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A156906
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Transform of F(n+1) with Hankel transform (-1)^C(n+1,2)*F(n+1).
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0
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1, 0, 1, -1, 0, 0, 2, -2, -3, 3, 11, -11, -31, 31, 101, -101, -328, 328, 1102, -1102, -3760, 3760, 13036, -13036, -45750, 45750, 162262, -162262, -580638, 580638, 2093802, -2093802, -7601043, 7601043, 27756627, -27756627, -101888163
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,7
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COMMENTS
| Hankel transform is F(n+1)*(-1)^C(n+1,2). Image of F(n+1) by the Riordan array
(1/(1+xc(-x^2)), xc(-x^2)/(1+xc(-x^2)))=(1/(1-x), x(1-x)/(1-2x))^{-1}=A055587^{-1}.
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FORMULA
| G.f.: (1+2x+sqrt(1+4x^2))/(2(1+x))=1+x^2c(-x^2)/(1+x) where c(x) is the g.f. of A000108;
G.f.: 1/(1-x^2/(1+x+2x^2/(1-x/2+3x^2/4/(1+x/6+10x^2/9/(1-x/15+24/25x^2/(1+.... (continued fraction);
In the continued fraction expansion of the g.f. the general term is
1+x*if(n=0,0,(-1)^n/(F(n)F(n+1))+x^2*(-0^n+F(n)F(n+2))/F(n+1)^2.
a(n)=sum{k=0..n, (-1)^(n-k)*b(k)} where b(n)=if(n=0,1,C(1,n)-A000108((n-2)/2)*(-1)^(n/2)*(1+(-1)^n)/2).
Conjecture: n*a(n) +n*a(n-1) +4*(n-3)*a(n-2) +4*(n-3)*a(n-3)=0. - R. J. Mathar, Nov 14 2011
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CROSSREFS
| Sequence in context: A095060 A094117 A097365 * A180000 A053094 A196080
Adjacent sequences: A156903 A156904 A156905 * A156907 A156908 A156909
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KEYWORD
| easy,sign
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AUTHOR
| Paul Barry (pbarry(AT)wit.ie), Feb 17 2009
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