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a(n) = (3^n + 1)^2/4.
3

%I #33 Sep 20 2024 06:31:35

%S 1,4,25,196,1681,14884,133225,1196836,10764961,96864964,871725625,

%T 7845353476,70607649841,635467254244,5719200505225,51472790198116,

%U 463255068736321,4169295489486724,37523659017960025,337712929999378756,3039416366507624401

%N a(n) = (3^n + 1)^2/4.

%C For n>0, 4*a(n) has the form 10...020...01 when it is converted to base 3. The number of zeros between 2 consecutive nonzero digits is n-1. - _Raul Prisacariu_, Aug 27 2024

%H Colin Barker, <a href="/A051500/b051500.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 (13,-39,27).

%F a(n) = A007051(n)^2. - _Michel Marcus_, Jun 08 2013

%F From _Colin Barker_, Feb 10 2016: (Start)

%F a(n) = 13*a(n-1) - 39*a(n-2) + 27*a(n-3) for n > 2.

%F G.f.: (1-9*x+12*x^2) / ((1-x)*(1-3*x)*(1-9*x)). (End)

%F E.g.f.: (1/4)*(exp(x) + 2*exp(3*x) + exp(9*x)). - _G. C. Greubel_, May 22 2023

%t (3^Range[0,20]+1)^2/4 (* or *) LinearRecurrence[{13,-39,27},{1,4,25},30] (* _Harvey P. Dale_, Nov 05 2016 *)

%o (PARI) Vec((1-9*x+12*x^2)/((1-x)*(1-3*x)*(1-9*x)) + O(x^30)) \\ _Colin Barker_, Feb 10 2016

%o (Python)

%o def A051500(n): return (3**n+1)**2>>2 # _Chai Wah Wu_, Nov 14 2022

%o (Magma) [(3^n+1)^2/4: n in [0..40]]; // _G. C. Greubel_, May 22 2023

%o (SageMath) [((3^n+1)//2)^2 for n in range(41)] # _G. C. Greubel_, May 22 2023

%Y Cf. A007051.

%K nonn,easy

%O 0,2

%A _Colin Mallows_