A326389 Non-oblong numbers that are repdigits with length > 2 in exactly three bases.

32767, 65535, 67053, 2097151, 4381419, 7174453, 9808617, 13938267, 14348906, 19617234, 21523360, 29425851, 39234468, 43046720, 48686547, 49043085, 58851702, 68660319, 71270178, 78468936, 88277553, 98086170, 107894787, 115174101, 117703404, 134217727, 142540356

Offset: 1

Author

Bernard Schott, Jul 20 2019

Keywords

• nonn
• base

Comments

The number of Brazilian representations of a non-oblong number m with repdigits of length = 2 is beta'(m) = tau(m)/2 - 1. So, as here beta"(m) = 3, beta(m) = tau(m)/2 + 2 where beta(m) is the number of Brazilian representations of m. So, this sequence is the first subsequence of A326382.

As tau(m) = 2 * (beta(m) - 2) is even, the terms of this sequence are not squares.

Some Mersenne numbers belong to this sequence: M_15 = a(1), M_16 = a(2), M_21 = a(4), M_27 = a(26), ...

Examples

			tau(m) = 8 and beta(m) = 6 for m = 32767 with 32767 = R(15)_2 = 77777_8 = (31,31,31)_32.
tau(m) = 12 and beta(m) = 8 for m = 2097151 with 2097151 = R(21)_2 = 7777777_8 = (127,127,127)_128.
tau(m) = 16 and beta(m) = 10 with m = 67053 = (31,31,31)_46 = (21,21,21)_56 = 333_149.
		

Links

• Bernard Schott, Array of relations beta = f(tau)
• Index entries for sequences related to Brazilian numbers

Crossrefs

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

Subsequence of A167782, A167783, A290869, A308874 and A326382.

Cf. A326386 (non-oblongs with tau(m)/2 - 1), A326387 (non-oblongs with tau(m)/2), A326388 (non-oblongs with tau(m)/2 + 1), this sequence (non-oblongs with tau(m)/2 + 2), A326705 (non-oblongs with tau(m)/2 + k, k >=3).

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 + 2); \\ Jinyuan Wang, Aug 02 2019