A214706 a(n) = a(n-1)*a(n-2) with a(0)=1, a(1)=5.

1, 5, 5, 25, 125, 3125, 390625, 1220703125, 476837158203125, 582076609134674072265625, 277555756156289135105907917022705078125, 161558713389263217748322010169914619837072677910327911376953125

Offset: 0

Author

Vincenzo Librandi, Aug 01 2012

Keywords

• nonn
• easy

Comments

a(17) has 1117 digits.

From Peter Bala, Nov 01 2013: (Start)

Let phi = 1/2*(1 + sqrt(5)) denote the golden ratio A001622. This sequence is the simple continued fraction expansion of the constant c := 4*Sum_{n = 1..oo} 1/5^floor(n*phi) (= 16*Sum_{n = 1..oo} floor(n/phi)/5^n) = 0.83866 83869 91037 14262 ... = 1/(1 + 1/(5 + 1/(5 + 1/(25 + 1/(125 + 1/(3125 + 1/(390625 + ...))))))). The constant c is known to be transcendental (see Adams and Davison 1977). Cf. A014565.

Furthermore, for k = 0,1,2,... if we define the real number X(k) = sum {n >= 1} 1/5^(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..16
• 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, A001622, A010098, A010099, A010100, A014565, A214706, A214887, A215270, A215271, A215272.

Column k=5 of A244003.

Programs

Magma

[5^Fibonacci(n): n in [0..13]];

Maple

a:= n-> 5^(<<1|1>, <1|0>>^n)[1, 2]:
seq(a(n), n=0..12);  # Alois P. Heinz, Jun 17 2014

Mathematica

5^Fibonacci[Range[0,11]]
nxt[{a_,b_}]:={b,a*b}; NestList[nxt,{1,5},12][[All,1]] (* Harvey P. Dale, Oct 14 2018 *)

SageMath

[5^fibonacci(n) for n in range(15)] # G. C. Greubel, Jan 07 2024

Formula

a(n) = 5^Fibonacci(n).