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Linear recurrence a(n) = a(n-3) + 2a(n-5), starting from all-one initial conditions.
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%I #10 Oct 03 2016 11:48:10

%S 1,1,1,1,1,3,3,3,5,5,9,11,11,19,21,29,41,43,67,83,101,149,169,235,315,

%T 371,533,653,841,1163,1395,1907,2469,3077,4233,5259,6891,9171,11413,

%U 15357,19689,25195,33699,42515,55909

%N Linear recurrence a(n) = a(n-3) + 2a(n-5), starting from all-one initial conditions.

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (0, 0, 1, 0, 2).

%F O.g.f.: -x*(1+x+x^2)/(-1+x^3+2*x^5) . - _R. J. Mathar_, Jan 07 2008

%e a(14) = a(11) + 2a(9) = 9 + 2*5 = 19

%t LinearRecurrence[{0,0,1,0,2},{1,1,1,1,1},50] (* _Harvey P. Dale_, Jan 16 2012 *)

%o (PARI) a(n)=([0,1,0,0,0; 0,0,1,0,0; 0,0,0,1,0; 0,0,0,0,1; 2,0,1,0,0]^(n-1)*[1;1;1;1;1])[1,1] \\ _Charles R Greathouse IV_, Oct 03 2016

%K easy,nonn

%O 1,6

%A _David Eppstein_, Jan 04 2008