A215270 - OEIS

%I #29 Jan 05 2025 19:51:39

%S 1,6,6,36,216,7776,1679616,13060694016,21936950640377856,

%T 286511799958070431838109696,

%U 6285195213566005335561053533150026217291776,1800782593726645086383198950649858141454002621435149880441896326660096

%N a(n) = a(n-1)*a(n-2) with a(0)=1, a(1)=6.

%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 := 5*sum {n = 1..inf} 1/6^floor(n*phi) (= 25*sum {n = 1..inf} floor(n/phi)/6^n) = 0.86045 01626 86090 61353 ... = 1/(1 + 1/(6 + 1/(6 + 1/(36 + 1/(216 + 1/(7776 + 1/(1679616 + ...))))))). 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} 1/6^(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 Bruno Berselli, <a href="/A215270/b215270.txt">Table of n, a(n) for n = 0..15</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="https://web.archive.org/web/2024*/https://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) = 6^Fibonacci(n).

%p a:= n-> 6^(<<1|1>, <1|0>>^n)[1, 2]:

%p seq(a(n), n=0..12);  # _Alois P. Heinz_, Jun 17 2014

%t RecurrenceTable[{a[0] == 1, a[1] == 6, a[n] == a[n - 1] a[n - 2]}, a[n], {n, 0, 15}]

%o (Magma) [6^Fibonacci(n): n in [0..11]];

%Y Cf. A000045, A000301, A010098-A010100, A214706, A214887, A215271, A215272.

%Y Cf. A166470 (same recurrence with initial values 2, 6). A014565.

%Y Column k=6 of A244003.

%K nonn,easy

%O 0,2

%A _Bruno Berselli_, Aug 07 2012