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%I #45 Dec 17 2020 07:11:51
%S 1,5,15,37,83,177,367,749,1515,3049,6119,12261,24547,49121,98271,
%T 196573,393179,786393,1572823,3145685,6291411,12582865,25165775,
%U 50331597,100663243,201326537,402653127,805306309,1610612675,3221225409,6442450879,12884901821
%N a(n) = 6*2^n - 2*n - 5.
%C Previous name was: One half of second column (m=1) of triangle A142963.
%C Essentially a duplicate of A050488. - _Johannes W. Meijer_, Feb 20 2009
%D Eric Billault, Walter Damin, Robert Ferréol, Rodolphe Garin, MPSI Classes Prépas - Khôlles de Maths, Exercices corrigés, Ellipses, 2012, exercice 2.22 (1) pp 26, 43-44.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (4,-5,2).
%F a(n) = A142693(n+2,1)/2.
%F From _Johannes W. Meijer_, Feb 20 2009: (Start)
%F a(n) = 4a(n-1) - 5a(n-2) + 2a(n-3) for n > 2 with a(0) = 1, a(1) = 5, a(2) = 15.
%F G.f.: (1+z)/((1-z)^2*(1-2*z)). (End)
%F a(n) = Sum_{i=0..n} Sum_{j=0..n} 2^min(i,j) (Billault et al) (compare with A339771 that has max instead of min). - _Bernard Schott_, Dec 16 2020
%F a(n) = 2*A066524(n+1) - A339771(n). - _Kevin Ryde_, Dec 17 2020
%F E.g.f.: 6*exp(2*x) - exp(x)*(5 + 2*x). - _Stefano Spezia_, Dec 17 2020
%e a(3) = 6*2^3 - 2*3 - 5 = 37.
%p seq(6*2^n-2*n-5,n=0..40); # _Bernard Schott_, Dec 16 2020
%o (PARI) Vec((1+z)/((1-z)^2*(1-2*z)) + O(z^50)) \\ _Michel Marcus_, Jun 18 2017
%Y Cf. A142965 (m=2 column/4).
%Y Equals A050488(n+1).
%Y Equals A156920(n+1,1).
%Y Equals A156919(n+1,1)/2^n.
%Y Cf. A156925, A339771.
%Y Partial sums of A033484.
%K nonn,easy
%O 0,2
%A _Wolfdieter Lang_, Sep 15 2008
%E New name using a formula of _Bernard Schott_ by _Peter Luschny_, Dec 17 2020
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