A352160 Numbers m such that the smallest digit in the decimal expansion of 1/m is k = 6, ignoring leading and trailing 0's.
15, 150, 1500, 15000, 103125, 150000, 1031250, 1500000, 10312500, 15000000, 103125000, 130078125, 150000000, 1031250000, 1300781250, 1500000000, 10312500000
Offset: 1
Examples
m = 150 is a term since 1/150 = 0.0066666666... and the smallest digit after the leading 0's is 6. m = 103125 is a term since 1/103125 = 0.000009696969... and the smallest digit after the leading 0's is 6.
Crossrefs
Programs
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Mathematica
f[n_] := Union[ Flatten[ RealDigits[ 1/n][[1]] ]]; Select[ Range@ 1100, Min@ f@# == 6 &]
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Python
from itertools import count, islice from sympy import multiplicity, n_order def A352160_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) == '6': yield n A352160_list = list(islice(A352160_gen(),5)) # Chai Wah Wu, Mar 28 2022
Formula
A352153(a(n)) = 6.
Extensions
a(9)-a(17) from Jinyuan Wang, Mar 28 2022
Comments