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a(n) = p^n + q^n, p = (1 + sqrt(21))/2, q = (1 - sqrt(21))/2.
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%I #27 Sep 08 2022 08:45:11

%S 1,11,16,71,151,506,1261,3791,10096,29051,79531,224786,622441,1746371,

%T 4858576,13590431,37883311,105835466,295252021,824429351,2300689456,

%U 6422836211,17926283491,50040464546,139671882001,389874204731,1088233614736

%N a(n) = p^n + q^n, p = (1 + sqrt(21))/2, q = (1 - sqrt(21))/2.

%C p + q = 1, p*q = -5, p - q = sqrt(21).

%C The Lucas sequence V(1,-5) apart from the initial term a(0) = 2. - _Peter Bala_, Jun 23 2015

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

%H Vincenzo Librandi, <a href="/A085487/b085487.txt">Table of n, a(n) for n = 1..1000</a>

%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Lucas_sequence">Lucas sequence</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (1,5).

%F G.f.: (10*x^2+x)/(1-x-5*x^2).

%F a(n) = n*sum(k=1..n, (C(k,n-k)*1^(2*k-n)*(5)^(n-k))/k). - _Dmitry Kruchinin_, May 16 2011

%F a(n) = a(n-1) + 5a(n-2), n>1.

%F a(n) = [x^n] ( (1 + x + sqrt(1 + 2*x + 21*x^2))/2 )^n. - _Peter Bala_, Jun 23 2015

%e a(5) = 151 = p^5 + q^5, with p = 2.79128...; q = -1.79128...

%t CoefficientList[Series[(10 x + 1) / (1 - x - 5 x^2), {x, 0, 40}], x] (* _Vincenzo Librandi_, Jul 20 2013 *)

%o (Sage) [lucas_number2(n,1,-5) for n in range(1, 11)] # _Zerinvary Lajos_, May 14 2009

%o (Maxima) a(n):=n*sum((binomial(k,n-k)*1^(2*k-n)*(5)^(n-k))/k,k,1,n) /* _Dmitry Kruchinin_, May 16 2011 */

%o (Magma) I:=[ 1,11]; [n le 2 select I[n] else Self(n-1)+5*Self(n-2): n in [1..40]]; // _Vincenzo Librandi_, Jul 20 2013

%Y Cf. A015440.

%K nonn,easy

%O 1,2

%A _Gary W. Adamson_, Jul 02 2003