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A124297 a(n) = 5*F(n)^2 + 5*F(n) + 1, where F(n) = Fibonacci(n).

%I A124297 #22 Feb 16 2025 08:33:03
%S A124297 1,11,11,31,61,151,361,911,2311,5951,15401,40051,104401,272611,712531,
%T A124297 1863551,4875781,12760031,33398201,87424711,228859951,599129311,
%U A124297 1568486161,4106261531,10750188961,28144128251,73681909211,192901135711
%N A124297 a(n) = 5*F(n)^2 + 5*F(n) + 1, where F(n) = Fibonacci(n).
%C A124297 11 = Lucas(5) divides a(1+10k), a(2+10k), and a(9+10k). Last digit of a(n) is 1, or 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 A124297 John Cerkan, <a href="/A124297/b124297.txt">Table of n, a(n) for n = 0..2373</a>
%H A124297 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/AurifeuilleanFactorization.html">Aurifeuillean Factorization</a>
%H A124297 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (4,-2,-6,4,2,-1).
%F A124297 a(n) = 5*Fibonacci(n)^2 + 5*Fibonacci(n) + 1.
%F A124297 G.f.: -(11*x^5-21*x^4-15*x^3+31*x^2-7*x-1) / ((x-1)*(x+1)*(x^2-3*x+1)*(x^2+x-1)). [_Colin Barker_, Jan 03 2013]
%t A124297 Table[5*Fibonacci[n]^2+5*Fibonacci[n]+1,{n,0,50}]
%t A124297 LinearRecurrence[{4,-2,-6,4,2,-1},{1,11,11,31,61,151},30] (* _Harvey P. Dale_, Feb 23 2023 *)
%o A124297 (PARI) a(n)=subst(5*t*(t+1)+1,t,fibonacci(n)) \\ _Charles R Greathouse IV_, Jan 03 2013
%Y A124297 Cf. A000032, A000045, A121171, A001946, A124296.
%Y A124297 Bisections: A001604, A156095.
%K A124297 nonn,easy
%O A124297 0,2
%A A124297 _Alexander Adamchuk_, Oct 25 2006