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a(0)=1, a(1)=5, a(n+2)=4a(n), n>0.
1

%I #18 May 09 2024 04:43:11

%S 1,5,8,20,32,80,128,320,512,1280,2048,5120,8192,20480,32768,81920,

%T 131072,327680,524288,1310720,2097152,5242880,8388608,20971520,

%U 33554432,83886080,134217728,335544320,536870912,1342177280,2147483648

%N a(0)=1, a(1)=5, a(n+2)=4a(n), n>0.

%C Binomial transform is A080926 (without leading zero).

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (0,4)

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

%F E.g.f.: (9*exp(2*x)-exp(-2*x))/4-exp(0).

%F a(n) = (9*2^n-(-2)^n)/4-0^n.

%t Join[{1},LinearRecurrence[{0,4},{5,8},30]] (* _Harvey P. Dale_, Sep 12 2013 *)

%Y Cf. A080926, A104721.

%K nonn,easy

%O 0,2

%A _Paul Barry_, May 30 2003