%I #33 Aug 03 2024 19:24:00
%S 0,1,199,39602,7880997,1568358005,312111123992,62111682032413,
%T 12360536835574179,2459808941961294034,489514339987133086945,
%U 97415813466381445596089,19386236394149894806708656,3857958458249295447980618633,767753119428003944042949816623
%N a(n) = Fibonacci(11*n)/89.
%H S. Falcon, <a href="https://doi.org/10.5539/jmr.v4n2p97">Generalized Fibonacci Sequences Generated from a k-Fibonacci Sequence</a>, Journal of Mathematics Research, Vol. 4, No. 2 (2012), 97-100.
%H Shaoxiong Yuan, <a href="https://arxiv.org/abs/1907.12459">Generalized Identities of Certain Continued Fractions</a>, arXiv:1907.12459 [math.NT], 2019.
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (199,1).
%F G.f.: x/(1 - 199*x - x^2).
%F a(n) = 199*a(n-1) + a(n-2) for n>1, a(0)=0, a(1)=1.
%F a(n) = A167398(n)/89.
%F For n >= 1, a(n) equals the denominator of the continued fraction [199, 199, ..., 199] (with n copies of 199). The numerator of that continued fraction is a(n+1). - _Greg Dresden_ and _Shaoxiong Yuan_, Jul 29 2019
%t Fibonacci[11 Range[0, 20]]/89
%t LinearRecurrence[{199,1},{0,1},20] (* _Harvey P. Dale_, Aug 03 2024 *)
%o (Magma) [Fibonacci(11*n)/89: n in [0..30]];
%o (PARI) a(n) = fibonacci(11*n)/89 \\ _Felix Fröhlich_, Jul 30 2019
%Y Cf. A000045, A167398.
%Y Cf. similar sequences: F(3*n)/2 (A001076), F(4*n)/3 (A004187), F(5*n)/5 (A049666), F(6*n)/8 (A049660), F(7*n)/13 (A049667), F(8*n)/21 (A049668), F(9*n)/34 (A049669), F(10*n)/55 (A049670), F(11*n)/89 (this sequence), F(12*n)/144 (A253368).
%K nonn,easy
%O 0,3
%A _Vincenzo Librandi_, Jun 05 2018