%I #20 Feb 08 2017 09:02:36

%S 1,11,-29,-189,451,3011,-7229,-48189,115651,771011,-1850429,-12336189,

%T 29606851,197379011,-473709629,-3158064189,7579354051,50529027011,

%U -121269664829,-808464432189,1940314637251,12935430915011,-31045034196029,-206966894640189

%N Expansion of (24x^2-10x-1)/(16x^3-16x^2+x-1).

%C Related to base i-1 representation of integers (Khmelnik encoding): presumably a(0) is the most common first difference of A066321 (occurs with density 1/2), a(1) is the second most common difference (density 1/4), a(2) has density 1/8, and so on; in particular, A066322 consists entirely of the terms a(n) with n>3.

%H Colin Barker, <a href="/A282137/b282137.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,-16,16).

%F a(k+8) - 257 * a(k+4) + 256 * a(k) = 0, for k >= 0. - _Altug Alkan_, Feb 07 2017

%F G.f.: (24*x^2-10*x-1)/(16*x^3-16*x^2+x-1).

%F From _Colin Barker_, Feb 07 2017: (Start)

%F a(n) = (-13 + (15+25*i)*(-4*i)^n + (15-25*i)*(4*i)^n) / 17 where i=sqrt(-1).

%F a(n) = a(n-1) - 16*a(n-2) + 16*a(n-3) for n>2.

%F (End)

%t LinearRecurrence[{0,0,0,257,0,0,0,-256}, {1, 11, -29, -189, 451, 3011, -7229, -48189}, 24]

%t LinearRecurrence[{1, -16, 16}, {1, 11, -29}, 24]

%o (Python)

%o print([[1, 11, -29, -189][n%4] + [450, 3000, -7200, -48000][n%4]*(256**(n//4)-1)//255 for n in range(24)])

%o (PARI) Vec((1 - 2*x)*(1 + 12*x) / ((1 - x)*(1 + 16*x^2)) + O(x^30)) \\ _Colin Barker_, Feb 07 2017

%Y Cf. A066321, A066322, A218723.

%K sign,easy

%O 0,2

%A _Andrey Zabolotskiy_, Feb 06 2017