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%I #33 Sep 08 2022 08:45:01
%S 1,10,59,344,2005,11686,68111,396980,2313769,13485634,78600035,
%T 458114576,2670087421,15562409950,90704372279,528663823724,
%U 3081278570065,17959007596666,104672767009931,610077594462920
%N Expansion of (1+4*x)/(1-6*x+x^2).
%C Numbers n such that 8*n^2 + 41 is a square.
%C (x, y) = (a(n), a(n+1)) are solutions to x^2 + y^2 - 6*x*y = 41. - _John O. Oladokun_, Mar 17 2021
%D A. H. Beiler, Recreations in the Theory of Numbers, Dover, N. Y., 1964, pp. 122-125, 194-196.
%H G. C. Greubel, <a href="/A054489/b054489.txt">Table of n, a(n) for n = 0..1000</a>
%H I. Adler, <a href="http://www.fq.math.ca/Scanned/7-2/adler.pdf">Three Diophantine equations - Part II</a>, Fib. Quart., 7 (1969), pp. 181-193.
%H E. I. Emerson, <a href="http://www.fq.math.ca/Scanned/7-3/emerson.pdf">Recurrent Sequences in the Equation DQ^2=R^2+N</a>, Fib. Quart., 7 (1969), pp. 231-242.
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (6,-1).
%F a(n) = 6*a(n-1) - a(n-2), a(0)=1, a(1)=10.
%F a(n) = (10*((3+2*sqrt(2))^n - (3-2*sqrt(2))^n) - ((3+2*sqrt(2))^(n-1) - (3-2*sqrt(2))^(n-1)))/(4*sqrt(2)).
%F From _G. C. Greubel_, Jan 19 2020: (Start)
%F a(n) = ChebyshevU(n,3) + 4*ChebyshevU(n-1,3).
%F a(n) = (Pell(2*n+2) + 4*Pell(2*n))/2 = (Pell-Lucas(2*n+1) + 3*Pell(2*n))/2.
%F E.g.f.: exp(3*x)*( cosh(2*sqrt(2)*x) + 7*sinh(2*sqrt(2)*x)/(2*sqrt(2)) ). (End)
%p a[0]:=1: a[1]:=10: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n], n=0..19); # _Zerinvary Lajos_, Jul 26 2006
%t Table[(LucasL[2*n+1, 2] + 3*Fibonacci[2*n, 2])/2, {n,0,30}] (* _G. C. Greubel_, Jan 19 2020 *)
%o (PARI) vector(31, n, polchebyshev(n-1,2,3) +4*polchebyshev(n-2,2,3) ) \\ _G. C. Greubel_, Jan 19 2020
%o (Magma) I:=[1,10]; [n le 2 select I[n] else 6*Self(n-1) - Self(n-2): n in [1..30]]; // _G. C. Greubel_, Jan 19 2020
%o (Sage) [chebyshev_U(n,3) +4*chebyshev_U(n-1,3) for n in (0..30)] # _G. C. Greubel_, Jan 19 2020
%o (GAP) a:=[1,10];; for n in [3..30] do a[n]:=6*a[n-1]-a[n-2]; od; a; # _G. C. Greubel_, Jan 19 2020
%Y Cf. A000129, A002203, A038761, A054488.
%K easy,nonn
%O 0,2
%A _Barry E. Williams_, May 04 2000
%E More terms from _James A. Sellers_, May 05 2000
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