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a(n+5) = a(n+4)+a(n+3)+a(n+2)+a(n+1)+2*a(n) with a(0)=2, a(1)=1, a(2)=5, a(3)=10, a(4)=20.
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%I #29 Jan 31 2021 03:24:43

%S 2,1,5,10,20,40,77,157,314,628,1256,2509,5021,10042,20084,40168,80333,

%T 160669,321338,642676,1285352,2570701,5141405,10282810,20565620,

%U 41131240,82262477,164524957,329049914,658099828,1316199656,2632399309,5264798621,10529597242,21059194484,42118388968,84236777933,168473555869

%N a(n+5) = a(n+4)+a(n+3)+a(n+2)+a(n+1)+2*a(n) with a(0)=2, a(1)=1, a(2)=5, a(3)=10, a(4)=20.

%H Vincenzo Librandi, <a href="/A226311/b226311.txt">Table of n, a(n) for n = 0..1000</a>

%H Charles K. Cook and Michael R. Bacon, <a href="http://ami.ektf.hu/uploads/papers/finalpdf/AMI_41_from27to39.pdf">Some identities for Jacobsthal and Jacobsthal-Lucas numbers satisfying higher order recurrence relations</a>, Annales Mathematicae et Informaticae, 41 (2013) pp. 27-39.

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

%F G.f.: (2*x^4+2*x^3+2*x^2-x+2) / ((1-2*x)*(x^4+x^3+x^2+x+1)). - _Colin Barker_, Jun 08 2013

%p f:=proc(n) option remember;

%p if n=0 then 2 elif n=1 then 1 elif n=2 then 5 elif n=3 then 10 elif n=4 then 20 else

%p f(n-1)+f(n-2)+f(n-3)+f(n-4)+2*f(n-5); fi; end;

%p [seq(f(n),n=0..40)];

%t CoefficientList[Series[-(2 x^4 + 2 x^3 + 2 x^2 - x + 2) / ((2 x - 1) (x^4 + x^3 + x^2 + x + 1)), {x, 0, 40}], x] (* _Vincenzo Librandi_, Jun 19 2013 *)

%t LinearRecurrence[{1,1,1,1,2},{2,1,5,10,20},20] (* _Harvey P. Dale_, Jan 20 2015 *)

%K nonn,easy

%O 0,1

%A _N. J. A. Sloane_, Jun 08 2013