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%I #32 Feb 06 2017 10:17:50
%S 1,4,4,16,64,1024,65536,67108864,4398046511104,295147905179352825856,
%T 1298074214633706907132624082305024,
%U 383123885216472214589586756787577295904684780545900544
%N a(n) = a(n-1)*a(n-2) with a(0)=1, a(1)=4.
%C From _Peter Bala_, Nov 01 2013: (Start)
%C Let phi = 1/2*(1 + sqrt(5)) denote the golden ratio A001622. This sequence is the simple continued fraction expansion of the constant c := 3*sum {n = 1..inf} 1/4^floor(n*phi) (= 9*sum {n = 1..inf} floor(n/phi)/4^n) = 0.80938 42984 64421 90504 ... = 1/(1 + 1/(4 + 1/(4 + 1/(16 + 1/(64 + 1/(1024 + 1/(65536 + ...))))))). The constant c is known to be transcendental (see Adams and Davison 1977). Cf. A014565.
%C Furthermore, for k = 0,1,2,... if we define the real number X(k) = sum {n >= 1} 1/4^(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)
%H Indranil Ghosh, <a href="/A010099/b010099.txt">Table of n, a(n) for n = 0..17</a>
%H W. W. Adams and J. L. Davison, <a href="http://www.jstor.org/stable/2041889">A remarkable class of continued fractions</a>, Proc. Amer. Math. Soc. 65 (1977), 194-198.
%H P. G. Anderson, T. C. Brown, P. J.-S. Shiue, <a href="http://people.math.sfu.ca/~vjungic/tbrown/tom-28.pdf">A simple proof of a remarkable continued fraction identity</a>, Proc. Amer. Math. Soc. 123 (1995), 2005-2009.
%H D. Bowman, <a href="http://www.fq.math.ca/26-1.html">A new generalization of Davison's theorem</a>, Fib. Quart. Volume 26 (1988), 40-45
%F a(n) = 4^Fibonacci(n).
%p a[ -1]:=1:a[0]:=4: a[1]:=4: 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
%Y Cf. A000045, A000301, A010098, A010100, A014565, A214706, A214887, A215270, A215271, A215272.
%Y Column k=4 of A244003.
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_
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