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%I #23 Apr 04 2017 06:16:22
%S 1,1,1,11,31,101,281,781,2101,5611,14851,39161,102961,270281,708761,
%T 1857451,4865911,12744061,33372361,87382901,228792301,599019851,
%U 1568309051,4105974961,10749725281,28143378001,73680695281,192899171531
%N a(n) = 5*F(n)^2 - 5*F(n) + 1, where F(n) = Fibonacci(n).
%C 11 = Lucas(5) divides a(3+10k), a(7+10k), and a(8+10k). The last digit of a(n) is 1, so a(n) mod 10 = 1. For odd n there exists the so-called Aurifeuillian factorization A001946(n) = Lucas(5n) = Lucas(n)*A(n)*B(n) = A000032(n)*A124296(n)*A124297(n), where A(n) = A124296(n) = 5*F(n)^2 - 5*F(n) + 1 and B(n) = A124297(n) = 5*F(n)^2 + 5*F(n) + 1, where F(n) = Fibonacci(n).
%H John Cerkan, <a href="/A124296/b124296.txt">Table of n, a(n) for n = 0..2375</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/AurifeuilleanFactorization.html">Aurifeuillean Factorization</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (4,-2,-6,4,2,-1).
%F a(n) = 5*Fibonacci(n)^2 - 5*Fibonacci(n) + 1.
%F G.f.: -(x^5+9*x^4-15*x^3+x^2+3*x-1) / ((x-1)*(x+1)*(x^2-3*x+1)*(x^2+x-1)). [_Colin Barker_, Jan 03 2013]
%t Table[5*Fibonacci[n]^2-5*Fibonacci[n]+1,{n,0,50}]
%t 5#^2-5#+1&/@Fibonacci[Range[0,30]] (* _Harvey P. Dale_, Nov 29 2011 *)
%o (PARI) a(n)=subst(5*t*(t-1)+1, t, fibonacci(n)) \\ _Charles R Greathouse IV_, Jan 03 2013
%Y Cf. A000032, A000045, A121171, A001946, A124297.
%Y Bisections: A001604, A156094.
%K nonn,easy
%O 0,4
%A _Alexander Adamchuk_, Oct 25 2006
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