A247890 Number of digits in (R_n)^n.

1, 3, 7, 13, 21, 31, 43, 57, 73, 91, 111, 133, 157, 183, 211, 241, 273, 307, 343, 381, 421, 464, 508, 554, 602, 652, 704, 758, 814, 872, 932, 994, 1058, 1124, 1192, 1262, 1334, 1408, 1484, 1562, 1642, 1724, 1808, 1895, 1983, 2073, 2165, 2259, 2355, 2453, 2553, 2655, 2759, 2865

Offset: 1

Author

Derek Orr, Sep 25 2014

Keywords

• nonn
• easy
• base

Comments

R_n is the n-th repunit (i.e., R_n = 11...111 with n 1's).

From David A. Corneth, Jun 27 2016: (Start)

The number of digits of m is floor(log(m)/log(10)) + 1 for m > 0.

R_n = (10^n - 1) / 9 = (10 - 10^(1-n))/9 * 10^(n-1). Its number of digits is floor(n * log((10 - 10^(1-n))/9) / log(10)) + n * (n - 1) + 1. [corrected by Jason Yuen, Nov 11 2024] (End)

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A002061, A014206, A002275.

Cf. A055642, A245593.

Programs

Magma

[#Intseq(Floor((10^n-1)/9)^n): n in [1..50]]; // Marius A. Burtea, May 20 2019

Mathematica

Table[IntegerLength[((10^n - 1)/9)^n], {n, 54}] (* or *)
Table[IntegerLength[FromDigits[Table[1, {n}]]^n], {n, 54}] (* Michael De Vlieger, Jun 27 2016 *)

PARI

vector(100,n,#Str(((10^n-1)/9)^n))

PARI

a(n) = logint(((10 - 10^(1-n))/9)^n\1,10)+n^2-n+1 \\ David A. Corneth, Jun 27 2016

Python

def a(n): return len(str(int("1"*n)**n))
print([a(n) for n in range(1, 55)]) # Michael S. Branicky, Apr 20 2022

Formula

a(n) = n^2 - n + 1 = A002061(n), for 1 <= n <= 21.

a(n) = n^2 - n + 2 = A014206(n-1), for 22 <= n <= 43.

a(n) = A055642(A245593(n)). - Michel Marcus, Apr 20 2022

Extensions

Incorrect conjectures removed by Georg Fischer, May 19 2019