A215271 a(n) = a(n-1)*a(n-2) with a(0)=1, a(1)=8.

1, 8, 8, 64, 512, 32768, 16777216, 549755813888, 9223372036854775808, 5070602400912917605986812821504, 46768052394588893382517914646921056628989841375232, 237142198758023568227473377297792835283496928595231875152809132048206089502588928

Offset: 0

Author

Bruno Berselli, Aug 07 2012

Keywords

• nonn
• easy

Comments

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 := 7*sum {n = 1..inf} 1/8^floor(n*phi) (= 49*sum {n = 1..inf} floor(n/phi)/8^n) = 0.89040 80325 60827 28336 ... = 1/(1 + 1/(8 + 1/(8 + 1/(64 + 1/(512 + 1/(32768 + 1/(16777216 + ...))))))). 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} 1/8^(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

• Bruno Berselli, Table of n, a(n) for n = 0..15
• 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, 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, A010098-A010100, A214706, A214887, A215270, A215272, A014565.

Column k=8 of A244003.

Programs

Magma

[8^Fibonacci(n): n in [0..11]];

Maple

a:= n-> 8^(<<1|1>, <1|0>>^n)[1, 2]:
seq(a(n), n=0..12);  # Alois P. Heinz, Jun 17 2014

Mathematica

RecurrenceTable[{a[0] == 1, a[1] == 8, a[n] == a[n - 1] a[n - 2]}, a[n], {n, 0, 15}]

Formula

a(n) = 8^Fibonacci(n).