A326705 - OEIS

%I #30 Aug 03 2019 21:41:44

%S 4095,262143,265720,531440,1048575,5592405,11184810,16777215,

%T 122070312,183105468,193710244,244140624,268435455,387420488,435356467

%N Non-oblong numbers that are repdigits with length > 2 in more than three bases.

%C 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) = r with r >= 4, beta(m) = tau(m)/2 + k with k >= 3 where beta(m) is the number of Brazilian representations of m.

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

%C The terms which have exactly four Brazilian representations with three digits or more form the first subsequence of A326383. Indeed, for the given terms, the number of bases is 4, except for a(8) and a(15) where this number of bases is respectively 5 and 6 (see examples).

%C Some Mersenne numbers belong to this sequence: M_12 = a(1), M_18 = a(2), M_20 = a(5), M_24 = a(8), M_28 = a(13), M_32, ...

%H Bernard Schott, <a href="/A326705/a326705.pdf">Array for the relations 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 If beta"(m)is the number of Brazilian representations with three digits or more of the integer m, then:

%e 1) With beta"(m) = 4; tau(4095) = 24 and 4095 has exactly four Brazilian representations with three digits or more: [R(12)]_2 = 333333_4 = 7777_4 = (15,15,15)_16 and 11 representations with 2 digits, so beta(4095) = 15 and k = 3.

%e 2) With beta"(m) = 5; tau(435356467) = 64 and 435356467 has exactly five Brazilian representations with three digits or more: R(12)_6 = 777777_36 = (43,43,43)_216 = (259,259,259)_1296 = (31,31,31)_3747 and has 31 representations with 2 digits, so beta(435356467) = 36 and k = 4.

%e 3) With beta"(m)=6; tau(16777215)= 96 and 16777215 has exactly six Brazilian representations with three digits or more: [R(24)]_2 = 333333333333_4 = 7777777_8 = (15,15,15,15,15,15)_16 = (63,63,63,63)_64 = (255,255,255)_256 and 47 representations with 2 digits, so beta(16777215) = 53 and k = 5.

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

%Y Subsequence of A167782, A167783, A290869 and A308874.

%Y Cf. A326386 (non-oblongs with tau(m)/2 - 1), A326387 (non-oblongs with tau(m)/2), A326388 (non-oblongs with tau(m)/2 + 1), A326389 (non-oblongs with tau(m)/2 + 2), this sequence (non-oblongs with tau(m/2) + k, k >= 3).

%K nonn,base,more

%O 1,1

%A _Bernard Schott_, Jul 21 2019