A369841 n-th digit of the decimal expansion of 1/n, with the digit to the left of the decimal point counted as the first digit.

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

Offset: 1

Author

Itamar Zamir, Feb 03 2024

Keywords

• nonn
• base

Comments

The digits seem to occur with unequal frequency. In the limit as n increases, it seems that the digits, sorted in decreasing order of frequency of occurrence, are 0,5,1,3,7,9,2,6,8,4.

Examples

			a(2) = 5 since 1/2 = 0.5 and the 2nd digit of "0.5" (including the 0) is 5.
a(7) = 7 since 1/7 = 0.142857142857... and its 7th digit is 7.
From _Jon E. Schoenfield_, Feb 03 2024: (Start)
In each row of the following table, the n-th digit is surrounded by spaces:
.
   n         1/n         a(n)
  --  -----------------  ----
   1   1 .0000000000...    1
   2  0. 5 000000000...    5
   3  0.3 3 33333333...    3
   4  0.25 0 0000000...    0
   5  0.200 0 000000...    0
   6  0.1666 6 66666...    6
   7  0.14284 7 1428...    7
   8  0.125000 0 000...    0
   9  0.1111111 1 11...    1
  10  0.10000000 0 0...    0
(End)
		

Links

• Michael S. Branicky, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A061480.

Programs

Mathematica

Table[Mod[Floor[10^(n-1)/n],10],{n,100}] (* James C. McMahon, Feb 04 2024 *)

Python

def a(n): return (10**(n-1)//n)%10
print([a(n) for n in range(1, 101)]) # Michael S. Branicky, Feb 03 2024

Formula

a(n) = floor((10^(n-1))/n) mod 10.