login
A326380
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.
10
7, 13, 15, 21, 26, 40, 43, 57, 62, 73, 80, 85, 86, 91, 93, 111, 114, 124, 127, 129, 133, 146, 157, 170, 171, 172, 183, 211, 215, 219, 222, 228, 241, 242, 259, 266, 285, 292, 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
OFFSET
1,1
COMMENTS
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):
1) Non-oblong composites which have only one Brazilian representation with three digits or more, they form A326387.
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.
3) Brazilian primes for which beta(p) = tau(p)/2 = 1, they are in A085104 \ {31, 8191}.
EXAMPLE
One example for each type:
15 = 1111_2 = 33_4 with tau(15) = 4 and beta(15) = 2.
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.
43 = 111_6 is Brazilian prime, so tau(43) = 2 and beta(43) = 1.
PROG
(PARI) beta(n) = sum(i=2, n-2, #vecsort(digits(n, i), , 8)==1); \\ A220136
isok(n) = beta(n) == numdiv(n)/2; \\ Michel Marcus, Jul 03 2019
CROSSREFS
Cf. A000005 (tau), A220136 (beta).
Cf. A085104 (Brazilian primes).
Subsequence of A167782.
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).
Sequence in context: A349751 A349752 A167782 * A257521 A053696 A090503
KEYWORD
nonn,base
AUTHOR
Bernard Schott, Jul 03 2019
STATUS
approved