OFFSET
0,1
COMMENTS
a(n) is also a sequence with the property that the difference between the sum and product of two consecutive terms is equal to the square of the difference between those terms, i.e., a(n)*a(n+1) - (a(n)+ a(n+1)) = (a(n) - a(n + 1))^2. The difference between those two terms, a(n + 1) - a(n) = F(2n -2), the (2n - 2)th Fibonacci number. - John Baker, May 18 2010
Conjecture: consecutive terms of this sequence and consecutive terms of A101265 provide all the positive integer solutions of (a+b)*(a+b+1) == 0 (mod (a*b)). - Robert Israel, Aug 26 2015
Conjecture is true: see Mathematics Stack Exchange link. - Robert Israel, Sep 06 2015
Consecutive terms of this sequence and consecutive terms of A101879 provide all the positive integer pairs for which K = (a+1)/b + (b+1)/a is integer. For this sequence, K = 3. - Andrey Vyshnevyy, Sep 18 2015
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
M. 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.
R. Israel, W. Jagy et al., Diophantine equation (x+y)(x+y+1)-kxy=0, Mathematics Stack Exchange, Sep 1 2015.
H. D. Nguyen and D. Taggart, Mining the OEIS: Ten Experimental Conjectures, 2013. Mentions this sequence. - From N. J. A. Sloane, Mar 16 2014
Index entries for linear recurrences with constant coefficients, signature (4,-4,1).
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
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
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 *)
PROG
(PARI) Vec((2-6*x+3*x^2)/(1-4*x+4*x^2-x^3)+O(x^66)) \\ Joerg Arndt, Jul 02 2013
(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
CROSSREFS
KEYWORD
eigen,nonn,easy
AUTHOR
Michele Elia (elia(AT)polito.it)
EXTENSIONS
More terms from Ralf Stephan, Mar 10 2003
Index for Chebyshev polynomials and cross reference added by Wolfdieter Lang, Aug 27 2014
STATUS
approved