A099364 An inverse Chebyshev transform of (1-x)^2.

1, -2, 2, -4, 5, -10, 14, -28, 42, -84, 132, -264, 429, -858, 1430, -2860, 4862, -9724, 16796, -33592, 58786, -117572, 208012, -416024, 742900, -1485800, 2674440, -5348880, 9694845, -19389690, 35357670, -70715340, 129644790, -259289580, 477638700, -955277400, 1767263190, -3534526380

Offset: 0

Author

Paul Barry, Oct 13 2004

Keywords

• easy
• sign

Comments

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)).

Links

• G. Levy, Solution of second order recurrence equations (2010) PhD Thesis, Florida State University, page 2

Crossrefs

Cf. A089408, A002057.

Formula

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).

D-finite with recurrence: (n+4)*a(n) +2*a(n-1) -4*n*a(n-2)=0. - R. J. Mathar, Nov 09 2012