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A340796
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a(n) is the smallest number with exactly n divisors that are Brazilian.
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3
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1, 7, 14, 24, 40, 48, 60, 84, 140, 144, 120, 168, 252, 700, 240, 336, 560, 360, 420, 672, 1120, 2304, 960, 720, 1008, 1080, 840, 2184, 1800, 1260, 2016, 5376, 8960, 2160, 1680, 2880, 4032, 3600, 7056, 19600, 3960, 2520, 3360, 6480, 9072, 9900, 6300, 11520, 16128
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OFFSET
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0,2
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COMMENTS
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Primes can be partitioned into Brazilian primes and non-Brazilian primes. If two distinct primes each larger than 11 are in the same category then the larger one has a multiplicity that is smaller than or equal to that of the smaller prime. - David A. Corneth, Jan 24 2021
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LINKS
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EXAMPLE
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Of the eight divisors of 24, three are Brazilian numbers: 8, 12 and 24, and there is no smaller number with three Brazilian divisors, hence a(3) = 24.
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MATHEMATICA
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brazQ[n_] := Module[{b = 2, found = False}, While[b < n - 1 && Length[Union[IntegerDigits[n, b]]] > 1, b++]; b < n - 1]; d[n_] := DivisorSum[n, 1 &, brazQ[#] &]; m = 30; s = Table[0, {m}]; c = 0; n = 1; While[c < m, i = d[n]; If[i < m && s[[i + 1]] == 0, c++; s[[i + 1]] = n]; n++]; s (* Amiram Eldar, Jan 21 2021 *)
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PROG
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(PARI) isokb(n) = for(b=2, n-2, d=digits(n, b); if(vecmin(d)==vecmax(d), return(1))); \\ A125134
isok(k, n) = sumdiv(k, d, isokb(d)) == n;
a(n) = my(k=1); while (!isok(k, n), k++); k; \\ Michel Marcus, Jan 23 2021
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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