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 A032908 One of 4 3rd-order recurring sequences for which the first derived sequence and the Galois transformed sequence coincide. 8
 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 (list; graph; refs; listen; history; text; internal format)
 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 Math 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 Guo-Niu Han, Enumeration of Standard Puzzles [Cached copy] INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 919 R. Israel, W. Jagy et al, Diophantine equation (x+y)(x+y+1)-kxy=0, Math StackExchange, Sep 1 2015. H. D. Nguyen, 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 a(n) = 1 + S(n-1, 3) - S(n-2, 3) =  1 + A001519(n), with Chebyshev S-polynomials (see A049310). S(-1, x) = 0 and   S(-2, x) = -1. 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. - Wolfdieter Lang, Aug 27 2014 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 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)  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 Cf. A001519, A049310, A101265. Sequence in context: A095902 A103687 A166678 * A192366 A318039 A060631 Adjacent sequences:  A032905 A032906 A032907 * A032909 A032910 A032911 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

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Last modified May 21 11:54 EDT 2019. Contains 323443 sequences. (Running on oeis4.)