A326386 Non-oblong composites 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.

8, 10, 14, 18, 22, 24, 27, 28, 32, 33, 34, 35, 38, 39, 44, 45, 46, 48, 50, 51, 52, 54, 55, 58, 60, 65, 66, 68, 69, 70, 74, 75, 76, 77, 78, 82, 84, 87, 88, 92, 94, 95, 96, 98, 99, 102, 104, 105, 106, 108, 112, 115, 116, 117, 118, 119, 120, 122, 123, 125, 126, 128, 130, 134, 135, 136

Offset: 1

Author

Bernard Schott, Jul 12 2019

Keywords

• nonn
• base

Comments

As tau(m) = 2 * (1 + beta(m)), the terms of this sequence are not squares.

The number of Brazilian representations of a non-oblong number m with repdigits of length = 2 is beta'(n) = tau(n)/2 - 1.

This sequence is the first subsequence of A326379: non-oblong composites which have no Brazilian representation with three digits or more.

Examples

			tau(m) = 4 and beta(m)=1 for m = 8, 10, 14, 22, 27, 33, 34, 35, 38, ... 8 = 22_3,
tau(m) = 6 and beta(m)=2 for m = 18, 28, 32, 44, 45, 50, ... 18 = 33_5 = 22_8,
tau(m) = 8 and beta(m)=3 for m = 24, 54, 66, 70, ... 24 = 44_5 = 33_7 = 22_11,
tau(m) = 10 and beta(m) = 4: 48, 112, ... 48 = 66_7 = 44_11 = 33_15 = 22_23.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000
• Bernard Schott, beta = f(tau) - Abstract.
• Index entries for sequences related to Brazilian numbers.

Crossrefs

Cf. A000005 (tau), A220136 (beta).

Subsequence of A308874 and of A326379.

Cf. A326387 (non-oblongs with tau(m)/2), A326388 (non-oblongs with tau(m)/2 + 1), A326389 (non-oblongs with tau(m)/2 + 2).

Programs

PARI

isoblong(n) = my(m=sqrtint(n)); m*(m+1)==n; \\ A002378
beta(n) = sum(i=2, n-2, #vecsort(digits(n, i), , 8)==1); \\ A220136
isok(m) = !isprime(m) && !isoblong(m) && (beta(m) == numdiv(m)/2 - 1); \\ Michel Marcus, Jul 15 2019