A094057 Number of decimal digits of Lucas(2^n).

1, 1, 1, 2, 4, 7, 14, 27, 54, 108, 215, 429, 857, 1713, 3425, 6849, 13697, 27393, 54785, 109570, 219140, 438279, 876558, 1753116, 3506231, 7012462, 14024924, 28049847, 56099693, 112199386, 224398771, 448797541, 897595081, 1795190161, 3590380321, 7180760642, 14361521283

Offset: 0

Author

Matthijs Coster, Apr 29 2004

Keywords

• nonn
• base

Comments

a(n+1) is the number of decimal digits of A001566(n).

From Hans J. H. Tuenter, Jul 24 2025: (Start)

This sequence can be constructed by taking the first n digits of the binary expansion of

alpha = log_10(phi) = 0.00110 10110 00000 00011 ...

For example, expressing a(n)-1 in binary notation, gives

a(0)-1 = 0,

a(1)-1 = 0,

a(2)-1 = 0,

a(3)-1 = 1,

a(4)-1 = 11,

a(5)-1 = 110,

a(6)-1 = 1101,

a(7)-1 = 11010.

Another way of deriving the sequence is by the recurrence a(n+1)=2a(n)-1+d(n+1), with initial value a(0)=1, and d(n) the n-th digit in the binary expansion of alpha.

a(0) = 1,

a(1) = 2*1-1+0=1,

a(2) = 2*1-1+0=1,

a(3) = 2*1-1+1=2,

a(4) = 2*2-1+1=4,

a(5) = 2*4-1+0=7,

a(6) = 2*7-1+1=14,

a(7) = 2*14-1+0=27.

Alternatively, a(n) provides a encoding of the digits in the binary expansion of alpha,

d(n) = a(n)+1-2a(n-1). (End)

Examples

			a(5)=7, as L(2^5)=L(32)=4870847 and has seven digits.
		

Links

• Hans J. H. Tuenter, Table of n, a(n) for n = 0..100

Crossrefs

Cf. A000032, A001566, A001622, A055642, A114469.

Programs

PARI

a(n) = length(Str(fibonacci(2^(n+1))/fibonacci(2^n))); \\ adapted to new name by Michel Marcus, Jul 24 2025

Formula

a(n) = 1+floor(2^n*log_10(phi)), where phi=(1+sqrt(5))/2, the golden ratio. - Hans J. H. Tuenter, Jul 23 2025

a(n) = 1 + Sum_{i=0..n} d(i)*2^(n-i), where d(i) is the i-th digit in the binary expansion of log_10(phi). - Hans J. H. Tuenter, Jul 24 2025

Extensions

More terms from Jason Earls, Apr 30 2004

a(23)-a(36) from Arkadiusz Wesolowski, Jul 20 2012

Name edited and a(0)=1 inserted by Hans J. H. Tuenter, Jul 23 2025