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An inverse Chebyshev transform of (1-x)^2.
1

%I #9 Jan 30 2020 21:29:15

%S 1,-2,2,-4,5,-10,14,-28,42,-84,132,-264,429,-858,1430,-2860,4862,

%T -9724,16796,-33592,58786,-117572,208012,-416024,742900,-1485800,

%U 2674440,-5348880,9694845,-19389690,35357670,-70715340,129644790,-259289580,477638700,-955277400,1767263190,-3534526380

%N An inverse Chebyshev transform of (1-x)^2.

%C Second binomial transform of the expansion of c(-x)^4 (i.e. of (-1)^n*4C(2n+3,n)/(n+4)). The g.f. is transformed to (1-x)^2 under the Chebyshev transformation A(x)->(1/(1+x^2))A(x/(1+x^2)).

%H G. Levy, <a href="http://purl.flvc.org/fsu/fd/FSU_migr_etd-3099">Solution of second order recurrence equations</a> (2010) PhD Thesis, Florida State University, page 2

%F G.f.: (c(x^2)-1)(1-2x)/x^2 with c(x) the g.f. of A000108; a(n)=sum{k=0..n, (k+1)C(n, (n-k)/2)(-1)^k*C(2, k)(1+(-1)^(n-k))/(n+k+2)}; a(n)=sum{k=0..n, (k+1)C(n, (n-k)/2)b(k)(1+(-1)^(n-k))/(n+k+2)} where b(n)=0^n+sum{k=0..n, C(n, k)(-1)^(n-k)(-3k+k(k+1)/2)}; a(2n)=C(n+1); a(2n+1)=-2*C(n+1).

%F D-finite with recurrence: (n+4)*a(n) +2*a(n-1) -4*n*a(n-2)=0. - _R. J. Mathar_, Nov 09 2012

%Y Cf. A089408, A002057.

%K easy,sign

%O 0,2

%A _Paul Barry_, Oct 13 2004