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.

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

Author

Bernard Schott, Jul 07 2019

Keywords

• nonn
• base

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.

Examples

			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.
		

Links

• Wikipedia, Goormaghtigh conjecture.
• Index entries for sequences related to Brazilian numbers.

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).

Programs

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