A010098 - OEIS

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

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OFFSET

0,2

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

FORMULA

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

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

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 *)

PROG

(Haskell)a010098 n = a010098_list !! n

a010098_list = 1 : 3 : zipWith (*) a010098_list (tail a010098_list)