A305413 a(n) = Fibonacci(11*n)/89.

0, 1, 199, 39602, 7880997, 1568358005, 312111123992, 62111682032413, 12360536835574179, 2459808941961294034, 489514339987133086945, 97415813466381445596089, 19386236394149894806708656, 3857958458249295447980618633, 767753119428003944042949816623

Offset: 0

Author

Vincenzo Librandi, Jun 05 2018

Keywords

• nonn
• easy

Links

• S. Falcon, Generalized Fibonacci Sequences Generated from a k-Fibonacci Sequence, Journal of Mathematics Research, Vol. 4, No. 2 (2012), 97-100.
• Shaoxiong Yuan, Generalized Identities of Certain Continued Fractions, arXiv:1907.12459 [math.NT], 2019.
• Index entries for linear recurrences with constant coefficients, signature (199,1).

Crossrefs

Cf. A000045, A167398.

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).

Programs

Magma

[Fibonacci(11*n)/89: n in [0..30]];

Mathematica

Fibonacci[11 Range[0, 20]]/89
LinearRecurrence[{199,1},{0,1},20] (* Harvey P. Dale, Aug 03 2024 *)

PARI

a(n) = fibonacci(11*n)/89 \\ Felix Fröhlich, Jul 30 2019

Formula

G.f.: x/(1 - 199*x - x^2).

a(n) = 199*a(n-1) + a(n-2) for n>1, a(0)=0, a(1)=1.

a(n) = A167398(n)/89.

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