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