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a(0)=1, a(1)=5, a(n) = 20*a(n-2) - a(n-1).
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%I #20 Sep 08 2022 08:45:47

%S 1,5,15,85,215,1485,2815,26885,29415,508285,80015,10085685,-8485385,

%T 210199085,-379906785,4583888485,-12182024185,103859793885,

%U -347500277585,2424696155285,-9374701706985,57868624812685,-245362658952385,1402735155206085,-6309988334253785

%N a(0)=1, a(1)=5, a(n) = 20*a(n-2) - a(n-1).

%C a(n)/a(n-1) tends to -5.

%H G. C. Greubel, <a href="/A165470/b165470.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (-1,20).

%F G.f.: (1+6*x)/(1+x-20*x^2).

%F a(n) = Sum_{k=0..n} A112555(n,k)*4^k.

%F a(n) = (10*4^n-(-5)^n)/9. - _Klaus Brockhaus_, Sep 25 2009

%F E.g.f.: (10*exp(4*x) - exp(-5*x))/9. - _G. C. Greubel_, Oct 20 2018

%p seq(coeff(series((1+6*x)/(1+x-20*x^2),x,n+1), x, n), n = 0 .. 25); # _Muniru A Asiru_, Oct 21 2018

%t LinearRecurrence[{-1,20}, {1,5}, 40] (* _G. C. Greubel_, Oct 20 2018 *)

%o (PARI) vector(40, n, n--; (10*4^n-(-5)^n)/9) \\ _G. C. Greubel_, Oct 20 2018

%o (Magma) [(10*4^n-(-5)^n)/9: n in [0..40]]; // _G. C. Greubel_, Oct 20 2018

%o (Python) for n in range(0, 30): print(int((10*4**n-(-5)**n)/9), end=', ') # _Stefano Spezia_, Oct 21 2018

%o (GAP) a:=[1,5];; for n in [3..25] do a[n]:=20*a[n-2]-a[n-1]; od; a; # _Muniru A Asiru_, Oct 21 2018

%K easy,sign

%O 0,2

%A _Philippe Deléham_, Sep 20 2009