A333402 Numbers m such that the largest digit in the decimal expansion of 1/m is 1.
1, 9, 10, 90, 99, 100, 900, 909, 990, 999, 1000, 9000, 9009, 9090, 9900, 9990, 9999, 10000, 90000, 90009, 90090, 90900, 90909, 99000, 99900, 99990, 99999, 100000, 900000, 900009, 900090, 900900, 909000, 909090, 990000, 990099, 999000, 999900, 999990, 999999, 1000000
Offset: 1
Examples
As 1/101 = 0.009900990099..., 101 is not a term. As 1/909 = 0.001100110011..., 909 is a term. As 1/9099 = 0.000109902187..., 9099 is not a term. As 1/9999 = 0.000100010001..., 9999 is also a term.
Crossrefs
Programs
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Mathematica
Select[Range[10^4], Max @ RealDigits[1/#][[1]] == 1 &] (* Amiram Eldar, Mar 19 2020 *)
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Python
from itertools import count, islice def A333402_gen(startvalue=1): # generator of terms >= startvalue for m in count(max(startvalue,1)): k = 1 while k <= m: k *= 10 rset = {0} while True: k, r = divmod(k, m) if max(str(k)) > '1': break else: if r in rset: yield m break rset.add(r) k = r while k <= m: k *= 10 A333402_list = list(islice(A333402_gen(),30)) # Chai Wah Wu, Feb 17 2022
Formula
A333236(a(n))= 1.
Extensions
More terms from Jinyuan Wang, Mar 19 2020
Comments