A032908 One of four 3rd-order recurring sequences for which the first derived sequence and the Galois transformed sequence coincide.
2, 2, 3, 6, 14, 35, 90, 234, 611, 1598, 4182, 10947, 28658, 75026, 196419, 514230, 1346270, 3524579, 9227466, 24157818, 63245987, 165580142, 433494438, 1134903171, 2971215074, 7778742050, 20365011075, 53316291174, 139583862446, 365435296163, 956722026042
Offset: 0
References
- L. E. Dickson, History of the Theory of Numbers, Dover, New York, 1971.
Links
- Robert Israel, Table of n, a(n) for n = 0..2154
- Michele Elia, A Note on derived linear recurring sequences, pp. 83-92 of Proceedings Seventh Int. Conference on Fibonacci Numbers and their Applications (Austria, 1996), Applications of Fibonacci Numbers, Volume 7.
- Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
- Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
- INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 919.
- Robert Israel, Will Jagy et al., Diophantine equation (x+y)(x+y+1)-kxy=0, Mathematics Stack Exchange, Sep 1 2015.
- Hieu D. Nguyen and Douglas Taggart, Mining the OEIS: Ten Experimental Conjectures, 2013. Mentions this sequence. - From _N. J. A. Sloane_, Mar 16 2014
- Index entries for sequences related to Chebyshev polynomials.
- Index entries for linear recurrences with constant coefficients, signature (4,-4,1).
Programs
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Magma
[2] cat [n le 1 select 2 else Floor((3*Self(n-1) + Sqrt(5*Self(n-1)^2 - 10*Self(n-1) + 1) - 1)/2): n in [1..30]]; // Vincenzo Librandi, Aug 27 2015
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Maple
f:= proc(n) option remember; local x; x:= procname(n-1); (3*x + sqrt(5*x^2 - 10*x + 1) - 1)/2 end proc: f(0):= 2: f(1):= 2: map(f, [$0..30]); # Robert Israel, Aug 26 2015
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Mathematica
Table[Fibonacci[2 n - 1] + 1, {n, 0, 27}] (* Michael De Vlieger, Aug 26 2015 *) LinearRecurrence[{4,-4,1},{2,2,3},40] (* Harvey P. Dale, Apr 11 2018 *)
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PARI
Vec((2-6*x+3*x^2)/(1-4*x+4*x^2-x^3)+O(x^66)) \\ Joerg Arndt, Jul 02 2013
Formula
G.f.: (2 - 6*x + 3*x^2)/((1 - x)*(1 - 3*x + x^2)).
a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3).
a(n) = Fibonacci(2*n-1) + 1 = A001519(n) + 1. - Vladeta Jovovic, Mar 19 2003
a(n) = 3*a(n - 1) - a(n - 2) - 1. - N. Sato, Jan 21 2010
From Wolfdieter Lang, Aug 27 2014: (Start)
a(n) = 1 + S(n-1, 3) - S(n-2, 3) = 1 + A001519(n), with Chebyshev S-polynomials (see A049310). For n < 0, we have S(-1, x) = 0 and S(-2, x) = -1.
This follows from the partial fraction decomposition of the g.f., 1/(1 - x) + (1 - 2*x)/ (1 - 3*x + x^2), using the recurrence for S, or from A001519. (End)
From Robert Israel, Aug 26 2015: (Start)
(a(n) + a(n+1))*(a(n) + a(n+1) + 1) = 5*a(n)*a(n+1).
a(n+1) = (3*a(n) + sqrt(5*a(n)^2 - 10*a(n) + 1) - 1)/2 for n >= 1. (End)
a(n) = 1 + (2^(-1-n) * ((3 - sqrt(5))^n * (1 + sqrt(5)) + (-1 + sqrt(5)) * (3 + sqrt(5))^n)) / sqrt(5). - Colin Barker, Nov 02 2016
Sum_{n>=0} 1/a(n) = phi (A001622). - Amiram Eldar, Oct 05 2020
Product_{n>=1} (1 + 1/a(n)) = (3+sqrt(5))/2 = phi^2 (A104457). - Amiram Eldar, Nov 28 2024
Extensions
More terms from Ralf Stephan, Mar 10 2003
Index for Chebyshev polynomials and cross reference added by Wolfdieter Lang, Aug 27 2014
Comments