A185424 Numerators of generalized Bernoulli numbers associated with the zigzag numbers A000111.

1, -1, 1, -1, 19, -5, 253, -61, 3319, -1385, 222557, -50521, 422152729, -2702765, 59833795, -199360981, 439264083023, -19391512145, 76632373664299, -2404879675441, 4432283799315809, -370371188237525

Offset

0,5

Comments

DEFINITION

Let E(t) = sec(t)+tan(t) denote the generating function for the zigzag numbers A000111. The zigzag Bernoulli numbers, denoted ZB(n), are defined by means of the generating function

(1)... log E(t)/(E(t)-1) = sum {n = 0..inf} ZB(n)*t^n/n!.

Notice that if we were to take E(t) equal to exp(t) then (1) would be the defining function for the classical Bernoulli numbers B_n. The first few even-indexed values of ZB(n) are

....n..|..0...2.....4.......6........8..........10...........12....

===================================================================

.ZB(n).|..1..1/6..19/30..253/42..3319/30..222557/66..422152729/2730

while the odd-indexed values begin

....n..|. ..1......3......5.......7........9.........11..

=========================================================

.ZB(n).|. -1/2...-1/2...-5/2...-61/2...-1385/2...-50521/2

The present sequence gives the numerators of the zigzag Bernoulli numbers. It is not difficult to show that the odd-indexed value ZB(2*n+1) equals -1/2*A000364(n). The numerators of the even-indexed values ZB(2*n) are shown separately in A185425.

VON STAUDT-CLAUSEN THEOREM

The following analog of the von Staudt-Clausen theorem holds:

(2)... ZB(2*n) + 1/2 + S(1) + (-1)^(n+1)*S(3) equals an integer, where

... S(1) = sum {prime p, p = 1 (mod 4), p-1|2*n} 1/p,

... S(3) = sum {prime p, p = 3 (mod 4), p-1|2*n} 1/p.

For example,

(3)... ZB(12) + 1/2 + (1/5+1/13) - (1/3+1/7) = 154635.

Further examples are given below.

Links

Table of n, a(n) for n=0..21.

Formula

SEQUENCE ENTRIES

a(n) = numerator of the rational number ZB(n) where

(1)... ZB(n) = (-1)^(n*(n-1)/2)*sum {k = 0..n} binomial(n,k)/(k+1)* Bernoulli(n- k)*Euler(k).

For odd indices this simplifies to

(2)... ZB(2*n+1) = (-1)^n*Euler(2*n)/2, where Euler(2*n) = A028296(n).

For even indices we have

(3)... ZB(2*n) = (-1)^n*sum {k = 0..n} binomial(2*n,2*k)/(2*k+1)* Bernoulli(2*n- 2*k)*Euler(2*k).

GENERATING FUNCTION

E.g.f:

(4)... log(sec(t)+tan(t))/(sec(t)+tan(t)-1) =

1 -1/2*t +1/6*t^2/2! -1/2*t^3/3! + ....

RELATION WITH ZIGZAG POLYNOMIALS OF A147309

The classical Bernoulli numbers B_n are given by the double sum

(5)... B_n = sum {k=0..n} sum {j=0..k} (-1)^j*binomial(k,j)*j^n/(k+1).

The corresponding formula for the zigzag Bernoulli numbers is

(6)... ZB(n) = sum {k=0..n} sum {j=0..k}(-1)^j*binomial(k,j)*Z(n,j)/(k+1), where Z(n,x) is a zigzag polynomial as defined in A147309. Umbrally, we can express this as

(7)... ZB(n) = Z(n,B), where on the lhs the understanding is that in the expansion of the zigzag polynomial Z(n,x) a term such as c_k*x^k is to be replaced with c_k*B_k. For example, Z(6,x) = 40*x^2+20*x^4+x^6 and so ZB(6) = 40*B_2+20*B_4+B_6 = 40*(1/6)+20*(-1/30)+(1/42) = 253/42.

Example

Examples of von Staudt and Clausen's theorem for ZB(2*n):
ZB(2) = 1/6 = 1 - 1/2 - 1/3;
ZB(4) = 19/30 = 1 - 1/2 + 1/3 - 1/5;
ZB(6) = 253/42 = 7 - 1/2 - 1/3 - 1/7;
ZB(8) = 3319/30 = 111 - 1/2 + 1/3 - 1/5;
ZB(10) = 222557/66 = 3373 - 1/2 - 1/3 - 1/11.

Maple

#A185424
a:= n-> numer((-1)^(n*(n-1)/2)*add(binomial(n, k)/(k+1)* bernoulli(n-k) *euler(k), k = 0..n)):
seq(a(n), n = 0..20);

Mathematica

Numerator[ Range[0, 30]! CoefficientList[ Series[Log(Sec[x]+Tan[x])/(Sec[x] +Tan[x] - 1), {x, 0, 30}], x]]

Crossrefs

Cf. A000111, A027641, A027642, A147309, A185425.

Sequence of denominators is A141056.

Sequence in context: A040348 A128050 A365815 * A040347 A040346 A364786

Adjacent sequences: A185421 A185422 A185423 * A185425 A185426 A185427

Keyword

easy,sign,frac

Author

Peter Bala, Feb 18 2011

Status

approved