Az Eszterházy Károly Tanárképző Főiskola Tudományos Közleményei. 1995-1996. Sectio Mathematicae. (Acta Academiae Paedagogicae Agriensis : Nova series ; Tom. 23)

MÁTYÁS F., Two problems related to the Bernoulli numbers

Two Problems Related to the Bernoulli Numbers 25 and from (5) by 0 and (3) a

2 = a

4 = ... + a2/_2 = 0 follows. But 2Ö2/«I = k^[{

k

k^2f)

Bk-2f í 0, that is B

k.

2

f = ^/-1-2/ = £

2/-I / 0. It implies that 2/ — 1 = 1, and so / = 1, j = 2 and k — 3. Thus we have got the only solution m = 2, k = 3, h{n) = |n

2 + \n and 63(71) = (

~n

2 -f |n)

2 In [2], using the definition i\(n, c), one can find the proof of the equality -P2{n

, 3) = n — 3+^;-In our Theorem 2. we generalize this result and the proof will be similar to that proof which was sent for the original problem by the author of this paper. Theorem 2. a) If m > 2 is an integer then there exists a polynomial f(n) such that P

2(^c) = (f(

n))

m f°

r every integer n > 2 if and only if m = 2, c = 1 db i2\[2 and f(n) = n-l± b) If m > 2 and k — 3 or 4 then the equahty Pjt(n,c) — (f(n))"

1 can not be solved by any polynomial f(n) and parameter c. Proof. One can easily verify the following equality: j — c (2n — 1\ (2n — 1\ c f 2n ^ 2n - j V j J \2n - jJ 2n \2n - j Using (6), we have 2^

2 /

2

n - 1\ c 2^ f 2n \ (?) />*(», c) = n E (

2

n _ j) B

— - 2 E U - J ^ j — k j — k By the recursive definition (1) of the Bernoulli number we can write that 2n —

2 (8) and (9) j—k £ (ÍI-ÍJ^-^-U-i

1)

5*1 2 n-j í:,V r,->-C

,.,'i* £ (2n-j)^ = -(iT-i)

0

2"1 " G-M

0*1 j — k 2n \ /2rc\ „ /2n\ „ i-2 í

w-- l i

l" UK