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a(0)=a(1)=1, a(2)=6, a(3)=11; a(n+4) = 10*a(n+2) - a(n).
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%I #22 Oct 05 2024 03:10:43

%S 1,1,6,11,59,109,584,1079,5781,10681,57226,105731,566479,1046629,

%T 5607564,10360559,55509161,102558961,549484046,1015229051,5439331299,

%U 10049731549,53843828944,99482086439,532998958141,984771132841,5276145752466,9748229241971,52228458566519

%N a(0)=a(1)=1, a(2)=6, a(3)=11; a(n+4) = 10*a(n+2) - a(n).

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,10,0,-1).

%F a(n) = ((1/48)*sqrt(3)*sqrt(2) + 1/4 + (1/8)*sqrt(2))*(sqrt(3) + sqrt(2))^n + (-(1/48)*sqrt(3)*sqrt(2) + 1/4 - (1/8)*sqrt(2))*(sqrt(3) - sqrt(2))^n + ((1/48)*sqrt(3)*sqrt(2) + 1/4 - (1/8)*sqrt(2))*(-sqrt(3) - sqrt(2))^n + (1/4 - (1/48)*sqrt(3)*sqrt(2) + (1/8)*sqrt(2))*(-sqrt(3) + sqrt(2))^n.

%F From _R. J. Mathar_ and _Philippe Deléham_, Dec 05 2008: (Start)

%F a(2n) = A004189(n+1) - 4*A004189(n).

%F a(2n+1) = A004189(n) + A004189(n+1).

%F G.f.: (1+x-4x^2+x^3) / (1-10x^2+x^4). (End)

%t LinearRecurrence[{0,10,0,-1},{1,1,6,11},30] (* _Harvey P. Dale_, Nov 10 2018 *)

%Y Cf. A054320 (bisection).

%K easy,nonn

%O 0,3

%A _Richard Choulet_, Dec 04 2008