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%I #116 Feb 25 2023 03:07:51
%S 1,17,305,5473,98209,1762289,31622993,567451585,10182505537,
%T 182717648081,3278735159921,58834515230497,1055742538989025,
%U 18944531186571953,339945818819306129,6100080207560938369
%N a(n) = Fibonacci(6*n + 3)/2.
%C Hypotenuse (z) of Pythagorean triples (x,y,z) with |2x-y|=1.
%C x(n) := 2*A049629(n) and y(n) := a(n), n >= 0, give all positive solutions of the Pell equation x^2 - 5*y^2 = -1. See the Gregory V. Richardson formula, where his x is the y here and A075796(n+1) = x(n). - _Wolfdieter Lang_, Jun 20 2013
%C Positive numbers n such that 5*n^2 - 1 is a square (A075796(n+1)^2). - _Gregory V. Richardson_, Oct 13 2002
%H G. C. Greubel, <a href="/A007805/b007805.txt">Table of n, a(n) for n = 0..795</a> (terms 0..100 from T. D. Noe)
%H A. J. C. Cunningham, <a href="https://archive.org/details/binomialfactoris01cunn/page/n46/mode/1up">Binomial Factorisations</a>, Vols. 1-9, Hodgson, London, 1923-1929. See Vol. 1, page xxxv.
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H Giovanni Lucca, <a href="http://forumgeom.fau.edu/FG2019volume19/FG201902index.html">Integer Sequences and Circle Chains Inside a Hyperbola</a>, Forum Geometricorum (2019) Vol. 19, 11-16.
%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/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (18,-1).
%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>
%F G.f.: (1-x)/(1-18*x+x^2).
%F a(n) = 18*a(n-1) - a(n-2), n > 1, a(0)=1, a(1)=17.
%F a(n) = A134495(n)/2 = A001076(2n+1).
%F a(n+1) = 9*a(n) + 4*sqrt(5*a(n)^2-1). - _Richard Choulet_, Aug 30 2007, Dec 28 2007
%F a(n) = ((2+sqrt(5))^(2*n+1) - (2-sqrt(5))^(2*n+1))/(2*sqrt(5)). - _Dean Hickerson_, Dec 09 2002
%F a(n) ~ (1/10)*sqrt(5)*(sqrt(5) + 2)^(2*n+1). - Joe Keane (jgk(AT)jgk.org), May 15 2002
%F Limit_{n->infinity} a(n)/a(n-1) = 8*phi + 5 = 9 + 4*sqrt(5). - _Gregory V. Richardson_, Oct 13 2002
%F Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then a(n) = q(n, 16). - _Benoit Cloitre_, Dec 06 2002
%F a(n) = 19*a(n-1)- 19*a(n-2) + a(n-3); f(x) = (sqrt(5)/10)*((2+sqrt(5))*(9+4*sqrt(5))^(x-1) - (2-sqrt(5))*(9-4*sqrt(5))^(x-1)). - _Antonio Alberto Olivares_, May 15 2008
%F a(n) = 17*a(n-1) + 17*a(n-2) - a(n-3). - _Antonio Alberto Olivares_, Jun 19 2008
%F a(n) = b(n+1) - b(n), n >= 0, with b(n) := F(6*n)/F(6) = A049660(n). First differences. See the o.g.f.s. - _Wolfdieter Lang_, 2012
%F a(n) = S(n,18) - S(n-1,18) with the Chebyshev S-polynomials (A049310). - _Wolfdieter Lang_, Jun 20 2013
%F Sum_{n >= 1} 1/( a(n) - 1/a(n) ) = 1/4^2. Compare with A001519 and A097843. - _Peter Bala_, Nov 29 2013
%F a(n) = 9*a(n-1) + 8*A049629(n-1), n >= 1, a(0) = 1. This is just the rewritten Chebyshev S(n, 18) recurrence. - _Wolfdieter Lang_, Aug 26 2014
%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.
%F The aerated sequence (b(n)) n>=1 = [1, 0, 17, 0, 305, 0, 5473, 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 = -20, 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) = sqrt(2 + (9-4*sqrt(5))^(1+2*n) + (9+4*sqrt(5))^(1+2*n))/(2*sqrt(5)). - _Gerry Martens_, Jun 04 2015
%p seq(combinat:-fibonacci(6*n+3)/2, n=0..30); # _Robert Israel_, Sep 10 2014
%t LinearRecurrence[{18, -1}, {1, 17}, 50] (* _Sture Sjöstedt_, Nov 29 2011 *)
%t Table[Fibonacci[6n+3]/2, {n, 0, 20}] (* _Harvey P. Dale_, Dec 17 2011 *)
%t CoefficientList[Series[(1-x)/(1-18*x+x^2), {x, 0, 50}], x] (* _G. C. Greubel_, Dec 19 2017 *)
%o (Haskell)
%o a007805 = (`div` 2) . a000045 . (* 3) . (+ 1) . (* 2)
%o -- _Reinhard Zumkeller_, Mar 26 2013
%o (PARI) a(n)=fibonacci(6*n+3)/2 \\ _Edward Jiang_, Sep 09 2014
%o (PARI) x='x+O('x^30); Vec((1-x)/(1-18*x+x^2)) \\ _G. C. Greubel_, Dec 19 2017
%o (Magma) I:=[1, 17]; [n le 2 select I[n] else 18*Self(n-1) - Self(n-2): n in [1..30]]; // _G. C. Greubel_, Dec 19 2017
%Y Cf. A000045.
%Y Row 18 of array A094954.
%Y Row 2 of array A188647.
%Y Cf. similar sequences listed in A238379.
%K nonn,nice,easy
%O 0,2
%A _James A. Raymond_, _Clark Kimberling_
%E Better description and more terms from _Michael Somos_
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