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A258165
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Odd non-Brazilian numbers > 1.
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2
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3, 5, 9, 11, 17, 19, 23, 25, 29, 37, 41, 47, 49, 53, 59, 61, 67, 71, 79, 83, 89, 97, 101, 103, 107, 109, 113, 131, 137, 139, 149, 151, 163, 167, 169, 173, 179, 181, 191, 193, 197, 199, 223, 227, 229, 233, 239, 251, 257, 263, 269, 271, 277, 281, 283, 289, 293, 311, 313, 317, 331, 337
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OFFSET
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1,1
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COMMENTS
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The terms are only odd primes or squares of odd primes.
Most odd primes are present except those in A085104.
All terms which are not primes are squares of odd primes except 121 = 11^2.
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LINKS
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EXAMPLE
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11 is present because there is no base b < 11 - 1 = 10 such that the representation of 11 in base b is a repdigit (all digits are equal). In fact, we have: 11 = 1011_2 = 102_3 = 23_4 = 21_5 = 15_6 = 14_7 = 13_8 = 12_9, and none of these representations are repdigits. - Bernard Schott, Jun 21 2017
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MATHEMATICA
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fQ[n_] := Block[{b = 2}, While[b < n - 1 && Length@ Union@ IntegerDigits[n, b] > 1, b++]; b+1 == n]; Select[1 + 2 Range@ 170, fQ]
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PROG
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(PARI) forstep(n=3, 300, 2, c=1; for(b=2, n-2, d=digits(n, b); if(vecmin(d)==vecmax(d), c=0; break)); if(c, print1(n, ", "))) \\ Derek Orr, May 27 2015
(Python)
from sympy.ntheory.factor_ import digits
l=[]
for n in range(3, 301, 2):
c=1
for b in range(2, n - 1):
d=digits(n, b)[1:]
if max(d)==min(d):
c=0
break
if c: l.append(n)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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