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%I #32 Jul 10 2019 12:11:27
%S 7,13,15,21,26,40,43,57,62,73,80,85,86,91,93,111,114,124,127,129,133,
%T 146,157,170,171,172,183,211,215,219,222,228,241,242,259,266,285,292,
%U 307,312,314,333,341,343,365,366,381,399,421,422,438,444,455,463,468,471,482,507,518,532,549,553,555,585,601,614,624
%N Numbers m such that beta(m) = tau(m)/2 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 * 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 only one Brazilian representation with three digits or more, they form A326387.
%C 2) Oblong numbers that have exactly two Brazilian representations with three digits or more; these oblong integers are a subsequence of A167783 and form A326385.
%C 3) Brazilian primes for which beta(p) = tau(p)/2 = 1, they are in A085104 \ {31, 8191}.
%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 15 = 1111_2 = 33_4 with tau(15) = 4 and beta(15) = 2.
%e 3906 = 62 * 63 = 111111_5 = 666_25 = (42,42)_86 = (31,31)_125 = (21,21)_185 = (18,18)_216 = (14,14)_278 = 99_433 = 77_557 = 66_650 = 33_1301 = 22_1952, so tau(3906) = 24 with beta(3906) = 12.
%e 43 = 111_6 is Brazilian prime, so tau(43) = 2 and beta(43) = 1.
%o (PARI) beta(n) = sum(i=2, n-2, #vecsort(digits(n, i), , 8)==1); \\ A220136
%o isok(n) = beta(n) == numdiv(n)/2; \\ _Michel Marcus_, Jul 03 2019
%Y Cf. A000005 (tau), A220136 (beta).
%Y Cf. A085104 (Brazilian primes).
%Y Subsequence of A167782.
%Y Cf. A326378 (tau(m)/2 - 2), A326379 (tau(m)/2 - 1), 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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