%I #40 May 31 2024 22:05:08
%S 31,63,255,273,364,511,546,728,777,931,1023,1365,1464,2730,3280,3549,
%T 3783,4557,6560,7566,7812,8191,9114,9331,9841,10507,11349,11718,13671,
%U 14043,14763,15132,15624,16383,18291,18662,18915,19608,19682,21845,22351,22698
%N 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.
%C As tau(m) = 2 * (beta(m) - 1), the terms of this sequence are not squares.
%C There are 3 subsequences which realize a partition of this sequence (see examples):
%C 1) Non-oblong composites which have exactly two Brazilian representations with three digits or more, they form A326388.
%C 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.
%C 3) The two Brazilian primes 31 and 8191 of the Goormaghtigh conjecture (A119598) for which beta(p) = tau(p)/2 + 1 = 2.
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Goormaghtigh_conjecture">Goormaghtigh conjecture</a>.
%H <a href="https://oeis.org/wiki/Index_to_OEIS:_Section_Br#Brazilian_numbers">Index entries for sequences related to Brazilian numbers</a>.
%e One example for each type:
%e 1) 63 = 111111_2 = 333_4 = 77_8 = 33_20 with tau(63) = 6 and beta(63) = 4.
%e 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.
%e 3) 31 = 11111_2 = 111_5 and 8191 = 1111111111111_2 = 11_90 with beta(p) = tau(p)/2 + 1 = 2.
%o (PARI) beta(n) = sum(i=2, n-2, #vecsort(digits(n, i), , 8)==1); \\ A220136
%o isok(n) = beta(n) == numdiv(n)/2 + 1; \\ _Michel Marcus_, Jul 08 2019
%Y Cf. A000005 (tau), A220136 (beta).
%Y Cf. A119598 (Goormaghtigh conjecture).
%Y Subsequence of A167783.
%Y 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).
%K nonn,base
%O 1,1
%A _Bernard Schott_, Jul 07 2019