A049669 a(n) = Fibonacci(9*n)/34.
0, 1, 76, 5777, 439128, 33379505, 2537281508, 192866774113, 14660412114096, 1114384187445409, 84707858657965180, 6438911642192799089, 489441992665310695944, 37204030354205805690833
Offset: 0
Keywords
Links
- G. C. Greubel, Table of n, a(n) for n = 0..530
- S. Falcon, Generalized Fibonacci Sequences Generated from a k-Fibonacci Sequence, Journal of Mathematics Research Vol. 4, No. 2; April 2012. - From _N. J. A. Sloane_, Sep 22 2012
- Tanya Khovanova, Recursive Sequences
- Shaoxiong Yuan, Generalized Identities of Certain Continued Fractions, arXiv:1907.12459 [math.NT], 2019.
- Index entries for linear recurrences with constant coefficients, signature (76, 1).
Programs
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Magma
[Fibonacci(9*n)/(34): n in [0..30]]; // G. C. Greubel, Dec 02 2017
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Maple
with (combinat):seq(fibonacci(3*n,4)/17, n=0..13); # Zerinvary Lajos, Apr 20 2008
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Mathematica
Fibonacci[9Range[0,20]]/34 (* or *) LinearRecurrence[{76,1},{0,1},20] (* Harvey P. Dale, Jan 20 2013 *) -
MuPAD
numlib::fibonacci(9*n)/34 $ n = 0..25; // Zerinvary Lajos, May 09 2008
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PARI
for(n=0,30, print1(fibonacci(9*n)/34, ", ")) \\ G. C. Greubel, Dec 02 2017
Formula
G.f.: x/(1-76*x-x^2), 76=L(9)=A000032(9) (Lucas).
a(n) = 76*a(n-1) + a(n-2), n>1, a(0)=0, a(1)=1. - Philippe Deléham, Nov 23 2008
a(n) = (9*F(n) + (-1)^n*30*5*F(n)^3 + 27*5^2*F(n)^5 + (-1)^n*9*5^3*F(n)^7 + 5^4*F(n)^9)/34, n >= 0. See the general D. Jennings formula given in a comment on the triangle A111125, where also the reference is given. Here the fifth row (k=4) applies. - Wolfdieter Lang, Sep 01 2012
For n >= 1, a(n) equals the denominator of the continued fraction [76, 76, ..., 76] (with n copies of 76). The numerator of that continued fraction is a(n+1). - Greg Dresden and Shaoxiong Yuan, Jul 26 2019
E.g.f.: exp(38*x)*sinh(17*sqrt(5)*x)/(17*sqrt(5)). - Stefano Spezia, Aug 05 2019
Extensions
More terms from James Sellers, Jan 20 2000