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%I #26 Feb 15 2015 16:18:26
%S 1,-1,1,-1,19,-5,253,-61,3319,-1385,222557,-50521,422152729,-2702765,
%T 59833795,-199360981,439264083023,-19391512145,76632373664299,
%U -2404879675441,4432283799315809,-370371188237525
%N Numerators of generalized Bernoulli numbers associated with the zigzag numbers A000111.
%C DEFINITION
%C 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
%C (1)... log E(t)/(E(t)-1) = sum {n = 0..inf} ZB(n)*t^n/n!.
%C 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
%C ....n..|..0...2.....4.......6........8..........10...........12....
%C ===================================================================
%C .ZB(n).|..1..1/6..19/30..253/42..3319/30..222557/66..422152729/2730
%C while the odd-indexed values begin
%C ....n..|. ..1......3......5.......7........9.........11..
%C =========================================================
%C .ZB(n).|. -1/2...-1/2...-5/2...-61/2...-1385/2...-50521/2
%C 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.
%C VON STAUDT-CLAUSEN THEOREM
%C The following analog of the von Staudt-Clausen theorem holds:
%C (2)... ZB(2*n) + 1/2 + S(1) + (-1)^(n+1)*S(3) equals an integer, where
%C ... S(1) = sum {prime p, p = 1 (mod 4), p-1|2*n} 1/p,
%C ... S(3) = sum {prime p, p = 3 (mod 4), p-1|2*n} 1/p.
%C For example,
%C (3)... ZB(12) + 1/2 + (1/5+1/13) - (1/3+1/7) = 154635.
%C Further examples are given below.
%F SEQUENCE ENTRIES
%F a(n) = numerator of the rational number ZB(n) where
%F (1)... ZB(n) = (-1)^(n*(n-1)/2)*sum {k = 0..n} binomial(n,k)/(k+1)* Bernoulli(n- k)*Euler(k).
%F For odd indices this simplifies to
%F (2)... ZB(2*n+1) = (-1)^n*Euler(2*n)/2, where Euler(2*n) = A028296(n).
%F For even indices we have
%F (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).
%F GENERATING FUNCTION
%F E.g.f:
%F (4)... log(sec(t)+tan(t))/(sec(t)+tan(t)-1) =
%F 1 -1/2*t +1/6*t^2/2! -1/2*t^3/3! + ....
%F RELATION WITH ZIGZAG POLYNOMIALS OF A147309
%F The classical Bernoulli numbers B_n are given by the double sum
%F (5)... B_n = sum {k=0..n} sum {j=0..k} (-1)^j*binomial(k,j)*j^n/(k+1).
%F The corresponding formula for the zigzag Bernoulli numbers is
%F (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
%F (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.
%e Examples of von Staudt and Clausen's theorem for ZB(2*n):
%e ZB(2) = 1/6 = 1 - 1/2 - 1/3;
%e ZB(4) = 19/30 = 1 - 1/2 + 1/3 - 1/5;
%e ZB(6) = 253/42 = 7 - 1/2 - 1/3 - 1/7;
%e ZB(8) = 3319/30 = 111 - 1/2 + 1/3 - 1/5;
%e ZB(10) = 222557/66 = 3373 - 1/2 - 1/3 - 1/11.
%p #A185424
%p a:= n-> numer((-1)^(n*(n-1)/2)*add(binomial(n,k)/(k+1)* bernoulli(n-k) *euler(k), k = 0..n)):
%p seq(a(n), n = 0..20);
%t Numerator[ Range[0, 30]! CoefficientList[ Series[Log(Sec[x]+Tan[x])/(Sec[x] +Tan[x] - 1), {x, 0, 30}], x]]
%Y Cf. A000111, A027641, A027642, A147309, A185425.
%Y Sequence of denominators is A141056.
%K easy,sign,frac
%O 0,5
%A _Peter Bala_, Feb 18 2011
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