OFFSET
0,2
COMMENTS
Hypotenuse (z) of Pythagorean triples (x,y,z) with |2x-y|=1.
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
Positive numbers n such that 5*n^2 - 1 is a square (A075796(n+1)^2). - Gregory V. Richardson, Oct 13 2002
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..795 (terms 0..100 from T. D. Noe)
A. J. C. Cunningham, Binomial Factorisations, Vols. 1-9, Hodgson, London, 1923-1929. See Vol. 1, page xxxv.
Tanya Khovanova, Recursive Sequences
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum (2019) Vol. 19, 11-16.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume
Index entries for linear recurrences with constant coefficients, signature (18,-1).
FORMULA
G.f.: (1-x)/(1-18*x+x^2).
a(n) = 18*a(n-1) - a(n-2), n > 1, a(0)=1, a(1)=17.
a(n+1) = 9*a(n) + 4*sqrt(5*a(n)^2-1). - Richard Choulet, Aug 30 2007, Dec 28 2007
a(n) = ((2+sqrt(5))^(2*n+1) - (2-sqrt(5))^(2*n+1))/(2*sqrt(5)). - Dean Hickerson, Dec 09 2002
a(n) ~ (1/10)*sqrt(5)*(sqrt(5) + 2)^(2*n+1). - Joe Keane (jgk(AT)jgk.org), May 15 2002
Limit_{n->infinity} a(n)/a(n-1) = 8*phi + 5 = 9 + 4*sqrt(5). - Gregory V. Richardson, Oct 13 2002
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
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
a(n) = 17*a(n-1) + 17*a(n-2) - a(n-3). - Antonio Alberto Olivares, Jun 19 2008
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
a(n) = S(n,18) - S(n-1,18) with the Chebyshev S-polynomials (A049310). - Wolfdieter Lang, Jun 20 2013
Sum_{n >= 1} 1/( a(n) - 1/a(n) ) = 1/4^2. Compare with A001519 and A097843. - Peter Bala, Nov 29 2013
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
From Peter Bala, Mar 23 2015: (Start)
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.
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.
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)
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
MAPLE
seq(combinat:-fibonacci(6*n+3)/2, n=0..30); # Robert Israel, Sep 10 2014
MATHEMATICA
LinearRecurrence[{18, -1}, {1, 17}, 50] (* Sture Sjöstedt, Nov 29 2011 *)
Table[Fibonacci[6n+3]/2, {n, 0, 20}] (* Harvey P. Dale, Dec 17 2011 *)
CoefficientList[Series[(1-x)/(1-18*x+x^2), {x, 0, 50}], x] (* G. C. Greubel, Dec 19 2017 *)
PROG
(Haskell)
a007805 = (`div` 2) . a000045 . (* 3) . (+ 1) . (* 2)
-- Reinhard Zumkeller, Mar 26 2013
(PARI) a(n)=fibonacci(6*n+3)/2 \\ Edward Jiang, Sep 09 2014
(PARI) x='x+O('x^30); Vec((1-x)/(1-18*x+x^2)) \\ G. C. Greubel, Dec 19 2017
(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
CROSSREFS
KEYWORD
nonn,nice,easy
AUTHOR
EXTENSIONS
Better description and more terms from Michael Somos
STATUS
approved