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A131885 a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) for n >= 4 starting with a(0) = 1, a(1) = 2, a(2) = 4, and a(3) = 6.

%I A131885 #35 Sep 01 2020 17:07:36
%S A131885 1,2,4,6,8,12,24,56,128,272,544,1056,2048,4032,8064,16256,32768,65792,
%T A131885 131584,262656,524288,1047552,2095104,4192256,8388608,16781312,
%U A131885 33562624,67117056,134217728,268419072,536838144,1073709056,2147483648,4295032832,8590065664
%N A131885 a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) for n >= 4 starting with a(0) = 1, a(1) = 2, a(2) = 4, and a(3) = 6.
%H A131885 Karl Dilcher and Maciej Ulas, <a href="https://arxiv.org/abs/2008.13475">Divisibility and Arithmetic Properties of a Class of Sparse Polynomials</a>, arXiv:2008.13475 [math.NT], 2020. See Table 1, 2nd column, p. 3.
%H A131885 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4).
%F A131885 Binomial transform of 1, 1, 1, -1.
%F A131885 G.f.: (-1 + 2*x - 2*x^2 + 2*x^3)/(2*x - 1)/(2*x^2 - 2*x + 1). - _R. J. Mathar_, Nov 14 2007
%F A131885 a(n) = 2*A038504(n) for n > 0. - _R. J. Mathar_, Jul 17 2009
%F A131885 G.f.: 1/2*(1 - 1/(2*x-1) + x*Q(0)/(1-x)), where Q(k) = 1 + 1/(1 - x*(k+1)/(x*(k+2) + 1/Q(k+1) )) (continued fraction). - _Sergei N. Gladkovskii_, Sep 27 2013
%F A131885 a(n) = Sum_{j=0..n} binomial(n, j)*(-1)^binomial(j, 3); this is the case m=3 and z=-1 of f(m,n)(z) = Sum_{j=0..n} binomial(n, j)*z^binomial(j, m). See Dilcher and Ulas. - _Michel Marcus_, Sep 01 2020
%t A131885 Join[{1},LinearRecurrence[{4,-6,4},{2,4,6},60]] (* _Harvey P. Dale_, Jul 07 2011 *)
%K A131885 nonn
%O A131885 0,2
%A A131885 _Paul Curtz_, Oct 25 2007
%E A131885 More terms from _Harvey P. Dale_, Jul 07 2011