A385248 Number of digits in the decimal expansion of Fibonacci(2^n).

1, 1, 1, 2, 3, 7, 14, 27, 54, 107, 214, 428, 856, 1712, 3424, 6848, 13696, 27393, 54785, 109570, 219140, 438279, 876558, 1753116, 3506231, 7012462, 14024923, 28049846, 56099693, 112199385, 224398770, 448797540, 897595080, 1795190160, 3590380321, 7180760641

Offset: 0

Author

Juande Santander-Vela, Jul 28 2025

Keywords

• nonn
• base

Comments

Binet's formula is Fibonacci(k) = (phi^k - psi^k)/sqrt(5), with phi being the golden ratio (1 + sqrt(5))/2, and psi = (1 - sqrt(5))/2. For even values of k, Fibonacci(k) = floor((phi^k)/sqrt(5)) since psi^(2*k)/sqrt(5) < 0.17^k for all k > 0, and from which the formula below.

Links

• Paolo Xausa, Table of n, a(n) for n = 0..3000

Crossrefs

Cf. A000079, A000045, A001622, A055642, A058635, A060384, A068070.

Programs

Maple

a:= n-> `if`(n=0, 1, floor(2^n*log[10]((1+sqrt(5))/2)-log[10](5)/2)+1):
seq(a(n), n=0..35);  # Alois P. Heinz, Jul 30 2025

Mathematica

a[n_] := IntegerLength[Fibonacci[2^n]]; Array[a, 30, 0] (* Amiram Eldar, Jul 30 2025 *)
A385248[n_] := If[n == 0, 1, Floor[2^n*Log10[GoldenRatio] - Log10[5]/2] + 1];
Array[A385248, 50, 0] (* Paolo Xausa, Aug 07 2025 *)

PARI

a(n) = #Str(fibonacci(2^n)); \\ Michel Marcus, Jul 30 2025

Python

from sympy import Rational, log, sqrt # uses symbolic computation
phi = (1+sqrt(5))/2
def a(n): return 1 if n==0 or n==1 else int(2**n *log(phi)/log(10)-Rational(1,2)*log(5)/log(10))+1

Formula

a(n) = A055642(A058635(n)).

a(n) = A060384(A000079(n)).

a(n) = floor(2^n * log10(phi) - (1/2) * log10(5)) + 1, for n >= 1.

Limit_{n->oo} a(n+1)/a(n) = 2.

Extensions

More terms from Michel Marcus, Jul 30 2025

a(29)-a(35) from Amiram Eldar, Jul 30 2025