

A027991


a(n)=Sum{T(n,k)*T(n,2nk)}, 0<=k<=n1, T given by A027926.


3



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
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OFFSET

1,2


COMMENTS

Contribution from Wolfdieter Lang, Jan 02 2012 (Start)
a(n)=A024458(2*n1), n>=1 (bisection, odd arguments).
chate(n):=a(n+1), n>=0, is the even part of the bisection of the halfconvolution of the sequence A000045(n+1), n>=0, with itself. See a comment on A201204 for the definition of halfconvolution. 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)=(1x+x^2)/(13*x+x^2)^2, the o.g.f. of A054444(n), and
U2(x)=(1x)/((1+x)*(13*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)


LINKS

Table of n, a(n) for n=1..26.


FORMULA

(1/5)[n*F(2n+2)  n*F(2n2) + F(2n1)  (1)^n], F(n)=A000045(n).
O.g.f.: x*(12*x+2*x^2)/((13*x+x^2)^2*(1+x)). See the comment above.  From Wolfdieter Lang, Jan 02 2012.


CROSSREFS

Cf. A027926, A024458, A201204, A007598.
Sequence in context: A052482 A061136 A247002 * A120304 A289652 A026071
Adjacent sequences: A027988 A027989 A027990 * A027992 A027993 A027994


KEYWORD

nonn


AUTHOR

Clark Kimberling


STATUS

approved



