%I #24 Sep 08 2022 08:46:19
%S 0,1,2,3,4,3,4,5,6,7,8,7,8,9,10,11,12,11,12,13,14,15,16,15,16,17,18,
%T 19,20,19,20,21,22,23,24,23,24,25,26,27,28,27,28,29,30,31,32,31,32,33,
%U 34,35,36,35,36,37,38,39,40,39,40,41,42,43,44,43,44,45
%N a(n) is the number of zeros of the Chebyshev S(n, x) polynomial (A049310) in the open interval (-sqrt(3), +sqrt(3)).
%C See a May 06 2017 comment on A049310 where these problems are considered which originated in a conjecture by _Michel Lagneau_ (see A008611) on Fibonacci polynomials.
%H G. C. Greubel, <a href="/A285872/b285872.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,0,0,1,-1).
%F a(n) = 2*b(n) if n is even and 1 + 2*b(n) if n is odd with b(n) = floor(n/2) - floor((n+1)/6) = A285870(n). See the g.f. for {b(n)}_{n>=0} there.
%F From _Colin Barker_, May 18 2017: (Start)
%F G.f.: x*(1 + x + x^2 + x^3 - x^4 + x^5) / ((1 - x)^2*(1 + x)*(1 - x + x^2)*(1 + x + x^2)).
%F a(n) = a(n-1) + a(n-6) - a(n-7) for n>6.
%F (End)
%e n = 3: S(3, x) = x*(-2 + x^2), with all three zeros (-sqrt(2), 0, +sqrt(2)) in the interval (-sqrt(3), +sqrt(3)).
%e n = 4: S(4, x) = 1 - 3*x^2 + x^4, all four zeros (-phi, -1/phi, 1/phi, phi) with phi = (1 + sqrt(5))/2, approximately 1.618, lie in the interval.
%e n = 6, two zeros of S(6, x) = -1 + 6*x^2 - 5*x^4 + x^6 are out of the interval (-sqrt(3), +sqrt(3)), namely - 1.8019... and +1.8019... .
%t CoefficientList[Series[x*(1+x+x^2+x^3-x^4+x^5)/((1-x)^2*(1+x)*(1-x+x^2)*(1+x+x^2)), {x, 0, 50}], x] (* _G. C. Greubel_, Mar 08 2018 *)
%o (PARI) concat(0, Vec(x*(1 + x + x^2 + x^3 - x^4 + x^5) / ((1 - x)^2*(1 + x)*(1 - x + x^2)*(1 + x + x^2)) + O(x^100))) \\ _Colin Barker_, May 18 2017
%o (Magma) m:=80; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!(x*(1+x+x^2+x^3-x^4+x^5)/((1-x)^2*(1+x)*(1-x+x^2)*(1+x+x^2)))); // _G. C. Greubel_, Mar 08 2018
%Y Cf. A008611(n-1) (1), A049310, A285869 (sqrt(2)), A285870.
%K nonn
%O 0,3
%A _Wolfdieter Lang_, May 12 2017