%I #24 Mar 11 2021 03:23:02
%S 7,8,10,12,13,15,18,22,27,31,33,34,38,43,46,49,51,55,57,58,69,73,74,
%T 82,85,87,94,95,106,111,115,118,121,122,123,125,127,134,141,142,145,
%U 157,158,159,166,169,177,178,183,185,187,194,201,202,205,206,209,211,213,214,218
%N Numbers that have only one divisor that is Brazilian.
%C m is a term iff m is a Brazilian prime (A085104), or m is the square of a Brazilian prime, or m = 121, the only square of prime that is Brazilian, or m = p*q >= 10 with p>q are non-Brazilian primes, or m is the cube of a Brazilian prime, or m = 12 or 18 (see corresponding examples).
%F A340795(a(n)) = 1.
%e One example for each type of terms that has k divisors:
%e -> k=2: 7 is a Brazilian prime, hence 7 = 111_2 is a term.
%e -> k=3: 169 has three divisors {1, 13, 169} and 13 = 111_3 is the only divisor of 169 that is Brazilian, hence 169 is a term.
%e -> k=3: 121 has three divisors {1, 11, 121} and 121 = 11111_3, hence, 121 that is the only square of prime that is Brazilian, is a term.
%e -> k=4: 34 has four divisors {1, 2, 17, 34} and 34 = 22_16 is the only divisor of 34 that is Brazilian, hence 34 is a term.
%e -> k=4: 27 has four divisors {1, 3, 9, 27} and 27 = 33_8 is the only divisor of 27 that is Brazilian, hence 27 is a term.
%e -> k=6: only two cases: 12 and 18, these integers have each 6 divisors and only 12 = 22_5 and 18 = 33_5 are Brazilian.
%t brazQ[n_] := Module[{b = 2, found = False}, While[b < n - 1 && Length @ Union[IntegerDigits[n, b]] > 1, b++]; b < n - 1]; Select[Range[200], DivisorSum[#, 1 &, brazQ[#1] &] == 1 &] (* _Amiram Eldar_, Feb 16 2021 *)
%Y Cf. A125134, A220627, A308851, A340795, A341057.
%Y Subsequence: A085104.
%K nonn,base
%O 1,1
%A _Bernard Schott_, Feb 15 2021