

A309493


Highly Brazilian numbers (A329383) that are not highly composite numbers (A002182).


3




OFFSET

1,1


COMMENTS

Is this sequence finite or infinite?
Indeed, from 6486480 to 321253732800, that is, during 41 successive terms (maybe more?), highly Brazilian numbers are the same as highly composite numbers.
Why are these six numbers HB (highly Brazilian) and not HC (highly composite)? (See link Why HB and not HC? for more details)
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).
2) For 336, 1440 and 5405400, it is because each of these three terms HB r is nonoblong, belong to A326386 and the greatest HC m less than r is oblong with the same number of divisors.


LINKS



EXAMPLE

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.
a(2) = 15 because 15 is the smallest integer 2Brazilian with 15 = 1111_2 = 33_4 and beta(15) = 2, as tau(15) = tau(6) = 4, 15 is highly Brazilian but not highly composite.
a(3) = 40 because 40 is the smallest integer 4Brazilian 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.
a(4) = 336 because beta(336) = 9 and tau(336) = tau(240) = 20.
a(5) = 1440 because beta(1440) = 17 and tau(1440) = tau(1260) = 36.
a(6) = 5405400 because beta(5405400) = 191 and tau(5405400) = tau(4324320) = 384.


CROSSREFS



KEYWORD

nonn,base,more


AUTHOR



STATUS

approved



