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%I #63 Sep 08 2022 08:44:58
%S 0,1,123,15128,1860621,228841255,28145613744,3461681649257,
%T 425758697244867,52364858079469384,6440451785077489365,
%U 792123204706451722511,97424713727108484379488,11982447665229637126954513,1473743638109518258131025611,181258485039805516112989195640
%N a(n) = Fibonacci(10*n)/55.
%C Chebyshev polynomials S(n-1,123).
%C Used for all positive integer solutions of Pell equation x^2 - 5*(5*y)^2 = -4. See A097842 with A097843.
%C This is the k = 10 member of the k-family of sequences {F(k*n)/F(k)}, n >= 0 for k >= 1, with o.g.f. x/(1 - L(k)*x + (-1)^k*x^2). Proof: Binet-de Moivre formula for F and L. See also A028412. - _Wolfdieter Lang_, Aug 26 2012
%H Robert Israel, <a href="/A049670/b049670.txt">Table of n, a(n) for n = 0..383</a>
%H Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL23/Nemeth/nemeth7.html">Ellipse Chains and Associated Sequences</a>, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (123,-1).
%F G.f.: x/(1-123*x+x^2), 123=L(10)=A000032(10) (Lucas).
%F a(n+1) = S(n, 123) = U(n, 123/2) = S(2*n+1, 5*sqrt(5))/(5*sqrt(5)), n>=0, with S(n, x) = U(n, x/2) Chebyshev's polynomials of the second kind, A049310. S(-1, x)= 0 = U(-1, x).
%F a(n) = 123*a(n-1) - a(n-2), n >= 2; a(0)=0, a(1)=1.
%F a(n) = (ap^n - am^n)/(ap-am) with ap := (123+55*sqrt(5))/2 and am := (123-55*sqrt(5))/2 = 1/ap.
%F From _Peter Bala_, Nov 29 2013: (Start)
%F a(n) = 1/(11*55)*(F(10*n + 5) - F(10*n - 5)).
%F Sum_{n >= 1} 1/( 11*a(n) + 1/(11*a(n)) ) = 1/11. Compare with A001906 and A049660. (End)
%F From _Peter Bala_, Apr 03 2015: (Start)
%F For integer k, 1 + k*(22 - k)*Sum_{n >= 1} a(n)*x^(2*n) = ( 1 + k/5*Sum_{n >= 1} Fibonacci(5*n)*x^n )*( 1 + k/5*Sum_{n >= 1} Fibonacci(5*n)*(-x)^n ).
%F 1 + 4*Sum_{n >= 1} a(n)*x^(2*n) = ( 1 + 2/5*Sum_{n >= 1} Fibonacci(5*n+5)*x^n )*( 1 + 2/5*Sum_{n >= 1} Fibonacci(5*n+5)*(-x)^n ) = ( 1 + 2/5*Sum_{n >= 1} Fibonacci(5*n-5)*x^n )*( 1 + 2/5*Sum_{n >= 1} Fibonacci(5*n-5)*(-x)^n ).
%F 1 + 25*Sum_{n >= 1} a(n)*x^(2*n) = ( 1 + Sum_{n >= 1} Fibonacci(5*n+3)*x^n )*( 1 + Sum_{n >= 1} Fibonacci(5*n+3)*(-x)^n ) = ( 1 + Sum_{n >= 1} Fibonacci(5*n-3)*x^n )*( 1 + Sum_{n >= 1} Fibonacci(5*n-3)*(-x)^n ).
%F 1 + 100*Sum_{n >= 1} a(n)*x^(2*n) = ( 1 + 2*Sum_{n >= 1} Fibonacci(5*n+1)*x^n )*( 1 + 2*Sum_{n >= 1} Fibonacci(5*n+1)*(-x)^n ) = ( 1 + 2*Sum_{n >= 1} Fibonacci(5*n-1)*x^n )*( 1 + 2*Sum_{n >= 1} Fibonacci(5*n-1)*(-x)^n ).
%F 1 + 125*Sum_{n >= 1} a(n)*x^(2*n) = ( 1 + Sum_{n >= 1} Lucas(5*n)*x^n )*( 1 + Sum_{n >= 1} Lucas(5*n)*(-x)^n ). (End)
%p seq(combinat:-fibonacci(10*n)/55, n=0..20); # _Robert Israel_, Apr 03 2015
%t Table[Fibonacci[10 n]/55, {n, 12}] (* _Michael De Vlieger_, Apr 03 2015 *)
%t LinearRecurrence[{123,-1},{0,1},20] (* _Harvey P. Dale_, Dec 03 2019 *)
%o (MuPAD) numlib::fibonacci(10*n)/55 $ n = 0..25; // _Zerinvary Lajos_, May 09 2008
%o (PARI) a(n)=fibonacci(10*n)/55 \\ _Charles R Greathouse IV_, Oct 07 2016
%o (Magma) [ Fibonacci(10*n)/55: n in [0..30]]; // _G. C. Greubel_, Dec 02 2017
%Y A column of array A028412.
%Y Cf. A000045.
%K nonn,easy
%O 0,3
%A _Clark Kimberling_
%E More terms from _James A. Sellers_, Jan 20 2000
%E Chebyshev and Pell comments from _Wolfdieter Lang_, Sep 10 2004
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