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a(n) = (3/5)*2^(4n+1) - (1/5).
1

%I #26 Mar 28 2024 16:12:02

%S 1,19,307,4915,78643,1258291,20132659,322122547,5153960755,

%T 82463372083,1319413953331,21110623253299,337769972052787,

%U 5404319552844595,86469112845513523,1383505805528216371,22136092888451461939,354177486215223391027,5666839779443574256435,90669436471097188102963,1450710983537555009647411

%N a(n) = (3/5)*2^(4n+1) - (1/5).

%C Bisection of A112627.

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (17,-16).

%F a(n) = (3/5)*2^(4*n+1) - (1/5).

%F a(n) = 16*a(n-1) + 3 for n > 0.

%F a(n) = (1/5)*A153893(4*n+1).

%F a(n) = A016029(4*n+2).

%F a(n) = A112627(2*n+1).

%F G.f.: (1+2*x)/((1-x)*(1-16*x)). - _Colin Barker_, May 06 2012

%t (3*2^(4*Range[0,20]+1)-1)/5 (* or *) LinearRecurrence[{17,-16},{1,19},30] (* _Harvey P. Dale_, Jul 21 2021 *)

%o (Magma) [(3/5)*2^(4*n+1) - (1/5): n in [0..20]];

%Y Cf. A112627, A016029, A153893.

%K nonn,easy

%O 0,2

%A _Brad Clardy_, Apr 30 2012