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a(n+5) = a(n+3) + a(n+2) + a(n), with a(1) = a(2) = a(3) = a(4) = a(5) = 1.
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%I #19 Sep 08 2022 08:45:52

%S 1,1,1,1,1,3,3,5,7,9,15,19,29,41,57,85,117,171,243,345,499,705,1015,

%T 1447,2065,2961,4217,6041,8625,12323,17627,25165,35991,51417,73479,

%U 105035,150061,214505,306513,438045,626053,894619,1278603,1827185

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

%C Lim_{n -> infinity} a(n+1)/a(n) = s = 1.4291083198381..., where s is the root of the characteristic equation s^5 = s^3 + s^2 + 1.

%H G. C. Greubel, <a href="/A176513/b176513.txt">Table of n, a(n) for n = 1..1000</a>

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

%F G.f.: (x^4 + x^3 - x - 1)/(x^5 + x^3 + x^2 - 1).

%e a(7) = a(5) + a(4) + a(2) = 3.

%p seq(coeff(series((x^4+x^3-x-1)/(x^5+x^3+x^2-1),x,n+1), x, n), n = 0 .. 50); # _Muniru A Asiru_, Feb 17 2019

%t LinearRecurrence[{0,1,1,0,1}, {1,1,1,1,1}, 50] (* _G. C. Greubel_, Feb 17 2019 *)

%o (PARI) a(n) = my(v=vector(n)); for(k=1, n, v[k]=if(k<=5, 1, v[k-2]+v[k-3]+v[k-5])); v[n] \\ _Jianing Song_, Feb 04 2019

%o (PARI) my(x='x+O('x^50)); Vec((x^4+x^3-x-1)/(x^5+x^3+x^2-1)) \\ _G. C. Greubel_, Feb 17 2019

%o (Magma) m:=50; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!( (x^4+x^3-x-1)/(x^5+x^3+x^2-1) )); // _G. C. Greubel_, Feb 17 2019

%o (Sage) ((x^4+x^3-x-1)/(x^5+x^3+x^2-1)).series(x, 50).coefficients(x, sparse=False) # _G. C. Greubel_, Feb 17 2019

%o (GAP) a:=[1,1,1,1,1];; for n in [6..30] do a[n]:=a[n-2]+a[n-3]+a[n-5]; od; a; # _G. C. Greubel_, Feb 17 2019

%Y Cf. A176621.

%K nonn

%O 1,6

%A _Carmine Suriano_, Apr 19 2010