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%I #28 Aug 22 2020 16:18:01
%S 1,11,51,211,851,3411,13651,54611,218451,873811,3495251,13981011,
%T 55924051,223696211,894784851,3579139411,14316557651,57266230611,
%U 229064922451,916259689811,3665038759251,14660155037011,58640620148051,234562480592211,938249922368851
%N a(n) = 10*(4^n - 1)/3 + 1.
%H Colin Barker, <a href="/A321421/b321421.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 (5,-4).
%F a(n) = 4*a(n-1) + 7, a(0) = 1 for n > 0.
%F a(n) = 5*a(n-1) - 4*a(n-2), a(0) = 1, a(1) = 11, n > 1.
%F a(n) = a(n-1) + 10*4^(n-1), a(0) = 1, n > 0.
%F a(n) = A086462(n) + 1 for n > 0. - _Michel Marcus_, Nov 09 2018
%F G.f.: (1 + 6*x) / ((1 - x)*(1 - 4*x)). - _Colin Barker_, Nov 10 2018
%F E.g.f.: (-7*exp(x) + 10*exp(4*x))/3. - _Stefano Spezia_, Nov 10 2018
%F a(n) = 10*A002450(n) + 1. - _Omar E. Pol_, Nov 10 2018
%p seq(coeff(series((1+6*x)/((1-x)*(1-4*x)),x,n+1), x, n), n = 0 .. 25); # _Muniru A Asiru_, Nov 10 2018
%t a[n_]:=10*(4^n - 1)/3 + 1 ; Array[a, 20, 0] (* or *)
%t CoefficientList[Series[-((7 E^x)/3) + (10 E^(4 x))/3 , {x, 0, 20}], x]*Table[n!, {n, 0, 20}] (* _Stefano Spezia_, Nov 10 2018 *)
%t LinearRecurrence[{5,-4},{1,11},30] (* _Harvey P. Dale_, Aug 22 2020 *)
%o (PARI) Vec((1 + 6*x) / ((1 - x)*(1 - 4*x)) + O(x^30)) \\ _Colin Barker_, Nov 10 2018
%o (GAP) List([0..25],n->10*(4^n-1)/3+1); # _Muniru A Asiru_, Nov 10 2018
%Y Cf. A000302, A002450, A010727.
%K nonn,easy
%O 0,2
%A _Paul Curtz_, Nov 09 2018
%E More terms from _Colin Barker_, Nov 10 2018
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