A352158 Numbers m such that the smallest digit in the decimal expansion of 1/m is 4, ignoring leading and trailing 0's.
22, 25, 144, 220, 225, 250, 1056, 1184, 1440, 2184, 2200, 2250, 2500, 10560, 11840, 14400, 15625, 20625, 21024, 21840, 22000, 22500, 25000, 104192, 105600, 115625, 118400, 144000, 156250, 168192, 179712, 206250, 210240, 213312, 218400, 220000
Offset: 1
Examples
m = 22 is a term since 1/22 = 0.045454545... and the smallest digit after the leading 0 is 4. m = 1184 is a term since 1/1184 = 0.00084459459... and the smallest digit after the leading 0's is 4.
Crossrefs
Programs
-
Mathematica
f[n_] := Union[ Flatten[ RealDigits[ 1/n][[1]] ]]; Select[ Range@ 1100, Min@ f@# == 4 &]
-
Python
from itertools import count, islice from sympy import multiplicity, n_order def A352158_gen(startvalue=1): # generator of terms >= startvalue for n in count(max(startvalue,1)): m2, m5 = multiplicity(2,n), multiplicity(5,n) k, m = 10**max(m2,m5), 10**(t := n_order(10,n//2**m2//5**m5))-1 c = k//n s = str(m*k//n-c*m).zfill(t) if s == '0' and min(str(c)) == '4': yield n elif '0' not in s and min(str(c).lstrip('0')+s) == '4': yield n A352158_list = list(islice(A352158_gen(),20)) # Chai Wah Wu, Mar 28 2022
Formula
A352153(a(n)) = 4.
Comments