A084219 Inverse binomial transform of A053088.

1, -1, 4, -8, 20, -44, 100, -220, 484, -1052, 2276, -4892, 10468, -22300, 47332, -100124, 211172, -444188, 932068, -1951516, 4077796, -8505116, 17709284, -36816668, 76429540, -158451484, 328087780, -678545180, 1401829604

Offset: 0

Author

Paul Barry, May 20 2003

Keywords

• easy
• sign

Comments

Contribution from Gary W. Adamson, Jan 05 2009: (Start)

Unsigned, starting with offset 1: generated from iterates of M * [1,1,1,...]

where M = a tridiagonal matrix with [0,1,1,1,...] as the main diagonal,

[1,1,1,...] as the uperdiagonal and [2,0,0,0,...] as the subdiagonal. (End)

Define a triangle via T(n,0) = T(n,n) = A001045(n) and T(r,c) = T(r-1,c-1) + T(r-1,c). The row sums of the triangle are s(n) = 0, 2, 4, 12, ... = 2*A059570(n), and their first differences are s(n+1)-s(n) = 2*|a(n)|. J. M. Bergot, May 15 2013

Links

• Roland Bacher, Chebyshev polynomials, quadratic surds and a variation of Pascal's triangle, arXiv:1509.09054 [math.CO], 2015.
• Index entries for linear recurrences with constant coefficients, signature (-3,0,4).

Programs

Mathematica

LinearRecurrence[{-3,0,4},{1,-1,4},30] (* Harvey P. Dale, Dec 16 2016 *)

Formula

a(n) = (4 - 3*n*(-2)^(n-1) + 5*(-2)^n)/9.

a(n) = (1/4) + Sum_{k=0..n} (-2)^k*(k+3)/4.

G.f.: (1+x)^2/((1-x)(1+2x)^2).