OFFSET
0,1
COMMENTS
First n terms give number of digits of Fibonacci(10^n), except that it can be off by 1. This is a highly compressed sequence. As a result, it can be off by one. The uncompressed version goes like this: 2, 21, 209, 2090, 20899, 208988, 2089877, 20898764, 208987640, 2089876403, ... (see A068070). Fibonacci(10) = 55 has 2 digits, Fibonacci(100) = 354224848179261915075 has 21 digits and so on.
Considering the very good approximation F(n) = 5^(-1/2)*phi^n, the number of digits of F(10^n) is given by floor(log_10(F(10^n))) = floor(-(1/2)*log_10(5) + 10^n*log_10(phi)). Similarly L(n) tends to phi^n, so the number of digits of L(10^n) is given by floor(10^n*log_10(phi)). Both numbers can differ at most by 1. F(n) and L(n) denote the Fibonacci and Lucas numbers, resp. - Christoph Pacher (christoph.pacher(AT)arcs.ac.at), Nov 22 2006
Decimal expansion of log_10(phi) = log(phi) / log(10), where phi = golden ratio = (1 + sqrt(5))/2 = A001622. - Jaroslav Krizek, Dec 23 2013
LINKS
Eric Weisstein's World of Mathematics, Fibonacci Number
Eric Weisstein's World of Mathematics, Lucas Number
EXAMPLE
0.20898764024997873376...
Fibonacci(10^9) has 208987640 decimal digits;
Fibonacci(10^21) has 208987640249978733769 decimal digits;
Fibonacci(10^27) has 208987640249978733769272089 decimal digits.
MAPLE
phi := (1+sqrt(5))/2 ; evalf( log(phi)/log(10)) ; # R. J. Mathar, Oct 17 2012
MATHEMATICA
FibonacciDigits[n_] := Ceiling[(2*n*ArcCsch[2] - Log[5])/Log[100]]
RealDigits[ArcCsch[2]/Log[10], 10, 105][[1]] (* Vaclav Kotesovec, Aug 09 2015 *)
PROG
(PARI) solve(x=.1, 1, sinh(x)-.5)/log(10) \\ Charles R Greathouse IV, Aug 04 2020
(PARI) log((1+sqrt(5))/2)/log(10) \\ Charles R Greathouse IV, Aug 04 2020
CROSSREFS
KEYWORD
AUTHOR
Ed Pegg Jr, Aug 06 2004
EXTENSIONS
Offset corrected by Lee A. Newberg, Oct 13 2022
STATUS
approved