A333442 For any n > 0, let Sum_{k >= 0} d_k / 10^k be the decimal representation of 1/n; a(n) is the least m such that d_m = max_{k >= 0} d_k.

0, 1, 1, 2, 1, 2, 4, 3, 1, 1, 2, 2, 4, 6, 2, 2, 9, 2, 10, 2, 6, 3, 12, 4, 2, 3, 3, 8, 15, 2, 15, 5, 2, 3, 3, 3, 3, 9, 4, 3, 5, 6, 10, 4, 2, 6, 8, 4, 22, 2, 3, 3, 7, 3, 3, 4, 9, 9, 4, 3, 5, 6, 4, 4, 5, 3, 4, 13, 5, 5, 35, 4, 5, 4, 3, 8, 4, 4, 6, 4, 9, 5, 8, 4

Offset: 1

Author

Rémy Sigrist, Mar 21 2020

Keywords

• nonn
• base

Comments

In other words, a(n) is the position of the first occurrence of the largest digit in the decimal representation of 1/n (A333236).

Examples

			The first terms, alongside 1/n with the first occurrence of A333236(n) in parentheses, are:
  n   a(n)  1/n
  --  ----  ---------------
   1     0  (1)
   2     1    0.(5)
   3     1    0.(3)33333...
   4     2    0.2(5)
   5     1    0.(2)
   6     2    0.1(6)6666...
   7     4    0.142(8)57...
   8     3    0.12(5)
   9     1    0.(1)11111...
  10     1    0.(1)
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 1..10000
• Rémy Sigrist, PARI program for A333442
• Index entries for sequences related to decimal expansion of 1/n

Crossrefs

Cf. A333236.

Programs

PARI

See Links section.

Python

from sympy import n_order, multiplicity
def A333442(n):
    if n == 1: return 0
    m2, m5 = multiplicity(2,n), multiplicity(5,n)
    r = max(m2,m5)+n_order(10,n//2**m2//5**m5)
    s = str(10**r//n).zfill(r)
    return s.index(max(s))+1 # Chai Wah Wu, Feb 07 2022

Formula

a(10*n) = a(n) + 1.