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A085601 a(n) = 2 * (4^n + 2^n) + 1. 25

%I

%S 5,13,41,145,545,2113,8321,33025,131585,525313,2099201,8392705,

%T 33562625,134234113,536903681,2147549185,8590065665,34360000513,

%U 137439477761,549756862465,2199025352705,8796097216513,35184380477441

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

%C 1. Begin with a square tile.

%C 2. Place square tiles on each edge to form a diamond shape.

%C 3. Count the tiles: a(0) = 5.

%C 4. Add tiles to fill the enclosing square.

%C 5. Go to step 2.

%H Iain Fox, <a href="/A085601/b085601.txt">Table of n, a(n) for n = 0..1660</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (7,-14,8)

%F From _R. J. Mathar_, Apr 20 2009: (Start)

%F a(n) = 7*a(n-1) - 14*a(n-2) + 8*a(n-3).

%F G.f.: -(5 - 22*x + 20*x^2)/((x - 1)*(2*x - 1)*(4*x - 1)).

%F (End)

%F E.g.f.: e^x + 2*(e^(2*x) + e^(4*x)). - _Iain Fox_, Dec 30 2017

%t Table[2(4^n+2^n)+1,{n,0,30}] (* or *) LinearRecurrence[{7,-14,8},{5,13,41},30] (* _Harvey P. Dale_, Dec 30 2017 *)

%o (PARI) first(n) = Vec((5 - 22*x + 20*x^2)/(1 - 7*x + 14*x^2 - 8*x^3) + O(x^n)) \\ _Iain Fox_, Dec 30 2017

%Y Cf. A028403, A046142, A056640, A064412.

%K nonn,easy

%O 0,1

%A Jun Mizuki (suzuki32(AT)sanken.osaka-u.ac.jp), Jul 07 2003

%E Edited by _Franklin T. Adams-Watters_ and _Don Reble_, Aug 15 2006

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Last modified October 20 05:42 EDT 2019. Contains 328247 sequences. (Running on oeis4.)