|
|
A010099
|
|
a(n) = a(n-1)*a(n-2) with a(0)=1, a(1)=4.
|
|
8
|
|
|
1, 4, 4, 16, 64, 1024, 65536, 67108864, 4398046511104, 295147905179352825856, 1298074214633706907132624082305024, 383123885216472214589586756787577295904684780545900544
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,2
|
|
COMMENTS
|
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.
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)
|
|
LINKS
|
|
|
FORMULA
|
a(n) = 4^Fibonacci(n).
|
|
MAPLE
|
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
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|