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A002283 a(n) = 10^n - 1. 149
0, 9, 99, 999, 9999, 99999, 999999, 9999999, 99999999, 999999999, 9999999999, 99999999999, 999999999999, 9999999999999, 99999999999999, 999999999999999, 9999999999999999, 99999999999999999, 999999999999999999, 9999999999999999999, 99999999999999999999, 999999999999999999999, 9999999999999999999999 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

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)

LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..100

P. Bala, A002283 and some continued fraction expansions

Rick Regan, Nines in quinary, September 8th, 2009.

Index entries for linear recurrences with constant coefficients, signature (11,-10).

FORMULA

G.f.: 1/(1-10*x)-1/(1-x). E.g.f.: e^(10*x)-e^x. - Mohammad K. Azarian, Jan 14 2009

a(n) = 10*a(n-1) + 9, with a(0)=0. - Paolo P. Lava, Jan 23 2009

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

EXAMPLE

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)

MATHEMATICA

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

PROG

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

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

(Maxima) A002283(n):=10^n-1$

makelist(A002283(n), n, 0, 20); /* Martin Ettl, Nov 08 2012 */

(Haskell)

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

CROSSREFS

Cf. A000533, A003020, A007138, A066138, A168624, A276352.

Sequence in context: A108908 A116260 A103456 * A155157 A264005 A232943

Adjacent sequences:  A002280 A002281 A002282 * A002284 A002285 A002286

KEYWORD

nonn,easy

AUTHOR

N. J. A. Sloane

EXTENSIONS

More terms from Michael De Vlieger, Sep 27 2015

STATUS

approved

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Last modified September 24 22:31 EDT 2017. Contains 292441 sequences.