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Generalized Pellian with 2nd term equal to 6.
4

%I #27 Jun 13 2015 00:50:00

%S 1,6,13,32,77,186,449,1084,2617,6318,15253,36824,88901,214626,518153,

%T 1250932,3020017,7290966,17601949,42494864,102591677,247678218,

%U 597948113,1443574444,3485097001,8413768446

%N Generalized Pellian with 2nd term equal to 6.

%C Pisano period lengths: 1, 2, 8, 4, 12, 8, 6, 8, 24, 12, 24, 8, 28, 6, 24, 16, 16, 24, 40, 12, ... (is this A175181?). - _R. J. Mathar_, Aug 10 2012

%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,1)

%F a(n) = 2*a(n-1) + a(n-2); a(0)=1, a(1)=6.

%F G.f.: (1+4*x)/(1 - 2*x - x^2). - _Philippe Deléham_, Nov 03 2008

%F a(n) = 4*A000129(n) + A000129(n+1). - _R. J. Mathar_, Aug 10 2012

%e a(n)=[ (5+sqrt(2))(1+sqrt(2))^n-(5-sqrt(2))(1-sqrt(2))^n ]/2*sqrt(2)

%p with(combinat): a:=n->4*fibonacci(n-1,2)+fibonacci(n,2): seq(a(n), n=1..26); # _Zerinvary Lajos_, Apr 04 2008

%t a[n_]:=(MatrixPower[{{1,2},{1,1}},n].{{5},{1}})[[2,1]]; Table[a[n],{n,0,40}] (* _Vladimir Joseph Stephan Orlovsky_, Feb 20 2010 *)

%t LinearRecurrence[{2,1},{1,6},30] (* _Harvey P. Dale_, Mar 29 2013 *)

%o (Maxima)

%o a[0]:1$

%o a[1]:6$

%o a[n]:=2*a[n-1]+a[n-2]$

%o A048693(n):=a[n]$

%o makelist(A048693(n),n,0,30); /* _Martin Ettl_, Nov 03 2012 */

%Y Cf. A001333, A000129, A048654, A048655.

%K easy,nonn

%O 0,2

%A _Barry E. Williams_