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Inverse binomial transform of A053088.
2

%I #32 Feb 13 2022 04:18:24

%S 1,-1,4,-8,20,-44,100,-220,484,-1052,2276,-4892,10468,-22300,47332,

%T -100124,211172,-444188,932068,-1951516,4077796,-8505116,17709284,

%U -36816668,76429540,-158451484,328087780,-678545180,1401829604

%N Inverse binomial transform of A053088.

%C Contribution from _Gary W. Adamson_, Jan 05 2009: (Start)

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

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

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

%C 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

%H Roland Bacher, <a href="http://arxiv.org/abs/1509.09054">Chebyshev polynomials, quadratic surds and a variation of Pascal's triangle</a>, arXiv:1509.09054 [math.CO], 2015.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (-3,0,4).

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

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

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

%t LinearRecurrence[{-3,0,4},{1,-1,4},30] (* _Harvey P. Dale_, Dec 16 2016 *)

%K easy,sign

%O 0,3

%A _Paul Barry_, May 20 2003