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%I #25 Mar 02 2024 13:39:49
%S 0,1,2,6,13,32,72,169,386,894,2053,4736,10896,25105,57794,133110,
%T 306493,705824,1625304,3742777,8618690,19847022,45703093,105244160,
%U 242353440,558085921,1285146242,2959404006,6814842733,15693054752,36137582952
%N a(n) = 2*a(n-1) + 2*a(n-2) - 3*a(n-3).
%C Partial sums of A006130 (with leading zero).
%C Specify a triangle by T(n,0) = T(n+1,1) = A001045(n) and T(n,k) = T(n-1,k-1) + T(n-1,k-2) + T(n-2,k-2) otherwise. Then T(n,n)= a(n-1). - _J. M. Bergot_, May 24 2013
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (2,2,-3).
%F G.f.: x/((1-x)*(1-x-3*x^2)).
%F a(n) = Sum_{k=0..floor(n/2)} binomial(n-k, k+1)*3^k.
%F a(n) = (1/2 + sqrt(13)/2)^n*(1/6 + 7*sqrt(13)/78) + (1/6 - 7*sqrt(13)/78)*(1/2 - sqrt(13)/2)^n - 1/3.
%F a(n+1) = Sum_{k=0..n} C(k+1,n-k+1)*3^(n-k). - _Paul Barry_, May 21 2006
%F a(n) = a(n-1) + 3*a(n-2) + 1, n > 1. - _Gary Detlefs_, Jun 21 2010
%F G.f.: Q(0)*x/(2-2*x), where Q(k) = 1 + 1/(1 - x*(4*k+1 + 3*x)/( x*(4*k+3 + 3*x) + 1/Q(k+1) )); (continued fraction). - _Sergei N. Gladkovskii_, Sep 09 2013
%t LinearRecurrence[{2,2,-3},{0,1,2},40] (* _Harvey P. Dale_, Mar 02 2024 *)
%Y Cf. A001045, A006130.
%K easy,nonn
%O 0,3
%A _Paul Barry_, Oct 08 2004
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