A114469 -id:A114469 - cp's OEIS frontend

A097348 Decimal expansion of arccsch(2)/log(10).

2, 0, 8, 9, 8, 7, 6, 4, 0, 2, 4, 9, 9, 7, 8, 7, 3, 3, 7, 6, 9, 2, 7, 2, 0, 8, 9, 2, 3, 7, 5, 5, 5, 4, 1, 6, 8, 2, 2, 4, 5, 9, 2, 3, 9, 9, 1, 8, 2, 1, 0, 9, 5, 3, 5, 3, 9, 2, 8, 7, 5, 6, 1, 3, 9, 7, 4, 1, 0, 4, 8, 5, 3, 4, 9, 6, 7, 4, 5, 9, 6, 3, 2, 7, 7, 6, 5, 8, 5, 5, 6, 2, 3, 5, 1, 0, 3, 5, 3, 5, 1, 4, 5, 0

Offset: 0

Ed Pegg Jr, Aug 06 2004

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

			0.20898764024997873376...
Fibonacci(10^9)  has 208987640 decimal digits;
Fibonacci(10^21) has 208987640249978733769 decimal digits;
Fibonacci(10^27) has 208987640249978733769272089 decimal digits.

Eric Weisstein's World of Mathematics, Fibonacci Number
Eric Weisstein's World of Mathematics, Lucas Number

Cf. A000045, A001622, A068070, A114469.

phi := (1+sqrt(5))/2 ; evalf( log(phi)/log(10)) ; # R. J. Mathar, Oct 17 2012

FibonacciDigits[n_] := Ceiling[(2*n*ArcCsch[2] - Log[5])/Log[100]]
RealDigits[ArcCsch[2]/Log[10], 10, 105][[1]] (* Vaclav Kotesovec, Aug 09 2015 *)

solve(x=.1,1,sinh(x)-.5)/log(10) \\ Charles R Greathouse IV, Aug 04 2020

log((1+sqrt(5))/2)/log(10) \\ Charles R Greathouse IV, Aug 04 2020

Equals log_10(phi) where phi = (1+sqrt(5))/2. - Jaroslav Krizek, Dec 23 2013

Offset corrected by Lee A. Newberg, Oct 13 2022

A068070 Number of digits in 10^n-th Fibonacci number.

Original entry on oeis.org

1, 2, 21, 209, 2090, 20899, 208988, 2089877, 20898764, 208987640, 2089876403, 20898764025, 208987640250, 2089876402500, 20898764024998, 208987640249979, 2089876402499787, 20898764024997874, 208987640249978734, 2089876402499787338, 20898764024997873377, 208987640249978733769

Offset: 0

Robert G. Wilson v, Feb 26 2002

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:

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.

Cf. A000045, A000032, A001622, A055642, A097348, A114469, A250489.

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

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}]

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.
a(n) = A055642(A250489(n)). - Alois P. Heinz, Jul 09 2025

Corrected by Vladeta Jovovic, Sep 01 2004

a(19)-a(21) from Hans J. H. Tuenter, Jul 21 2025

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

Original entry on oeis.org

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

Matthijs Coster, Apr 29 2004

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)

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

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

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

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

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

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

A386758 Number of decimal digits in the n-th Lucas number.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 10, 10, 10, 10, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 14, 14, 14, 14, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 17, 17, 17, 17, 17

Offset: 0

Hans J. H. Tuenter, Aug 06 2025

As F(n)<=L(n), the number of decimal digits of the Lucas number L(n) is at least as large as the number of decimal digits of the Fibonacci number F(n). Furthermore, the difference is at most one. The indices for which the difference is one is A386760.

			L(0)=2 has one digit, so that a(0)=1; L(5)=11 has two digits, so that a(5)=2.

Hans J. H. Tuenter, Table of n, a(n) for n = 0..10000

Cf. A000032, A001622, A055642, A060384, A097348.

Number of digits of L(p^n): A094057 (p=2), A114469 (p=10).

a:= n-> 1+floor(n*log[10]((1+sqrt(5))/2)):
seq(a(n), n=0..81);

a[n_] := IntegerLength[LucasL[n]]; Array[a, 100, 0] (* Amiram Eldar, Aug 16 2025 *)

a(n) = A055642(A000032(n)).

a(n) = 1 + floor(n*log_10(phi)), where log_10(phi) = A097348, and phi = (1+sqrt(5))/2 = A001622.

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A097348 Decimal expansion of arccsch(2)/log(10).

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A068070 Number of digits in 10^n-th Fibonacci number.

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

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A386758 Number of decimal digits in the n-th Lucas number.

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