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A326383
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.
7
4095, 262143, 265720, 531440, 1048575, 5592405, 11184810, 122070312, 183105468, 193710244, 244140624, 268435455, 387420488
OFFSET
1,1
COMMENTS
As tau(m) = 2 * (beta(m) - 3), the terms of this sequence are not squares.
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.
EXAMPLE
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.
CROSSREFS
Cf. A000005 (tau), A220136 (beta).
Subsequence of A167782, A167783 and A290869.
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).
Sequence in context: A075953 A011562 A326705 * A160972 A038999 A022528
KEYWORD
nonn,base,more
AUTHOR
Bernard Schott, Jul 08 2019
STATUS
approved