A276352 -id:A276352 - cp's OEIS frontend

A002283 a(n) = 10^n - 1.

0, 9, 99, 999, 9999, 99999, 999999, 9999999, 99999999, 999999999, 9999999999, 99999999999, 999999999999, 9999999999999, 99999999999999, 999999999999999, 9999999999999999, 99999999999999999, 999999999999999999, 9999999999999999999, 99999999999999999999, 999999999999999999999, 9999999999999999999999

Offset: 0

N. J. A. Sloane

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

			From _Peter Bala_, Sep 27 2015: (Start)
Continued fraction expansions showing large partial quotients:
a(12)^(1/3) = [9999; 1, 299999998, 1, 9998, 1, 449999998, 1, 7998, 1, 535714284, 1, 2, 2, 142, 2, 2, 1, 599999999, 3, 1, 1,...].
Compare with a(30)^(1/3) = [9999999999; 1, 299999999999999999998, 1, 9999999998, 1, 449999999999999999998, 1, 7999999998, 1, 535714285714285714284, 1, 2, 2, 142857142, 2, 2, 1, 599999999999999999999, 3, 1, 1,...].
a(24)^(1/4) = [999999; 1, 3999999999999999998, 1, 666665, 1, 1, 1, 799999999999999999, 3, 476190, 7, 190476190476190476, 21, 43289, 1, 229, 1, 1864801864801863, 1, 4, 6,...].
Compare with a(48)^(1/4) = [999999999999; 1, 3999999999999999999999999999999999998, 1, 666666666665, 1, 1, 1, 799999999999999999999999999999999999, 3, 476190476190, 7, 190476190476190476190476190476190476, 21, 43290043289, 1, 229, 1, 1864801864801864801864801864801863, 1, 4, 6,...].
a(25)^(1/5) = [99999, 1, 499999999999999999998, 1, 49998, 1, 999999999999999999998, 1, 33332, 3, 151515151515151515151, 5, 1, 1, 1947, 1, 1, 38, 3787878787878787878, 1, 3, 5,...].
(End)

Vincenzo Librandi, Table of n, a(n) for n = 0..100
Peter Bala, A002283 and some continued fraction expansions.
Rick Regan, Nines in quinary, September 8th, 2009.
Amelia Carolina Sparavigna, The groupoids of Mersenne, Fermat, Cullen, Woodall and other Numbers and their representations by means of integer sequences, Politecnico di Torino, Italy (2019), [math.NT].
Amelia Carolina Sparavigna, Some Groupoids and their Representations by Means of Integer Sequences, International Journal of Sciences (2019) Vol. 8, No. 10.
Index entries for linear recurrences with constant coefficients, signature (11,-10).

Cf. A000533, A003020, A007138, A007953, A008591, A010888, A048379, A066138, A073668, A168624, A276352.
Cf. A002275, A002276, A002277, A002278, A002279, A002280, A002281, A002282.
Cf. A075412, A075415, A178630, A178631, A178632, A178633, A178634, A178635.
Cf. A010785.

a002283 = subtract 1 . (10 ^)  -- Reinhard Zumkeller, Feb 21 2014

[(10^n-1): n in [0..20]]; // Vincenzo Librandi, Apr 26 2011

Table[10^n - 1, {n, 0, 22}] (* Michael De Vlieger, Sep 27 2015 *)

A002283(n):=10^n-1$
makelist(A002283(n),n,0,20); /* Martin Ettl, Nov 08 2012 */

a(n)=10^n-1; \\ Charles R Greathouse IV, Jan 30 2012

def a(n): return 10**n-1 # Michael S. Branicky, Feb 25 2023

From Mohammad K. Azarian, Jan 14 2009: (Start)
G.f.: 1/(1-10*x)-1/(1-x).
E.g.f.: e^(10*x)-e^x. (End)
a(n) = A075412(n)/A002275(n) = A178630(n)/A002276(n) = A178631(n)/A002277(n) = A075415(n)/A002278(n) = A178632(n)/A002279(n) = A178633(n)/A002280(n) = A178634(n)/A002281(n) = A178635(n)/A002282(n). - Reinhard Zumkeller, May 31 2010
a(n) = a(n-1) + 9*10^(n-1) with a(0)=0; Also: a(n) = 11*a(n-1) - 10*a(n-2) with a(0)=0, a(1)=9. - Vincenzo Librandi, Jul 22 2010
For n>0, A007953(a(n)) = A008591(n) and A010888(a(n)) = 9. - Reinhard Zumkeller, Aug 06 2010
A048379(a(n)) = 0. - Reinhard Zumkeller, Feb 21 2014
a(n) = Sum_{k=1..n} 9*10^k. - Carauleanu Marc, Sep 03 2016
Sum_{n>=1} 1/a(n) = A073668. - Amiram Eldar, Nov 13 2020
From Elmo R. Oliveira, Jul 19 2025: (Start)
a(n) = 9*A002275(n).
a(n) = A010785(A008591(n)). (End)

More terms from Michael De Vlieger, Sep 27 2015

A138147 Concatenation of n digits 1 and n digits 0.

Original entry on oeis.org

10, 1100, 111000, 11110000, 1111100000, 111111000000, 11111110000000, 1111111100000000, 111111111000000000, 11111111110000000000, 1111111111100000000000, 111111111111000000000000, 11111111111110000000000000, 1111111111111100000000000000, 111111111111111000000000000000

Offset: 1

Omar E. Pol, Mar 29 2008

Also, a(n) = binary representation of A020522(n), for n>0 (see example).

			n ... A020522(n) ..... a(n)
1 ....... 2 ........... 10
2 ...... 12 .......... 1100
3 ...... 56 ......... 111000
4 ..... 240 ........ 11110000
5 ..... 992 ....... 1111100000
6 .... 4032 ...... 111111000000
7 ... 16256 ..... 11111110000000

J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 136, Ex. 4.2.2. - N. J. A. Sloane, Jul 27 2012

Vincenzo Librandi, Table of n, a(n) for n = 1..95
Index entries for linear recurrences with constant coefficients, signature (110,-1000).

Cf. A002275, A020522, A138144, A138145, A138720, A276352.

Cf. A109241, A138118, A138119, A138120, A138146, A138721, A138826. - Omar E. Pol, Nov 08 2008

[(10^(2*n) - 10^n)/9: n in [1..30]]; // Vincenzo Librandi, Apr 26 2011

Table[FromDigits[Join[PadRight[{},n,1],PadRight[{},n,0]]],{n,15}] (* Harvey P. Dale, Nov 20 2011 *)

Vec(10*x/((10*x-1)*(100*x-1)) + O(x^100)) \\ Colin Barker, Sep 16 2013

a(n) = (10^(2*n) - 10^n)/9 = A002275(n)*10^n. - Omar E. Pol, Apr 16 2008
a(n) = 10*A109241(n-1). - Omar E. Pol, Nov 08 2008
From Colin Barker, Sep 16 2013: (Start)
a(n) = 110*a(n-1) - 1000*a(n-2).
G.f.: 10*x/((10*x-1)*(100*x-1)). (End)
From Elmo R. Oliveira, Jun 13 2025: (Start)
E.g.f.: exp(10*x)*(exp(90*x) - 1)/9.
a(n) = A276352(n)/9. (End)

cp's OEIS Frontend

A002283 a(n) = 10^n - 1.

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