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a(n) = Sum_{k=0..n} (2^k + Fibonacci(k)).
1

%I #15 Jan 21 2019 19:01:55

%S 1,4,9,19,38,75,147,288,565,1111,2190,4327,8567,16992,33753,67131,

%T 133654,266323,531051,1059520,2114861,4222959,8434974,16852239,

%U 33675823,67305280,134535537,268949683,537702950,1075088091,2149661955,4298491872,8595637477

%N a(n) = Sum_{k=0..n} (2^k + Fibonacci(k)).

%H Colin Barker, <a href="/A101353/b101353.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-4,-1,2).

%F Fibonacci(n+2) + 2^(n+1) + 2. - _Ralf Stephan_, May 16 2007

%F a(n)= 4*a(n-1) -4*a(n-2) -a(n-3) +2*a(n-4). G.f.: (1-3*x^2)/((1-x) * (2*x-1) * (x^2+x-1)). - _R. J. Mathar_, Feb 06 2010

%F a(n) = (-2+2^(1+n)+(2^(-1-n)*((1-sqrt(5))^n*(-3+sqrt(5))+(1+sqrt(5))^n*(3+sqrt(5))))/sqrt(5)). - _Colin Barker_, Nov 03 2016

%p seq(sum(2^x+fibonacci(x),x=0..a),a=0..30);

%o (PARI) Vec((1-3*x^2)/((1-x)*(2*x-1)*(x^2+x-1)) + O(x^40)) \\ _Colin Barker_, Nov 03 2016

%Y Cf. A117591 (first differences). - _R. J. Mathar_, Feb 06 2010

%K nonn,easy

%O 0,2

%A _Jorge Coveiro_, Dec 25 2004