To get these numbers, expand the e.g.f. (3/2)/(1+exp(x)+exp(-x)), multiply coefficient of x^n by (n+1)! and take absolute values.
Or expand the e.g.f. (3/2)/(1+2*cos(x)) and multiply coefficient of x^n by (n+1)!. -
Herb Conn, Feb 25 2002
a(n) = Sum_{i=0, 2n} B(i)*C(2n+1, i)*3^i where B(i) are the Bernoulli numbers, C(2n, i) the binomial numbers. -
Benoit Cloitre, May 01 2002
a(n) = (-1)^n * (6*n + 3) * s(2*n), if n>0, where s(n) are the cubic Bernoulli numbers. -
Michael Somos, Feb 26 2004
E.g.f.: 3*x / (2 + 4*cos(x)) = Sum_{n>=0} a(n) * x^(2*n+1) / (2*n+1)!. -
Michael Somos, Feb 26 2004
E.g.f.: E(x) = (3/2)/(1+2*cos(x)) - 1/2 = x^2/(3*G(0)+x^2); G(k) = 2*(2*k+1)*(k+1) - x^2 + 2*x^2*(2*k+1)*(k+1)/G(k+1); (continued fraction Euler's kind, 1-step). Let f[n]:=coeftayl(E(x), x=0, n) then:
A002111[n]=f[2*n+2]*((2*n+3)!). -
Sergei N. Gladkovskii, Jan 14 2012
a(n) = Sum_{k=0..2n+1} Sum_{j=0..k} Sum_{v=0..j} ((-1)^(n-v+1)/(j+1))* binomial(2*n+1,k)*binomial(j,v)*(3*v)^k. -
Peter Luschny, Jun 03 2013
a(n) = (-1)^(n+1)*3^(2*n+1)*B(2*n+1,1/3), where B(n,x) denotes the n-th Bernoulli polynomial. Cf.
A009843,
A069852,
A069994.
Conjecturally, a(n) = the unsigned numerator of B(2*n+1,1/3). Cf.
A033470.
Essentially a bisection of |
A083007|.
G.f. for signed version of sequence: 1/2 + 1/2*Sum_{n >= 0} { 1/(n+1) * Sum_{k = 0..n} (-1)^(k+1)*binomial(n,k)/( (1 - (3*k + 1)*x)*(1 - (3*k + 2)*x) ) } = x^2 - 5*x^4 + 49*x^6 - .... (End)
