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%I #35 Jul 10 2019 12:11:18
%S 2,3,5,8,10,11,14,17,18,19,22,23,24,27,28,29,32,33,34,35,37,38,39,41,
%T 42,44,45,46,47,48,50,51,52,53,54,55,58,59,60,61,65,66,67,68,69,70,71,
%U 74,75,76,77,78,79,82,83,84,87,88,89,92,94,95,96,97,98,99,101,102,103,104,105,106,107,108,109,112,113,115,116
%N Numbers m such that beta(m) = tau(m)/2 - 1 where beta(m) is the number of Brazilian representations of m and tau(m) is the number of divisors of m.
%C As tau(m) = 2 * (1 + beta(m)), the terms of this sequence are not squares. Indeed, there are 3 subsequences which realize a partition of this sequence (see examples):
%C 1) Non-oblong composites which have no Brazilian representation with three digits or more, they form A326386.
%C 2) Oblong numbers that have only one Brazilian representation with three digits or more. These oblong integers are a subsequence of A167782 and form A326384.
%C 3) Non Brazilian primes, then beta(p) = tau(p)/2 - 1 = 0.
%H <a href="https://oeis.org/wiki/Index_to_OEIS:_Section_Br#Brazilian_numbers">Index entries for sequences related to Brazilian numbers</a>
%e One example for each type:
%e 10 = 22_4 and tau(10) = 4 with beta(10) = 1.
%e 42 = 6 * 7 = 222_4 = 33_13 = 22_20 and tau(42) = 8 with beta(42) = 3.
%e 17 is no Brazilian prime with tau(17) = 2 and beta(17) = 0.
%o (PARI) beta(n) = sum(i=2, n-2, #vecsort(digits(n, i), , 8)==1); \\ A220136
%o isok(n) = beta(n) == numdiv(n)/2 - 1; \\ _Michel Marcus_, Jul 03 2019
%Y Cf. A000005 (tau), A220136 (beta).
%Y Cf. A220627 (subsequence of non Brazilian primes).
%Y Cf. A326378 (tau(m)/2 - 2), A326380 (tau(m)/2), A326381 (tau(m)/2 + 1), A326382 (tau(m)/2 + 2), A326383 (tau(m)/2 + 3).
%K nonn,base
%O 1,1
%A _Bernard Schott_, Jul 03 2019
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