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A049629 a(n) = (F(6*n+5) - F(6*n+1))/4 = (F(6*n+4) + F(6*n+2))/4, where F = A000045. 24

%I

%S 1,19,341,6119,109801,1970299,35355581,634430159,11384387281,

%T 204284540899,3665737348901,65778987739319,1180356041958841,

%U 21180629767519819,380070979773397901,6820097006153642399,122381675130992165281,2196050055351705332659,39406519321199703822581

%N a(n) = (F(6*n+5) - F(6*n+1))/4 = (F(6*n+4) + F(6*n+2))/4, where F = A000045.

%C x(n) := 2*a(n) and y(n) := A007805(n), n >= 0, give all the positive solutions of the Pell equation x^2 - 5*y^2 = -1.

%C The Gregory V. Richardson formula follows from this. - _Wolfdieter Lang_, Jun 20 2013

%C From _Peter Bala_, Mar 23 2018: (Start)

%C Define a binary operation o on the real numbers by x o y = x*sqrt(1 + y^2) + y*sqrt(1 + x^2). The operation o is commutative and associative with identity 0. Then we have

%C 2*a(n) = 2 o 2 o ... o 2 (2*n+1 terms). For example, 2 o 2 = 4*sqrt(5) and 2 o 2 o 2 = 2 o 4*sqrt(5) = 38 = 2*a(1). Cf. A084068.

%C a(n) = U(2*n+1) where U(n) is the Lehmer sequence [Lehmer, 1930] defined by the recurrence U(n) = sqrt(20)*U(n-1) - U(n-2) with U(0) = 0 and U(1) = 1. The solution to the recurrence is U(n) = (1/4)*( (sqrt(5) + 2)^n - (sqrt(5) - 2)^n ). (End)

%H G. C. Greubel, <a href="/A049629/b049629.txt">Table of n, a(n) for n = 0..795</a>

%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>

%H D. H. Lehmer, <a href="http://www.jstor.org/stable/1968235">An extended theory of Lucas' functions</a>, Annals of Mathematics, Second Series, Vol. 31, No. 3 (Jul., 1930), pp. 419-448.

%H E. W. Weisstein, <a href="http://mathworld.wolfram.com/LehmerNumber.html">MathWorld: Lehmer Number</a>

%H H. C. Williams and R. K. Guy, <a href="http://dx.doi.org/10.1142/S1793042111004587">Some fourth-order linear divisibility sequences</a>, Intl. J. Number Theory 7 (5) (2011) 1255-1277.

%H H. C. Williams and R. K. Guy, <a href="http://www.emis.de/journals/INTEGERS/papers/a17self/a17self.Abstract.html">Some Monoapparitic Fourth Order Linear Divisibility Sequences</a> Integers, Volume 12A (2012) The John Selfridge Memorial Volume.

%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (18,-1).

%F a(n) ~ (1/4)*(sqrt(5) + 2)^(2*n+1). - Joe Keane (jgk(AT)jgk.org), May 15 2002

%F For all members x of the sequence, 20*x^2 + 5 is a square. Lim_{n -> inf} a(n)/a(n-1) = 9 + 2*sqrt(20) = 9 + 4*sqrt(5). The 20 can be seen to derive from the statement "20*x^2 + 5 is a square". - _Gregory V. Richardson_, Oct 12 2002

%F a(n) = (((9 + 4*sqrt(5))^n - (9 - 4*sqrt(5))^n) + ((9 + 4*sqrt(5))^(n-1) - (9 - 4*sqrt(5))^(n-1)) / (8*sqrt(5)). - _Gregory V. Richardson_, Oct 12 2002

%F From _R. J. Mathar_, Nov 04 2008: (Start)

%F G.f.: (1+x)/(1 - 18x + x^2).

%F a(n) = A049660(n) + A049660(n+1). (End)

%F a(n) = 18*a(n-1) - a(n-2) for n>1; a(0)=1, a(1)=19. - _Philippe Deléham_, Nov 17 2008

%F a(n) = S(n,18) + S(n-1,18) with the Chebyshev S-polynomials (A049310). - _Wolfdieter Lang_, Jun 20 2013

%F From _Peter Bala_, Mar 23 2015: (Start)

%F a(n) = ( Fibonacci(6*n + 6 - 2*k) + Fibonacci(6*n + 2*k) )/( Fibonacci(6 - 2*k) + Fibonacci(2*k) ), for k an arbitrary integer.

%F a(n) = ( Fibonacci(6*n + 6 - 2*k - 1) - Fibonacci(6*n + 2*k + 1) )/( Fibonacci(6 - 2*k - 1) - Fibonacci(2*k + 1) ), for k an arbitrary integer, k != 1.

%F The aerated sequence (b(n))n>=1 = [1, 0, 19, 0, 341, 0, 6119, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -16, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials. (End)

%F a(n) = (A188378(n)^3 + (A188378(n)-2)^3) / 8. - _Altug Alkan_, Jan 24 2016

%F a(n) = sqrt(5 * Fibonacci(3 + 6*n)^2 - 4)/4. - _Gerry Martens_, Jul 25 2016

%F a(n) = Lucas(6*n + 3)/4. - _Ehren Metcalfe_, Feb 18 2017

%F From _Peter Bala_, Mar 23 2018: (Start)

%F a(n) = 1/4*( (sqrt(5) + 2)^(2*n+1) - (sqrt(5) - 2)^(2*n+1) ).

%F a(n) = 9*a(n-1) + 2*sqrt(5 + 20*a(n-1)^2).

%F a(n) = (1/2)*sinh((2*n + 1)*arcsinh(2)). (End)

%e Pell, n=1: (2*19)^2 - 5*17^2 = -1.

%p with(numtheory): with(combinat):

%p seq((fibonacci(6*n+5)-fibonacci(6*n+1))/4,n=0..20); # _Muniru A Asiru_, Mar 25 2018

%t a[n_] := Simplify[(2 + Sqrt@5)^(2 n - 1) + (2 - Sqrt@5)^(2 n - 1)]/4; Array[a, 16] (* _Robert G. Wilson v_, Oct 28 2010 *)

%o (PARI) x='x+O('x^30); Vec((1+x)/(1 - 18x + x^2)) \\ _G. C. Greubel_, Dec 15 2017

%o (MAGMA) [(Fibonacci(6*n+5) - Fibonacci(6*n+1))/4: n in [0..30]]; // _G. C. Greubel_, Dec 15 2017

%Y Bisection of A001077 divided by 2.

%Y Cf. A000045, A007805, A049310, A049660, A100047, A188378.

%Y Cf. A005013, A033890, A049685, A084608.

%Y Cf. similar sequences of the type (1/k)*sinh((2*n+1)*arcsinh(k)) listed in A097775.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_

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Last modified December 13 16:00 EST 2018. Contains 318086 sequences. (Running on oeis4.)