OFFSET
0,2
COMMENTS
a(n)/10^n tends towards log_10((1+sqrt(5))/2) = 0.2089876402499787337692... = A097348. - Vladeta Jovovic, Mar 09 2002
This sequence is very close to A114469, the number of decimal digits in the Lucas number L(10^n). In fact, a(n) = A114469(n) or a(n) = A114469(n) - 1. The first difference occurs at a(1) as F(10) = 55 and L(10) = 123. The next differences occur at indices 8, 9, 16, 21, 23, 24, 27, 34, ... . - Hans J. H. Tuenter, Jul 21 2025
LINKS
Alois P. Heinz, Table of n, a(n) for n = 0..1000 (first 101 terms from Hans J. H. Tuenter)
Eric Weisstein's World of Mathematics, Fibonacci Number.
Eric Weisstein's World of Mathematics, Lucas Number.
FORMULA
Can be calculated easily using the fact that Fibonacci(n) is very close to GoldenRatio^n/(2*GoldenRatio-1) (cf. A000045). - Michael Taktikos, Aug 11 2004.
a(n) = 1+floor(10^n*log_10(phi)-log_10(5)/2), for n>0, where phi=(1+sqrt(5))/2, the golden ratio. - Hans J. H. Tuenter, Jul 08 2025.
MAPLE
a:= n-> `if`(n=0, 1, floor(10^n*log[10]((1+sqrt(5))/2)-log[10](5)/2)+1):
seq(a(n), n=0..21); # Alois P. Heinz, Jul 30 2025
MATHEMATICA
Table[ Floor[ Log[10, Fibonacci[10^n]] + 1], {n, 0, 7} ]
fib10[n_] := Block[{a = N[Log[10, Sqrt[5]/5], 64], b = N[Log[10, (1 + Sqrt[5])/2], 64]}, Floor[a + 10^n*b] + 1]; Table[ fib10[n], {n, 19}] (* Robert G. Wilson v, May 20 2005 *)
FibonacciDigits[n_] := Ceiling[(2*n*ArcCsch[2] - Log[5])/Log[100]]; Table[ FibonacciDigits[10^n], {n, 19}]
IntegerLength[Fibonacci[10^Range[0, 21]]] (* Harvey P. Dale, Nov 22 2025 *)
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Robert G. Wilson v, Feb 26 2002
EXTENSIONS
Corrected by Vladeta Jovovic, Sep 01 2004
a(19)-a(21) from Hans J. H. Tuenter, Jul 21 2025
STATUS
approved