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%I #34 Sep 08 2022 08:46:11
%S 1,3,4,3,1,0,1,3,4,3,1,0,1,3,4,3,1,0,1,3,4,3,1,0,1,3,4,3,1,0,1,3,4,3,
%T 1,0,1,3,4,3,1,0,1,3,4,3,1,0,1,3,4,3,1,0,1,3,4,3,1,0,1,3,4,3,1,0,1,3,
%U 4,3,1,0,1,3,4,3,1,0,1,3,4,3,1,0,1,3,4
%N Chebyshev polynomials of the second kind, U(n,x)^2, evaluated at x = sqrt(3)/2.
%C Period 6: repeat [1, 3, 4, 3, 1, 0].
%H Vincenzo Librandi, <a href="/A254745/b254745.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (2,-2,1).
%F Euler transform of length 6 sequence [3, -2, -1, 0, 0, 1].
%F G.f.: (1 + x) / ((1 - x) * (1 - x + x^2)) = (1 - x^2)^2 * (1 - x^3) / ((1 - x)^3 * (1 - x^6)).
%F a(n) = a(-2-n) = a(n+6) for all n in Z.
%F a(n) = (-1)^n*A078070(n) = A131027(n-1) for all n in Z.
%F a(n) = (n+1)*(Sum_{k=0..n} (-1)^k/(k+1)*binomial(n+k+1,2*k+1)) for n >= 0. - _Werner Schulte_, Jul 10 2017
%F Sum_{n>=0} a(n)/(n+1)*x^(n+1) = log(1-x+x^2)-2*log(1-x) for -1 < x < 1. - _Werner Schulte_, Jul 10 2017
%F a(n) = sqrt(3)*sin(Pi*n/3) - cos(Pi*n/3) + 2. - _Peter Luschny_, Jul 16 2017
%F a(n) = 2 + 2*cos(Pi/3*(n+4)) for n >= 0. - _Werner Schulte_, Jul 18 2017
%e G.f. = 1 + 3*x + 4*x^2 + 3*x^3 + x^4 + x^6 + 3*x^7 + 4*x^8 + 3*x^9 + ...
%t a[ n_] := {3, 4, 3, 1, 0, 1}[[Mod[n, 6, 1]]];
%t a[ n_] := ChebyshevU[ n, Sqrt[3] / 2]^2;
%t CoefficientList[Series[(1 + x) / ((1 - x) (1 - x + x^2)), {x, 0, 100}], x] (* _Vincenzo Librandi_, Jul 14 2017 *)
%o (PARI) {a(n) = [1, 3, 4, 3, 1, 0][n%6 + 1]};
%o (PARI) {a(n) = simplify( polchebyshev( n, 2, quadgen(12) / 2)^2)};
%o (Magma) m:=60; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!((1+x)/((1-x)*(1-x+x^2)))); // _G. C. Greubel_, Aug 03 2018
%Y Cf. A078070, A131027, A254744, A254612, A254707, A254708, A254875.
%K nonn,easy
%O 0,2
%A _Michael Somos_, Feb 07 2015
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