A027991 a(n) = Sum{T(n,k)*T(n,2n-k)}, 0<=k<=n-1, T given by A027926.

1, 3, 12, 40, 130, 404, 1227, 3653, 10720, 31090, 89316, 254568, 720757, 2029095, 5684340, 15855964, 44061862, 122032508, 336966015, 927953705, 2549229256, 6987648358, 19115124552, 52194037200, 142274514025, 387215773899

Offset: 1

Author

Clark Kimberling

Keywords

• nonn

Comments

From Wolfdieter Lang, Jan 02 2012: (Start)

a(n) = A024458(2*n-1), n>=1 (bisection, odd arguments).

chate(n):=a(n+1), n>=0, is the even part of the bisection of the half-convolution of the sequence A000045(n+1), n>=0, with itself. See a comment on A201204 for the definition of half-convolution. There one finds also the rule for the o.g.f.s of the bisection. Here the o.g.f. of the sequence chate(n), n>=0, is Chate(x):= (Ce(x)+U2(x))/2 with Ce(x)=(1-x+x^2)/(1-3*x+x^2)^2, the o.g.f. of A054444(n), and

U2(x)=(1-x)/((1+x)*(1-3*x+x^2)), the o.g.f. of A007598(n+1), n>=0. This results (after multiplying with x) in the o.g.f. given below in the formula section. It is equivalent to the explicit formula given there, as can be seen after a partial fraction decomposition of the o.g.f.

(End)

Crossrefs

Cf. A027926, A024458, A201204, A007598.

Formula

a(n) = (1/5)[n*F(2n+2) - n*F(2n-2) + F(2n-1) - (-1)^n], F(n)=A000045(n).

O.g.f.: x*(1-2*x+2*x^2)/((1-3*x+x^2)^2*(1+x)). See the comment above. - Wolfdieter Lang, Jan 02 2012