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a(n) = (5^n + 1)/2.
16

%I #61 Dec 10 2023 15:18:20

%S 1,3,13,63,313,1563,7813,39063,195313,976563,4882813,24414063,

%T 122070313,610351563,3051757813,15258789063,76293945313,381469726563,

%U 1907348632813,9536743164063,47683715820313,238418579101563

%N a(n) = (5^n + 1)/2.

%C Terms (with the offset changed to 1) are also the quotients arising from sequence A050621.

%C Partial sums of A020699. - _Paul Barry_, Sep 03 2003

%C Binomial transform of A081294. - _Paul Barry_, Jan 13 2005

%H Nathaniel Johnston, <a href="/A034478/b034478.txt">Table of n, a(n) for n = 0..250</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (6,-5).

%F E.g.f.: exp(3*x)*cosh(2*x). - _Paul Barry_, Mar 17 2003

%F G.f.: (1-3*x)/((1-x)*(1-5*x)). - _Paul Barry_, Sep 03 2003

%F a(n) = Sum_{k=0..n} Sum_{j=0..k} binomial(n, k)*binomial(2*k, 2*j). - _Paul Barry_, Jan 13 2005

%F a(n) = 6*a(n-1) - 5*a(n-2) for n>1, a(0)=1, a(1)=3. - _Philippe Deléham_, Jul 11 2005

%F a(n)^2 + (a(n) - 1)^2 = a(2*n). E.g., 63^2 + 62^2 = 7813 = a(6). - _Gary W. Adamson_, Jun 17 2006

%F a(n) = 5*a(n-1) - 2 for n>0, a(0)=1. - _Vincenzo Librandi_, Aug 01 2010

%F a(n) = A034474(n)/2. - _Elmo R. Oliveira_, Dec 10 2023

%e G.f. = 1 + 3*x + 13*x^2 + 63*x^3 + 313*x^4 + 1563*x^5 + 7813*x^6 + ...

%p seq((5^n + 1)/2, n=0..20); # _Zerinvary Lajos_, Jun 16 2007

%t LinearRecurrence[{6, -5},{1, 3},22] (* _Ray Chandler_, May 25 2021 *)

%o (Sage) [lucas_number2(n,6,5)/2 for n in range(0,22)] # _Zerinvary Lajos_, Jul 08 2008

%Y Cf. A020699, A034474, A050621, A081294.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_