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%I #36 Feb 11 2024 19:17:47
%S 3,9,27,69,171,405,939,2133,4779,10581,23211,50517,109227,234837,
%T 502443,1070421,2271915,4805973,10136235,21321045,44739243,93672789,
%U 195734187,408245589,850045611,1767200085,3668617899,7605671253,15748213419,32570168661,67287820971
%N a(n) = n*2^(n+1) + (2^(n+3)+(-1)^n)/3.
%C Another renewal type of sequence: Let X, X(1),X(2),... denote independent random variables with pdf P(X=1) = P(X=2) = P(X=4) = 1/3. Let N(x) denote the first value of k such that X(1)*X(2)...*X(k) > x, and let H(x) = E(N(x)). The sequence a(n) is given by a(n) = 2^(n+1)*H(2^n).
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,0,-4).
%F a(n) = n*2^(n+1) + (2^(n+3)+(-1)^n)/3.
%F a(n) = 3 * A045883(n+1).
%F G.f.: 3/((1 + x)*(1 - 2*x)^2). [_Bruno Berselli_, Oct 16 2014]
%t Table[n 2^(n + 1) + (2^(n + 3) + (-1)^n)/3, {n, 0, 70}] (* _Vincenzo Librandi_, Oct 16 2014 *)
%t LinearRecurrence[{3,0,-4},{3,9,27},40] (* _Harvey P. Dale_, Feb 11 2024 *)
%o (PARI) a(n) = n*2^(n+1) + (2^(n+3)+(-1)^n)/3; \\ _Michel Marcus_, Oct 16 2014
%o (Magma) [n*2^(n+1)+(2^(n+3)+(-1)^n)/3: n in [0..30]]; // _Vincenzo Librandi_, Oct 16 2014
%Y 3 times A045883.
%K nonn,easy,less
%O 0,1
%A _Edward Omey_, Jun 16 2011
%E Formula corrected and more terms from _Michel Marcus_, Oct 16 2014
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