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A341058
Numbers that have only one divisor that is Brazilian.
2
7, 8, 10, 12, 13, 15, 18, 22, 27, 31, 33, 34, 38, 43, 46, 49, 51, 55, 57, 58, 69, 73, 74, 82, 85, 87, 94, 95, 106, 111, 115, 118, 121, 122, 123, 125, 127, 134, 141, 142, 145, 157, 158, 159, 166, 169, 177, 178, 183, 185, 187, 194, 201, 202, 205, 206, 209, 211, 213, 214, 218
OFFSET
1,1
COMMENTS
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).
FORMULA
A340795(a(n)) = 1.
EXAMPLE
One example for each type of terms that has k divisors:
-> k=2: 7 is a Brazilian prime, hence 7 = 111_2 is a term.
-> 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.
-> 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.
-> 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.
-> 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.
-> 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.
MATHEMATICA
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 *)
CROSSREFS
Subsequence: A085104.
Sequence in context: A125134 A169876 A288783 * A120175 A291668 A175222
KEYWORD
nonn,base
AUTHOR
Bernard Schott, Feb 15 2021
STATUS
approved