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Partial sums of powers of 45 (A009989).

Partial sums of powers of 46 (A009990).

Partial sums of powers of 48 (A009992).

Let b(n) = A226542(n)-1. This sequence is a supersequence of b.

Conjecture 1: Let c(n) = A001220(n)-1. This sequence is a supersequence of c.

Conjecture 2: This is a supersequence of A240719.

From Bernard Schott, Sep 19 2019: (Start)

There are 3 distinct families of terms in this sequence:

1) Integers of the form: 2^q + 1 = 11_2^q with q >= 2.

First few terms: 5, 9, 17, 33, 65, 129, ...; this is A000051 \ {2, 3}. As 11_b is not a Brazilian representation, five of these terms are not Brazilian, they are 9 and the four known Fermat primes in A019434: 5, 17, 257 and 65537; all the other terms are composite and Brazilian but in a base that is not a power of two as 65 = 11_64 = 55_12.

2) Integers of the form: m * (2^q+1) = (mm)_2^q with q >= 2 and 1 < m < 2^q.

First few terms: 10, 15, 18, 27, 34, 36, ... These numbers are Brazilian with 2 digits in a base that is a power of two >= 4 as 10 = 22_4, 15 = 33_4 or 18 = 22_8.

3) Integers of the form: m * ((2^q)^s - 1)/(2^q - 1) = (mm...m)_ 2^q with q >= 2, s >= 1 and 1 <= m <= 2^q - 1.

First few terms: 21, 42, 63, 73, 85, ... These numbers are Brazilian repdigits with 3 digits or more in a base that is a power of two >= 4 as 42 = 222_4, 73 = 111_8 or 85 = 1111_4. The repunits (4^n-1)/3, (8^n-1)/7, (16^n-1)/15, (32^n-1)/31 respectively in A002450 (when >= 5), A023001 (when >=9), A131865 (when >=17), A132469 (when >=33) are subsequences of this last family.

Remark: there exist numbers that are in this sequence for two reasons as 63 = 77_8 = 333_4. (End)