A115243 - OEIS

%I #43 May 05 2023 07:44:29

%S 0,2,12,50,204,818,3276,13106,52428,209714,838860,3355442,13421772,

%T 53687090,214748364,858993458,3435973836,13743895346,54975581388,

%U 219902325554,879609302220,3518437208882,14073748835532,56294995342130,225179981368524,900719925474098

%N G.f.: (4*x^2 + 2*x)/(4*x^3 - x^2 - 4*x + 1).

%C Inverse Z-transform of polynomial in A112627.

%C a(n) is also the number of corners in the n-th approximation of the Hilbert Curve. The 1st Hilbert Curve approximation has 2 corners. To find a(n) given a(n - 1), look at how the n-th Hilbert Curve approximation is constructed: duplicate the (n-1)-th approximation 4 times and connect the duplicates with 3 line segments. a(n) will always be 4 * a(n - 1) corners from the 4 duplicates plus 4 new corners if n is even or 2 new corners if n is odd. - _Mikel Mcdaniel_, Jan 10 2019

%H G. C. Greubel, <a href="/A115243/b115243.txt">Table of n, a(n) for n = 0..500</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (4,1,-4).

%F a(n) = InverseZTransform[(1 + 2*x)/(1 - x - 16*x^2 + 16*x^3), x, n] * 2^(2*n).

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

%F a(n) = 4*a(n-1)+a(n-2)-4*a(n-3). - _Gary Detlefs_ Dec 17 2010

%F a(n) = (4^(n+1)+(-1)^n)/5 - 1. - _Robert Israel_, Mar 09 2016

%F a(n) = 4*a(n-1)+3+(-1)^n. - _Mikel Mcdaniel_, Jan 10 2019

%p seq((4^(n+1)+(-1)^n)/5 - 1, n=0..100); # _Robert Israel_, Mar 09 2016

%t Table[InverseZTransform[(1 + 2*x)/(1 - x - 16*x^2 + 16*x^3), x, n]*2^( 2*n), {n, 1, 25}]

%t LinearRecurrence[{4, 1, -4}, {0, 2, 12}, 50] (* _G. C. Greubel_, Feb 07 2016 *)

%o (Magma) [(4^(n+1)+(-1)^n)/5 - 1: n in [0..25]]; // _Vincenzo Librandi_, Jan 10 2019

%o (PARI) a(n) = (bitneg(0,2*n+2)-1)\5; \\ _Kevin Ryde_, May 05 2023

%Y Cf. A112627.

%K nonn,easy

%O 0,2

%A _Roger L. Bagula_, Mar 04 2006

%E Entry revised by _N. J. A. Sloane_, Dec 18 2010