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A049667
a(n) = Fibonacci(7*n)/13.
10
0, 1, 29, 842, 24447, 709805, 20608792, 598364773, 17373187209, 504420793834, 14645576208395, 425226130837289, 12346203370489776, 358465123875040793, 10407834795746672773, 302185674200528551210, 8773792386611074657863
OFFSET
0,3
LINKS
FORMULA
G.f.: x/(1 - 29*x - x^2).
a(n) = A134498(n)/13.
a(n) = F(n, 29), the n-th Fibonacci polynomial evaluated at x=29. - T. D. Noe, Jan 19 2006
a(n) = 29*a(n-1) + a(n-2), n > 1; a(0)=0, a(1)=1. - Philippe Deléham, Nov 22 2008
For n >= 1, a(n) equals the denominator of the continued fraction [29, 29, ..., 29] (with n copies of 29). The numerator of that continued fraction is a(n+1). - Greg Dresden and Shaoxiong Yuan, Jul 26 2019
a(n) = ((-1)^n*7*F(n) + 14*5*F(n)^3 + (-1)^n*7*5^2*F(n)^5 + 5^3*F(n)^7)/13, n >= 0. See the general D. Jennings formula given in comment on triangle A111125, where also the reference is given. Here the fourth row (k=3) applies. - Wolfdieter Lang, Sep 01 2012
G.f.: G(0)*x/(2-29*x), where G(k)= 1 + 1/(1 - (x*(845*k-841))/((x*(845*k+4)) - 58/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 15 2013
O.g.f.: x*exp(Sum_{n >= 1} Lucas(7*n)*x^n/n) = x + 29*x^2 + 842*x^3 + .... - Peter Bala, Oct 11 2019
MAPLE
a:= n-> (<<0|1>, <1|29>>^n)[1, 2]:
seq(a(n), n=0..20); # Alois P. Heinz, Sep 20 2017
MATHEMATICA
Fibonacci[(7*Range[0, 20])]/13 (* or *) LinearRecurrence[{29, 1}, {0, 1}, 20] (* Harvey P. Dale, Sep 17 2017 *)
PROG
(MuPAD) numlib::fibonacci(7*n)/13 $ n = 0..25; // Zerinvary Lajos, May 09 2008
(Sage) [fibonacci(7*n)/13 for n in range(0, 17)] # Zerinvary Lajos, May 15 2009
(PARI) a(n)=fibonacci(7*n)/13 \\ Charles R Greathouse IV, Oct 07 2016
(Magma) [Fibonacci(7*n)/13: n in [0..30]]; // G. C. Greubel, Dec 02 2017
CROSSREFS
A column of array A028412.
Sequence in context: A009973 A278475 A057687 * A042626 A157877 A158665
KEYWORD
nonn,easy
STATUS
approved