A352157 Numbers m such that the smallest digit in the decimal expansion of 1/m is 3, ignoring leading and trailing 0's.
3, 12, 30, 120, 264, 275, 296, 300, 1200, 1875, 2112, 2640, 2664, 2750, 2952, 2960, 3000, 10656, 11808, 12000, 18750, 21120, 22944, 26016, 26400, 26640, 27500, 28125, 29088, 29520, 29600, 30000, 103424, 106560, 106656, 118080, 120000, 156288, 187500, 211200, 229440
Offset: 1
Examples
m = 12 is a term since 1/12 = 0.08333333... and the smallest term after the leading 0 is 3. m = 264 is a term since 1/264 = 0.003787878... and the smallest term after the leading 0's is 3.
Crossrefs
Programs
-
Mathematica
f[n_] := Union[ Flatten[ RealDigits[ 1/n][[1]] ]]; Select[ Range@ 1100, Min@ f@# == 3 &]
-
Python
from itertools import count, islice from sympy import multiplicity, n_order def A352157_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 '0' not in s and min(str(c).lstrip('0')+s) == '3': yield n A352157_list = list(islice(A352157_gen(),20)) # Chai Wah Wu, Mar 28 2022
Formula
A352153(a(n)) = 3.
Comments