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A069006 Let M denote the 5 X 5 matrix with rows /1,1,1,1,1/1,1,1,1,0/1,1,1,0,0/1,1,0,0,0/1,0,0,0,0/ and A(n) = vector (x(n),y(n),z(n),t(n),u(n)) = M^n*A where A is the vector (1,1,1,1,1); then a(n) = t(n). 6

%I #20 Jun 13 2015 00:50:31

%S 1,2,9,29,105,365,1287,4516,15873,55759,195910,688286,2418195,8495917,

%T 29849041,104869718,368442700,1294463368,4547886208,15978257251,

%U 56137003923,197228218022,692929213991,2434493909304,8553197751125

%N Let M denote the 5 X 5 matrix with rows /1,1,1,1,1/1,1,1,1,0/1,1,1,0,0/1,1,0,0,0/1,0,0,0,0/ and A(n) = vector (x(n),y(n),z(n),t(n),u(n)) = M^n*A where A is the vector (1,1,1,1,1); then a(n) = t(n).

%C a(n-1) (with a(-1) = 0) appears in the formula for 1/rho(11)^n, n >= 0, with rho(11) = 2*cos(Pi/11) (the length ratio (smallest diagonal)/side in the regular 11-gon), when written in the power basis of the degree 5 number field Q(rho(11)): 1/rho(11)^n = A038342(n)*1 + A230080*rho(11) - A230081(n)*rho(11)^2 - a(n-1)*rho(11)^3 + A038342(n-1)* rho(11)^4, n >= 0, with A038342(-1) = 0. See A230080 with the example for n=4. - _Wolfdieter Lang_, Nov 04 2013

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (3,3,-4,-1,1).

%F G.f.:(1-x)/(1-x^5+x^4+4*x^3-3*x^2-3*x). - Maksym Voznyy (voznyy(AT)mail.ru), Aug 12 2009

%F a(n) = 3*a(n-1) + 3*a(n-2) - 4*a(n-3) - a(n-4) + a(n-5), n >= 0, with a(-5)=0, a(-4)=-1, a(-3)=a(-2)=a(-1)=0. - _Wolfdieter Lang_, Nov 04 2013

%t LinearRecurrence[{3,3,-4,-1,1},{1,2,9,29,105},30] (* _Harvey P. Dale_, Apr 16 2015 *)

%Y Cf. A006359, A069007, A069008, A069009, A070778, A006359(offset), for x(n), y(n), z(n), t(n), u(n), v(n).

%Y A038342, A230080, A230081 (for powers of 1/rho(11), see a comment above).

%K easy,nonn

%O 0,2

%A _Benoit Cloitre_, Apr 02 2002

%E Edited by _Henry Bottomley_, May 06 2002

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