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A085487 a(n) = p^n + q^n, p = (1 + sqrt(21))/2, q = (1 - sqrt(21))/2.

%I A085487 #30 Apr 15 2025 18:50:18
%S A085487 1,11,16,71,151,506,1261,3791,10096,29051,79531,224786,622441,1746371,
%T A085487 4858576,13590431,37883311,105835466,295252021,824429351,2300689456,
%U A085487 6422836211,17926283491,50040464546,139671882001,389874204731,1088233614736
%N A085487 a(n) = p^n + q^n, p = (1 + sqrt(21))/2, q = (1 - sqrt(21))/2.
%C A085487 p + q = 1, p*q = -5, p - q = sqrt(21).
%C A085487 The Lucas sequence V(1,-5) apart from the initial term a(0) = 2. - _Peter Bala_, Jun 23 2015
%D A085487 Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", Wiley, 2001, p. 471.
%H A085487 Vincenzo Librandi, <a href="/A085487/b085487.txt">Table of n, a(n) for n = 1..1000</a>
%H A085487 Wikipedia, <a href="http://en.wikipedia.org/wiki/Lucas_sequence">Lucas sequence</a>
%H A085487 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (1,5).
%F A085487 G.f.: (10*x^2+x)/(1-x-5*x^2).
%F A085487 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 A085487 a(n) = a(n-1) + 5a(n-2), n>1.
%F A085487 a(n) = [x^n] ( (1 + x + sqrt(1 + 2*x + 21*x^2))/2 )^n. - _Peter Bala_, Jun 23 2015
%e A085487 a(5) = 151 = p^5 + q^5, with p = 2.79128...; q = -1.79128...
%t A085487 CoefficientList[Series[(10 x + 1) / (1 - x - 5 x^2), {x, 0, 40}], x] (* _Vincenzo Librandi_, Jul 20 2013 *)
%o A085487 (Sage) [lucas_number2(n,1,-5) for n in range(1, 11)] # _Zerinvary Lajos_, May 14 2009
%o A085487 (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 A085487 (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 A085487 Cf. A015440.
%K A085487 nonn,easy
%O A085487 1,2
%A A085487 _Gary W. Adamson_, Jul 02 2003