%I #31 Jul 24 2020 02:48:21
%S 7,15,40,336,1440,5405400
%N Highly Brazilian numbers (A329383) that are not highly composite numbers (A002182).
%C Is this sequence finite or infinite?
%C Indeed, from 6486480 to 321253732800, that is, during 41 successive terms (maybe more?), highly Brazilian numbers are the same as highly composite numbers.
%C The data for this sequence comes from the new terms in the b-file of A066044 found by _Giovanni Resta_.
%C Why are these six numbers HB (highly Brazilian) and not HC (highly composite)? (See link Why HB and not HC? for more details)
%C 1) For 7, 15 and 40, it is because they have a Brazilian representation with 3 or 4 digits and belong to A326380 (see examples).
%C 2) For 336, 1440 and 5405400, it is because each of these three terms HB r is non-oblong, belong to A326386 and the greatest HC m less than r is oblong with the same number of divisors.
%C a(7) > A329383(91) = 321253732800.
%H Bernard Schott, <a href="/A309493/a309493.pdf">Why HB and not HC?</a>
%e a(1) = 7 because 7 is the smallest Brazilian number with 7 = 111_2 so beta(7) = 1, as tau(7) = tau(2) = 2, 7 is highly Brazilian but cannot be highly composite.
%e a(2) = 15 because 15 is the smallest integer 2-Brazilian with 15 = 1111_2 = 33_4 and beta(15) = 2, as tau(15) = tau(6) = 4, 15 is highly Brazilian but not highly composite.
%e a(3) = 40 because 40 is the smallest integer 4-Brazilian with 40 = 1111_3 = 55_7 = 44_9 = 22_19 so beta(40) = 4, as tau(40) = tau(24) = 8, 40 is highly Brazilian but not highly composite.
%e a(4) = 336 because beta(336) = 9 and tau(336) = tau(240) = 20.
%e a(5) = 1440 because beta(1440) = 17 and tau(1440) = tau(1260) = 36.
%e a(6) = 5405400 because beta(5405400) = 191 and tau(5405400) = tau(4324320) = 384.
%Y Cf. A002182, A279930, A309039, A329383.
%K nonn,base,more
%O 1,1
%A _Bernard Schott_, Aug 04 2019