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 A070998 a(n) = 9*a(n-1) - a(n-2), a(0)=1, a(-1)=1. 7

%I

%S 1,8,71,631,5608,49841,442961,3936808,34988311,310957991,2763633608,

%T 24561744481,218292066721,1940066856008,17242309637351,

%U 153240719880151,1361924169284008,12104076803675921

%N a(n) = 9*a(n-1) - a(n-2), a(0)=1, a(-1)=1.

%C A Pellian sequence.

%C In general, sum{k=0..n, binomial(2n-k,k)j^(n-k)}=(-1)^n*U(2n,I*sqrt(j)/2), I=sqrt(-1); - _Paul Barry_, Mar 13 2005

%C a(n) = L(n,9), where L is defined as in A108299; see also A057081 for L(n,-9). - _Reinhard Zumkeller_, Jun 01 2005

%C Number of 01-avoiding words of length n on alphabet {0,1,2,3,4,5,6,7,8} which do not end in 0. - _Tanya Khovanova_, Jan 10 2007

%C For positive n, a(n) equals the permanent of the (2n)X(2n) tridiagonal matrix with sqrt(7)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. [From John M. Campbell, Jul 08 2011]

%H <a href="/index/Rea#recLCC">Index entries for sequences related to linear recurrences with constant coefficients</a>

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

%F a(n) ~ 1/11*sqrt(11)*(1/2*(sqrt(11)+sqrt(7)))^(2*n+1)

%F Let q(n, x)=sum(i=0, n, x^(n-i)*binomial(2*n-i, i)); then q(n, 7)=a(n) - _Benoit Cloitre_, Nov 10 2002

%F a(n)a(n+3) = 63 + a(n+1)a(n+2). - R. Stephan, May 29 2004

%F a(n)=(-1)^n*U(2n, I*sqrt(7)/2), U(n, x) Chebyshev polynomial of second kind, I=sqrt(-1); - _Paul Barry_, Mar 13 2005

%F G.f.: (1-x)/1-9*x+x^2). [From _Philippe DELEHAM_, Nov 03 2008]

%F a(n)=(1/2)*[(9/2)+(1/2)*sqrt(77)]^(n+1)+(1/22)*[(9/2)-(1/2)*sqrt(77)]^(n+1)*sqrt(77)-(1/22)*[(9/2)+(1/2) *sqrt(77)]^(n+1)*sqrt(77)+(1/2)*[(9/2)-(1/2)*sqrt(77)]^(n+1), with n>=0 [From _Paolo P. Lava_, Nov 20 2008]

%o (Sage) [lucas_number1(n,9,1)-lucas_number1(n-1,9,1) for n in xrange(1, 19)]# [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Nov 10 2009]

%Y Cf. A057081, A056918.

%Y Row 9 of array A094954.

%K nonn

%O 0,2

%A Joe Keane (jgk(AT)jgk.org), May 18 2002

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