cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A032908 One of four 3rd-order recurring sequences for which the first derived sequence and the Galois transformed sequence coincide.

Original entry on oeis.org

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

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Author

Michele Elia (elia(AT)polito.it)

Keywords

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 an 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.

Crossrefs

Programs

  • 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
  • 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 *)
  • 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