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%I #8 Jan 12 2021 21:35:29
%S 1,12,113,1134,11227,112154,1112236,11111566,111123685,1111133874,
%T 11111178192,111111796422,1111111392823,11111112811396,
%U 111111112641445,1111111115954155,11111111158315794,111111111132821544,1111111111273944122,11111111111777673838,111111111113343756694
%N Least n-digit number m such that m and m^10 are zeroless.
%C a(n)^10 is converging to 2867971991..1 (1 repeated 10*n-18 times at end), or 10^(10*n-10) times the smallest rational greater than (10/9)^10 that contains no 0 digit. - _Michael S. Branicky_, Jan 12 2021
%H Michael S. Branicky, <a href="/A124651/b124651.txt">Table of n, a(n) for n = 1..25</a>
%e 12^10 is 61917364224 but 10 and 11^10 = 25937424601 have zeros. - _Michael S. Branicky_, Jan 12 2021
%o (Python)
%o from sympy import integer_nthroot
%o def a(n):
%o if n == 1: return 1
%o m, perfect = integer_nthroot(int('286797199' + '1'*(10*n-18)), 10)
%o strm = str(m)
%o # strm = "1"*n # slower than the foregoing for larger n
%o while strm.count('0') > 0 or str(m**10).count('0') > 0:
%o if '0' in strm:
%o ind0 = strm.find('0')
%o m = int(strm[:ind0] + '1'*(len(strm)-ind0))
%o elif strm[-1] == '9':
%o m += 2
%o else:
%o m += 1
%o strm = str(m)
%o return m
%o for n in range(1, 15):
%o print(a(n), end=", ") # _Michael S. Branicky_, Jan 12 2021
%K base,nonn
%O 1,2
%A _Zak Seidov_, Dec 22 2006
%E a(17) and beyond from _Michael S. Branicky_, Jan 12 2021
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