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A272931 a(n) = 2^(n+1)*cos(n*arctan(sqrt(15))). 2
2, 1, -7, -11, 17, 61, -7, -251, -223, 781, 1673, -1451, -8143, -2339, 30233, 39589, -81343, -239699, 85673, 1044469, 701777, -3476099, -6283207, 7621189, 32754017, 2269261, -128746807, -137823851, 377163377, 928458781, -580194727, -4294029851, -1973250943 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,1

COMMENTS

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

LINKS

Colin Barker, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (1,-4).

FORMULA

Let a(x) = x/2 - i*sqrt(15)*x/2 and b(x) = x/2 + i*sqrt(15)*x/2, then:

a(n) = a(1)^n + b(1)^n.

a(n) = n! [x^n] exp(a(x)) + exp(b(x)).

a(n) = [x^n] (2 - x)/(4*x^2 - x + 1).

a(n) = Sum_{k=0..floor(n/2)} (-4)^k*n*(n - k - 1)!/(k!*(n - 2*k)!) for n >= 1.

For n >= 1, 15*a(n)^2 + A106853(n-1)^2 = 4^(n+1). - Jianing Song, Jan 22 2019

a(n) = a(n-1) - 4*a(n-2) for n>1. - Colin Barker, Jan 22 2019

MAPLE

seq(simplify(((1-I*sqrt(15))^n + (1+I*sqrt(15))^n)/2^n), n=0..32);

MATHEMATICA

LinearRecurrence[{1, -4}, {2, 1}, 33]

PROG

(Sage)

[lucas_number2(i, 1, 4) for i in range(33)]

(PARI) Vec((2 - x) / (1 - x + 4*x^2) + O(x^40)) \\ Colin Barker, Jan 22 2019

CROSSREFS

Cf. A087204, A002249, A106853.

Sequence in context: A075118 A100245 A275320 * A095137 A239104 A260259

Adjacent sequences:  A272928 A272929 A272930 * A272932 A272933 A272934

KEYWORD

sign,easy

AUTHOR

Peter Luschny, May 11 2016

STATUS

approved

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Last modified May 27 21:12 EDT 2020. Contains 334671 sequences. (Running on oeis4.)