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Numbers without Brazilian divisors.
2

%I #29 Mar 11 2021 03:22:50

%S 1,2,3,4,5,6,9,11,17,19,23,25,29,37,41,47,53,59,61,67,71,79,83,89,97,

%T 101,103,107,109,113,131,137,139,149,151,163,167,173,179,181,191,193,

%U 197,199,223,227,229,233,239,251,257,263,269,271,277,281,283,289,293,311

%N Numbers without Brazilian divisors.

%C The first 16 terms are the first 16 terms of A220570 (non-Brazilian numbers), then a(17) = 53 while A220570(17) = 49.

%C m is a term iff m = 1, or m = 6, or m is a non-Brazilian prime (A220627) or m is the square of a non-Brazilian prime, except for 121 that is Brazilian (see examples).

%F A340795(a(n)) = 0.

%e One example for each type of terms that has k divisors:

%e -> k=1: 1 is the smallest number not Brazilian, hence 1 is the first term.

%e -> k=2: 17 is a prime non-Brazilian, hence 17 is a term.

%e -> k=3: 25 has three divisors {1, 5, 25} that are all not Brazilian, hence 25 is another term.

%e -> k=4: 6 has four divisors {1, 2, 3, 6} that are all not Brazilian, hence 6 is the term that has the largest number of divisors.

%t brazQ[n_] := Module[{b = 2, found = False}, While[b < n - 1 && Length @ Union @ IntegerDigits[n, b] > 1, b++]; b < n - 1]; q[n_] := AllTrue[Divisors[n], ! brazQ[#] &]; Select[Range[300], q] (* _Amiram Eldar_, Feb 04 2021 *)

%o (PARI) isb(n) = for(b=2, n-2, my(d=digits(n, b)); if(vecmin(d)==vecmax(d), return(1))); \\ A125134

%o isok(n) = fordiv(n, d, if (isb(d), return(0))); return(1); \\ _Michel Marcus_, Feb 07 2021

%Y Cf. A125134, A340795, A308851, A341058 (with 1 Brazilian divisor).

%Y Subsequence of A220570 (non-Brazilian numbers).

%Y Supersequence of A220627 (non-Brazilian primes).

%K nonn,base

%O 1,2

%A _Bernard Schott_, Feb 04 2021