0,2

This is the same as A059348 after the first term. [Proof by James Sellers, seqfan 19 May 2008: The generating functions are the same, ignoring the constant terms which cause the difference between the two sequences. If the g.f. in the formula here is expanded, the constant term ignored, we obtain ( 1 + 2x - x^2 - 4x^3 - (x+1)^2sqrt(1-4x^2) )/(2x^4).

From the Bernhart Reference in A059348 we see that A059348 originates from A000108 padded with zeros, 1 0 1 0 2 0 5 0 14 0 42 0 132 ... with g.f. C(x^2). Taking the sum of each pair of consecutive values we get the auxiliary sequence 1 1 1 2 2 5 5 14 14 42 42 132 132 .... with g.f. ((1+x)C(x^2) - 1)/x . Sum pairs of consecutive values once more to obtain 2 2 3 4 7 10 19 28 56 ... which is A059348.

So this generating function is (1+x)[((1+x)C(x^2) - 1)/x -1] / x again ignoring the constant term. Straightforward algebraic manipulations show that this quantity equals (1+2x-...)/(2x^4) above, again ignoring the constant term.]

G.f.: (1+x)^2*c(x^2)^2, c(x) the g.f. of the Catalan numbers A000108; Let b(n)=(binomial(n-1, (n-1)/2)/((n-1)/2+1))(1-(-1)^n)/2+ (binomial(n, n/2)/(n/2+1))(1+(-1)^n)/2; then a(n)=sum{k=0..n, b(k)b(n-k)}.

Cf. A104721.

Sequence in context: A053634 A094863 A094862 this_sequence A102282 A064933 A060731

Adjacent sequences: A104719 A104720 A104721 this_sequence A104723 A104724 A104725

easy,nonn

Paul Barry (pbarry(AT)wit.ie), Mar 20 2005