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%I #27 Aug 19 2022 05:58:56
%S 2,1,-7,-11,17,61,-7,-251,-223,781,1673,-1451,-8143,-2339,30233,39589,
%T -81343,-239699,85673,1044469,701777,-3476099,-6283207,7621189,
%U 32754017,2269261,-128746807,-137823851,377163377,928458781,-580194727,-4294029851,-1973250943
%N a(n) = 2^(n+1)*cos(n*arctan(sqrt(15))).
%C For n >= 1, |a(n)| is the unique odd positive solution y to 4^(n+1) = 15*x^2 + y^2. The value of x is |A106853(n-1)|. - _Jianing Song_, Jan 22 2019
%H Colin Barker, <a href="/A272931/b272931.txt">Table of n, a(n) for n = 0..1000</a>
%H Paul Barry, <a href="https://arxiv.org/abs/2004.04577">On a Central Transform of Integer Sequences</a>, arXiv:2004.04577 [math.CO], 2020.
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (1,-4).
%F Let a(x) = x/2 - i*sqrt(15)*x/2 and b(x) = x/2 + i*sqrt(15)*x/2, then:
%F a(n) = a(1)^n + b(1)^n.
%F a(n) = n! [x^n] exp(a(x)) + exp(b(x)).
%F a(n) = [x^n] (2 - x)/(4*x^2 - x + 1).
%F a(n) = Sum_{k=0..floor(n/2)} (-4)^k*n*(n - k - 1)!/(k!*(n - 2*k)!) for n >= 1.
%F For n >= 1, 15*a(n)^2 + A106853(n-1)^2 = 4^(n+1). - _Jianing Song_, Jan 22 2019
%F a(n) = a(n-1) - 4*a(n-2) for n>1. - _Colin Barker_, Jan 22 2019
%F a(n) = 2*A106853(n) - A106853(n-1). - _R. J. Mathar_, Aug 19 2022
%p seq(simplify(((1-I*sqrt(15))^n + (1+I*sqrt(15))^n)/2^n), n=0..32);
%t LinearRecurrence[{1, -4}, {2, 1}, 33]
%o (Sage)
%o [lucas_number2(i, 1, 4) for i in range(33)]
%o (PARI) Vec((2 - x) / (1 - x + 4*x^2) + O(x^40)) \\ _Colin Barker_, Jan 22 2019
%Y Cf. A087204, A002249, A106853.
%K sign,easy
%O 0,1
%A _Peter Luschny_, May 11 2016
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