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

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

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

Examples

			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)
		

Links

• 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).

Crossrefs

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.

Programs

Haskell

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

Magma

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

Mathematica

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

Maxima

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

PARI

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

Python

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

Formula

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)

Extensions

More terms from Michael De Vlieger, Sep 27 2015