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A326381
Numbers m such that beta(m) = tau(m)/2 + 1 where beta(m) is the number of Brazilian representations of m and tau(m) is the number of divisors of m.
7
31, 63, 255, 273, 364, 511, 546, 728, 777, 931, 1023, 1365, 1464, 2730, 3280, 3549, 3783, 4557, 6560, 7566, 7812, 8191, 9114, 9331, 9841, 10507, 11349, 11718, 13671, 14043, 14763, 15132, 15624, 16383, 18291, 18662, 18915, 19608, 19682, 21845, 22351, 22698
OFFSET
1,1
COMMENTS
As tau(m) = 2 * (beta(m) - 1), the terms of this sequence are not squares.
There are 3 subsequences which realize a partition of this sequence (see examples):
1) Non-oblong composites which have exactly two Brazilian representations with three digits or more, they form A326388.
2) Oblong numbers that have exactly three Brazilian representations with three digits or more; thanks to Michel Marcus, who found the smallest, 641431602. These oblong integers are a subsequence of A290869 and A309062.
3) The two Brazilian primes 31 and 8191 of the Goormaghtigh conjecture (A119598) for which beta(p) = tau(p)/2 + 1 = 2.
EXAMPLE
One example for each type:
1) 63 = 111111_2 = 333_4 = 77_8 = 33_20 with tau(63) = 6 and beta(63) = 4.
2) 641431602 = 25326 * 25327 is oblong with tau(641431602) = 256. The three Brazilian representations with three digits or more of 641431602 are 999999_37 = (342,342,342)_1369 = (54,54,54)_3446, so beta"(641431602) = 3 and beta(641431602) = tau(641431602)/2 + 1 = 129.
3) 31 = 11111_2 = 111_5 and 8191 = 1111111111111_2 = 11_90 with beta(p) = tau(p)/2 + 1 = 2.
PROG
(PARI) beta(n) = sum(i=2, n-2, #vecsort(digits(n, i), , 8)==1); \\ A220136
isok(n) = beta(n) == numdiv(n)/2 + 1; \\ Michel Marcus, Jul 08 2019
CROSSREFS
Cf. A000005 (tau), A220136 (beta).
Cf. A119598 (Goormaghtigh conjecture).
Subsequence of A167783.
Cf. A326378 (tau(m)/2 - 2), A326379 (tau(m)/2 - 1), A326380 (tau(m)/2), A326382 (tau(m)/2 + 2), A326383 (tau(m)/2 + 3).
Sequence in context: A042908 A042912 A167783 * A042914 A042916 A042918
KEYWORD
nonn,base
AUTHOR
Bernard Schott, Jul 07 2019
STATUS
approved