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Expansion of 3*x*(1+2*x)/(1-3*x-3*x^2).
4

%I #27 Jan 10 2021 12:26:05

%S 0,3,15,54,207,783,2970,11259,42687,161838,613575,2326239,8819442,

%T 33437043,126769455,480619494,1822166847,6908359023,26191577610,

%U 99299809899,376474162527,1427321917278,5411388239415,20516130470079

%N Expansion of 3*x*(1+2*x)/(1-3*x-3*x^2).

%C A Jacobsthal variation.

%C p - q = sqrt(21); p*q = -3; p + q = 3.

%D Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", Wiley, 2001, p. 471.

%H Harvey P. Dale, <a href="/A085480/b085480.txt">Table of n, a(n) for n = 1..1000</a>

%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (3,3).

%F a(n) = p^n + q^n, where p = (3 + sqrt(21))/2, q = (3 - sqrt 21)/2.

%F a(n) = 3*a(n-1) + 3*a(n-2), a(1)=3, a(2)=15. - _Philippe Deléham_, Nov 19 2008

%F G.f.: G(0)/x - 2/x, where G(k) = 1 + 1/(1 - x*(7*k-3)/(x*(7*k+4) - 2/G(k+1))); (continued fraction). - _Sergei N. Gladkovskii_, Jun 03 2013

%e a(4) = q^4 + q^4 = 207; p^5 + q^5 = 783, where p = (3 + sqrt(21))/2, q = (3 - sqrt(21))/2.

%t CoefficientList[Series[3x (1+2x)/(1-3x-3x^2),{x,0,30}],x] (* or *) LinearRecurrence[{3,3},{0,3,15},30] (* _Harvey P. Dale_, Jan 10 2021 *)

%Y Cf. A030195.

%K nonn,easy

%O 1,2

%A _Gary W. Adamson_, Jul 02 2003

%E Zero prepended by _Harvey P. Dale_, Jan 10 2021