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%I #12 Jul 10 2019 12:11:50
%S 4095,262143,265720,531440,1048575,5592405,11184810,122070312,
%T 183105468,193710244,244140624,268435455,387420488
%N Numbers m such that beta(m) = tau(m)/2 + 3 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) - 3), the terms of this sequence are not squares.
%C The current known terms are non-oblong composites that have exactly four Brazilian representations with three digits or more; but, maybe, there exist oblong integers that have exactly five Brazilian representations with three digits or more.
%H <a href="https://oeis.org/wiki/Index_to_OEIS:_Section_Br#Brazilian_numbers">Index entries for sequences related to Brazilian numbers</a>
%e The 24 divisors of 4095 = M_12 are {1, 3, 5, 7, 9, 13, 15, 21, 35, 39, 45, 63, 65, 91, 105, 117, 195, 273, 315, 455, 585, 819, 1365, 4095} and tau(4095) = 24; also, 4095 = R(12)_2 = 333333_4 = 7777_8 = (15,15,15)_16, so, beta(4095) = 15 with beta'(4095)= 11 and beta''(4095) = 4. The relation is beta(4095) = tau(4095)/2 + 3 = 15 and 4095 is a term.
%Y Cf. A000005 (tau), A220136 (beta).
%Y Subsequence of A167782, A167783 and A290869.
%Y Cf. A326378 (tau(m)/2 - 2), A326379 (tau(m)/2 - 1), A326380 (tau(m)/2), A326381 (tau(m)/2 + 1), A326382 (tau(m)/2 + 2).
%K nonn,base,more
%O 1,1
%A _Bernard Schott_, Jul 08 2019
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