|
%I #51 Jul 10 2019 12:11:09
%S 6,12,20,30,56,72,90,110,132,210,240,272,306,380,420,462,506,552,600,
%T 650,702,756,812,870,930,992,1056,1122,1190,1260,1332,1482,1560,1722,
%U 1806,1892,1980,2070,2162,2256,2352,2450,2550,2652,2756,2862,2970,3080,3192,3306,3422,3540,3660,3782
%N Numbers m such that beta(m) = tau(m)/2 - 2 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 * (2 + beta(m)), the terms of this sequence are not squares. Indeed, there exists only one family that satisfies this relation and these integers are exactly the oblong numbers that have no Brazilian representation with three digits or more.
%C There are no integers such as beta(m) = tau(m)/2 - q with q >= 3.
%H Bernard Schott, <a href="/A326378/a326378_2.pdf">Relation beta = f(tau)</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 1) tau(m) = 4 and beta(m) = 0: m = 6 which is not Brazilian.
%e 2) tau(m) = 6 and beta(m) = 1: m = 12, 20.
%e 12 = 3 * 4 = 22_5, 20 = 4 * 5 = 22_9.
%e 3) tau(m) = 8 and beta(m) = 2: m = 30, 56, 110, 506, 2162, 3422, ...
%e 30 = 5 * 6 = 33_9 = 22_14, 56 = 7 * 8 = 44_13 = 22_27.
%e 4) tau(m) = 10 and beta(m) = 3: m = 272, ...
%e 272 = 16 * 17 = 88_32 = 44_67 = 22_135.
%e 5) tau(m) = 12 and beta(m) = 4: m = 72, 90, 132, 306, 380, 650, 812, 992, ...
%e 72 = 8 * 9 = 66_11 = 44_17 = 33_23 = 22_35.
%o (PARI) beta(n) = sum(i=2, n-2, #vecsort(digits(n, i), , 8)==1); \\ A220136
%o isok(n) = beta(n) == numdiv(n)/2 - 2; \\ _Michel Marcus_, Jul 08 2019
%Y Cf. A000005 (tau), A220136 (beta).
%Y Subsequence of A002378 (oblong numbers).
%Y Cf. A326379 (tau(m)/2 - 1), A326380 (tau(m)/2), A326381 (tau(m)/2 + 1), A326382 (tau(m)/2 + 2), A326383 (tau(m)/2 + 3).
%Y Cf. A326384 (oblongs with tau(m)/2 - 1), A326385 (oblongs with tau(m)/2).
%K nonn,base
%O 1,1
%A _Bernard Schott_, Jul 02 2019
|
