (2+) n. ;(2n+)=(&) n (?2)2n+ 2(2n)! E 2n (4.0) (see, e.g., [, p. 807]). With this and (4.9) we can easily obtain an expression for j +}}}+j N =n ;(2j +) }}} ;(2j +). (4.) The right-hand side of (4.9) can be written in terms of ;( j) or `( j), depending on the parity of N. Indeed, the difference equation E & (x+)+e & (x)=2x & gives upon repeated application E &\ N [N2] 2+ =2 = &\ (&) N & 2 & +(&) + [N2] E &\ N 2 & _ N 2&+. (4.2)

15 BERNOULLI NUMBERS 37 Now for odd N, (4.8) and (4.0) give the appropriate ;( j), while for even N we use the formula E & (0)=&2 2&+ & B &+ &+, & (4.3) (see, e.g., [, p. 805]) and Euler's formula (.5). A further obvious consequence of (4.6) is obtained when y=0. Upon changing the order of summation we get the following Corollary 2. If x +x 2 +}}}+x N =0 then we have for nn, T N (n; x,..., x N )= (&2) ()! (&) j s(n, j+) E n+j (0). (4.4) This corollary allows us to deal with yet another class of series related to the Riemann zeta function, some results for which were proved in [3]. Using the notation of [, p. 807], we define *(s)= =0 With (.5) and (4.3) we immediately obtain In [3] it was shown that i+j+=n i, j, (2+) s=(&2&s ) `(s), s>. (4.5)? 2n *(2n)=(&) n 4(2n&)! E 2n&(0). (4.6) *(2i) *(2j) *(2)= (n&)(2n&) *(2n)&?2 *(2n&2). (4.7) 4 6 We will now prove an extension to an arbitrary number of factors. Theorem 7. For mn we have *(2 )} }}} }*(2 N )= 2&N (2m&N)! [()2] (&) j? 2j (2m&2j&)! _ { 2j =0\ N s(n&, N&2j) *(2m&2j) (4.8) + 2 (&)!= where the sum on the left-hand side is taen over all positive integers,..., N, with +}}}+ N =m.

16 38 KARL DILCHER Proof. We use (4.4) with x =}}}=x N =0. To simplify notation, we set T N (n) =T N (n; 0,..., 0) (see (4.3)), and denote by T N(n) the sum T N (n) with all those summands removed for which at least one of j,..., j N is zero. Since by (4.) we have E 0 (0)=, the inclusion-exclusion principle (see, e.g., [2, p. 76 ff]) gives T N(n)=T N (n)& \N + T (n)+ \N 2+ T N&2(n)+}}} = =0 (&) \ N + T N&(n). Using (4.4) and changing the order of summation, we obtain T N(n)= =0 = (&) &j (&) j \ N + (&2)N&& (N&&)! =0 (&) N&& (&) j s(n&, j+) E n+j (0) \ N + (&2)N&& (N&&)! s(n&, j+) E n+j(0). We reverse the order of summation in the first sum and obtain T N(n)= j (&) +j =0 (&) \ N + (&2)N&& (N&&)! _s(n&, N&j) E n+&j (0). (4.9) First we consider the left-hand side of (4.9). By (4.3) and the fact that B 2n+ =0 for n it is clear that E 2n (0)=0 for n. Let j i =2 i & for i=, 2,..., N. Then n= j +}}}+j N =2( +}}}+ N )&N or equivalently 2( +}}}+ N )=n+n. Now we use (4.6) with n replaced by,..., N. Then (n+n)2 n!22n T N(n)=(&)? *(2 )} }}} }*(2 n+n N ), where the sum is taen over all positive integers,..., N with +}}}+ N =(n+n)2. If we set n+n=2m, then m (2m&N)! 4N T N(n)=(&) *(2? 2m )} }}} }*(2 N ), (4.20) where +}}}+ N =m, i (i=,..., N ).

17 BERNOULLI NUMBERS 39 To deal with the right-hand side of (4.9), we recall that E 2n (0)=0 for n; then [()2] T N(n)=(&) 2j =0 (&) \ N + (&2)N&& (N&&)! _s(n&, N&2j) E 2m&2j& (0). (4.2) Finally we use (4.6) with n replaced by m& j; then (4.20) combined with (4.2) gives (4.8). K Examples. (i) With N=2 we get immediately m& = *(2) *(2m&2)=(m& 2 ) *(2m) (see also [3, p. 80]). For N=3 we obtain (4.7) if we note that s(3, )=2, s(2, )=&, s(, )=. (ii) N=4 With s(4, 2)=, s(3, 2)=&3, s(2, 2)= we get the first apparently new formula *(2 )}}}*(2 4 )= 8\ 2m& 3 + *(2m)&?2 (2m&3) *(2m&2), 24 where +}}}+ 4 =m, i. 5. ADDITIONAL REMARKS. Theorem with Lemma 3 can also be proved with (3.9), by careful use of the inclusion-exclusion principle to tae care of the case B =& 2. This would be similar to the proof of Theorem 7, and would be somewhat shorter than the proof in Section. However, Theorem 2 cannot be obtained in this way. On the other hand, the method of Section could probably be used to prove the results in Sections 24, including analogues to Theorem Sums similar to the ones treated in this paper have occured in connection with the study of the Riemann zeta function at odd positive integer arguments. In particular, the well-nown formula (.2) occurs in [4], and the sum n 2n (&) j\ 2j+ B 2j B 2n&2j (5.)

18 40 KARL DILCHER occurs in the main result of [4]. A closed form for this alternating sum does not appear to be nown. Finally, a sum very similar to (2.3) plays a fundamental role in [3]; for each integer r>2, `(r) is shown to be an infinite series involving A 2m (r&2), where 2m A 2m \ 2i,..., 2i (r) = r+ B 2i }}}}}B 2ir [2i +][2(i +i 2 )+] } }}} }[2(i +i 2 +}}}+i r& )+], with the summation as in (2.3). In spite of the similarlity to the sum (2.3), it does not appear to be possible to treat this sum with the methods of the present paper. 3. Both Bernoulli and Euler numbers are special cases of the generalized Bernoulli numbers B n / belonging to a residue class character /. It may be of interest to have formulas of the type (.2), and possibly of type (2.4), also for generalized Bernoulli numbers. We note that the corresponding alternating sum (of type (5.)) occurs in [5], again in connection with integer values of the Riemann zeta function. 4. It is clear (via (.5)) that Theorem gives a formula for the sum j +}}}+j N =n `(2j )} }}} }`(2j N ), (5.2) with nonnegative integers j,..., j N and with the convention `(0)=& 2. The product `(2j )} }}} }`(2j N ) is obviously related to the multiple sum A( j,..., j N )= n >n 2 >}}}>n N (see [8]) and the sum (5.2) corresponds to j +}}}+j N =n n j n j 2 }}}}}n j 2 A(j,..., j N ), (5.3) with positive integers j,..., j N, and j >. It was conjetured by C. Moen that the sum (5.3) is equal to `(n) for all n and N<n; see [8], where this was verified for n6. 5. The following interesting ``inverse problem'' was discussed in [7] Given an integer and a sequence [s n ] with s n {0 for some n, is there a sequence [r ] such that n! s n = i!i 2!}}}i! r i r i2 }}}r i? i +}}}+i =n

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