A010098 a(n) = a(n-1)*a(n-2) with a(0)=1, a(1)=3.

1, 3, 3, 9, 27, 243, 6561, 1594323, 10460353203, 16677181699666569, 174449211009120179071170507, 2909321189362570808630465826492242446680483, 507528786056415600719754159741696356908742250191663887263627442114881

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn

Comments

From Peter Bala, Nov 01 2013: (Start)

Let phi = (1/2)*(1 + sqrt(5)) denote the golden ratio A001622. This sequence gives the simple continued fraction expansion of the constant c := 2*Sum_{n>=1} 1/3^floor(n*phi) (= 4*Sum_{n>=1} floor(n/phi)/3^n) = 0.768597560593155198508 ... = 1/(1 + 1/(3 + 1/(3 + 1/(9 + 1/(27 + 1/(243 + 1/(6561 + ...))))))). The constant c is known to be transcendental (see Adams and Davison 1977). Cf. A014565.

Furthermore, for k = 0,1,2,... if we put X(k) = sum {n >= 1} 1/3^(n*Fibonacci(k) + Fibonacci(k+1)*floor(n*phi)) then the real number X(k+1)/X(k) has the simple continued fraction expansion [0; a(k+1), a(k+2), a(k+3), ...] (apply Bowman 1988, Corollary 1). (End)

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..17
• W. W. Adams and J. L. Davison, A remarkable class of continued fractions, Proc. Amer. Math. Soc. 65 (1977), 194-198.
• P. G. Anderson, T. C. Brown, and P. J.-S. Shiue, A simple proof of a remarkable continued fraction identity, Proc. Amer. Math. Soc. 123 (1995), 2005-2009.
• D. Bowman, A new generalization of Davison's theorem, Fib. Quart. Volume 26 (1988), 40-45

Crossrefs

Cf. A000045, A000301, A000304, A010099, A010100, A014565, A214706, A214887, A215270, A215271, A215272.

Column k=3 of A244003.

Programs

Haskell

a010098 n = a010098_list !! n
a010098_list = 1 : 3 : zipWith (*) a010098_list (tail a010098_list)
-- Reinhard Zumkeller, Jul 06 2014

Magma

[3^Fibonacci(n): n in [0..12]]; // G. C. Greubel, Jul 29 2024

Maple

a[-1]:=1: a[0]:=3: a[1]:=3: for n from 2 to 13 do a[n]:=a[n-1]*a[n-2] od: seq(a[n], n=-1..10); # Zerinvary Lajos, Mar 19 2009

Mathematica

3^Fibonacci[Range[0,13]] (* Vladimir Joseph Stephan Orlovsky, Jan 21 2012 *)
RecurrenceTable[{a[0]==1,a[1]==3,a[n]==a[n-1]a[n-2]},a,{n,15}] (* Harvey P. Dale, Jan 21 2021 *)

SageMath

[3^fibonacci(n) for n in range(13)] # G. C. Greubel, Jul 29 2024

Formula

a(n) = 3^Fibonacci(n).

a(n+1) = A000304(n+3) / A000301(n). - Reinhard Zumkeller, Jul 06 2014