cp's OEIS Frontend

a(1)=1, for n>1, a(n) is the concatenation of "1", 2(n-1)-1 digits "0" and "1". - Omar E. Pol, Dec 14 2008

A friend from Germany remarks that the sequence 9, 99, 999, 9999, 99999, 999999, ... might be called the grumpy German sequence: nein!, nein! nein!, nein! nein! nein!, ...

The Regan link shows that integers of the form 10^n -1 have binary representations with exactly n trailing 1 bits. Also those integers have quinary expressions with exactly n trailing 4's. For example, 10^4 -1 = (304444)5. The first digits in quinary correspond to the number 2^n -1, in our example (30)5 = 2^4 -1. A similar pattern occurs in the binary case. Consider 9 = (1001)2. - Washington Bomfim Dec 23 2010

a(n) is the number of positive integers with less than n+1 digits. - Bui Quang Tuan, Mar 09 2015

From Peter Bala, Sep 27 2015: (Start)

For n >= 1, the simple continued fraction expansion of sqrt(a(2*n)) = [10^n - 1; 1, 2*(10^n - 1), 1, 2*(10^n - 1), ...] has period 2. The simple continued fraction expansion of sqrt(a(2*n))/a(n) = [1; 10^n - 1, 2, 10^n - 1, 2, ...] also has period 2. Note the occurrence of large partial quotients in both expansions.

A theorem of Kuzmin in the measure theory of continued fractions says that large partial quotients are the exception in continued fraction expansions.

Empirically, we also see the presence of unexpectedly large partial quotients early in the continued fraction expansions of the m-th roots of the numbers a(m*n) for m >= 3. Some typical examples are given below. (End)

For n > 0, numbers whose smallest decimal digit is 9. - Stefano Spezia, Nov 16 2023

All members are palindromes A002113. The first five members are mentioned in A129868.

Also, binary representation of A129868.

a(A090748(n)) is equal to A138831(n), the n-th perfect number minus 1, written in base 2.

Except for the first term (replace 0 with 1) the binary representation of the n-th iteration of the elementary cellular automaton, Rule 219 starting with a single ON (black) cell. - Robert Price, Feb 21 2016

a(1) = 101 is only prime number in this sequence since a(n) = (10^(n+1)+1)*(10^n-1)/9. - Altug Alkan, May 11 2016

Inverse binomial transform of A005056.

Also, A000533 interpreted as binary numbers, written in base 10. Numbers whose representation in base 2 is has n+1 digits and the digit "1" is the initial and final digit and if n>1 then the internal digits are "0" (see example). - Omar E. Pol, Feb 24 2008

a(n) equals the number of ternary sequences of length n such that no two consecutive terms differ by 1. - David Nacin, May 31 2017

The first three terms (indices 0, 1 and 2) are the only known primes. Moreover, the terms not of the form a(2^k) are all composite, except for a(0). Indeed, for all n >= 0, a(2n+1) is divisible by 11, a(4n+2) is divisible by 101, a(8n+4) is divisible by 73, a(16n+8) is divisible by 17, a(32n+16) is divisible by 353, a(64n+32) is divisible by 19841, etc. - M. F. Hasler, Nov 03 2018 [Edited based on the comment by Jeppe Stig Nielsen, Oct 17 2019]

This sequence also results when each term is generated by converting the previous term into a Roman numeral, then replacing each letter with its corresponding decimal value, provided that the vinculum is used and numerals are written in a specific way for integers greater than 3999, e.g., IV with a vinculum over the I and V for 4000. - Jamie Robert Creasey, Apr 14 2021

By Mihăilescu's theorem, a(n) can never be a perfect power (see "Catalan's conjecture" in Links). - Marco Ripà, Mar 10 2025

Almost every prime appears in this sequence.

Among the first 10000 primes, only 2, 3, 5, 7, 37, 79, 239, 4649, and 62003 do not appear in the sequence. - Giovanni Resta, Mar 13 2020

The next primes not in the sequence are 538987, 35121409, and 265371653. - Robert Israel, Mar 18 2020

Leading 0's are not considered, otherwise every integer >= 11 would be a term (see examples).

Trailing 0's are also not considered, otherwise numbers of the form 2^i*5^j with i, j >= 0, apart 1 (A003592) would be terms.

If k is a term, 10*k is also a term; so, terms with no trailing zeros are all primitive.

Some subsequences:

{11, 111, 1111, ...} = A002275 \ {0, 1}

{33, 333, 3333, ...} = A002277 \ {0, 3}.

{77, 777, 7777, ...} = A002281 \ {0, 7}

{11, 101, 1001, 10001, ...} = A000533 \ {1}.

a(n) is also A145641(n) written in base 2. - Omar E. Pol, Oct 15 2008

a(n) has 3n digits. - Omar E. Pol, Nov 12 2008

a(n) is also A147596(n) written in base 2. - Omar E. Pol, Nov 08 2008

If m is a term, 10*m is also a term.

If m is a term then m has only digits {1}, {9}, {1,0} or {9,0} in its decimal representation, but this is not sufficient to be a term (see examples).

Some subsequences below (not exhaustive, see crossrefs):

m = 10^k, k >= 0, hence m is in A011557 = {1, 10, 100, 1000, 10000, ...};

m = 9*10^k, k >= 0, hence m is in A052268 = {9, 90, 900, 9000, 90000, ...};

m = 10^k-1, k >= 1, hence m is in A002283 = {9, 99, 999, 9999, 99999, ...};

m = 9*(10^k+1), k >= 1, hence m is in 9*A000533 = {99, 909, 9009, 90009, ...};

m = 9+100*(100^k-1)/11, k >= 0, hence m is in 9*A094028 = {9, 909, 90909, 9090909, ...}.