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a(n) = 10^n - 1.
182

%I #133 Feb 07 2024 01:13:47

%S 0,9,99,999,9999,99999,999999,9999999,99999999,999999999,9999999999,

%T 99999999999,999999999999,9999999999999,99999999999999,

%U 999999999999999,9999999999999999,99999999999999999,999999999999999999,9999999999999999999,99999999999999999999,999999999999999999999,9999999999999999999999

%N a(n) = 10^n - 1.

%C 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!, ...

%C 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

%C a(n) is the number of positive integers with less than n+1 digits. - _Bui Quang Tuan_, Mar 09 2015

%C From _Peter Bala_, Sep 27 2015: (Start)

%C 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.

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

%C 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)

%C For n > 0, numbers whose smallest decimal digit is 9. - _Stefano Spezia_, Nov 16 2023

%H Vincenzo Librandi, <a href="/A002283/b002283.txt">Table of n, a(n) for n = 0..100</a>

%H Peter Bala, <a href="/A002283/a002283.pdf">A002283 and some continued fraction expansions</a>.

%H Rick Regan, <a href="http://www.exploringbinary.com/nines-in-quinary/">Nines in quinary</a>, September 8th, 2009.

%H Amelia Carolina Sparavigna, <a href="https://doi.org/10.5281/zenodo.3471358">The groupoids of Mersenne, Fermat, Cullen, Woodall and other Numbers and their representations by means of integer sequences</a>, Politecnico di Torino, Italy (2019), [math.NT].

%H Amelia Carolina Sparavigna, <a href="https://doi.org/10.18483/ijSci.2188">Some Groupoids and their Representations by Means of Integer Sequences</a>, International Journal of Sciences (2019) Vol. 8, No. 10.

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (11,-10).

%F From _Mohammad K. Azarian_, Jan 14 2009: (Start)

%F G.f.: 1/(1-10*x)-1/(1-x).

%F E.g.f.: e^(10*x)-e^x. (End)

%F 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

%F 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

%F For n>0, A007953(a(n)) = A008591(n) and A010888(a(n)) = 9. - _Reinhard Zumkeller_, Aug 06 2010

%F A048379(a(n)) = 0. - _Reinhard Zumkeller_, Feb 21 2014

%F a(n) = Sum_{k=1..n} 9*10^k. - _Carauleanu Marc_, Sep 03 2016

%F Sum_{n>=1} 1/a(n) = A073668. - _Amiram Eldar_, Nov 13 2020

%e From _Peter Bala_, Sep 27 2015: (Start)

%e Continued fraction expansions showing large partial quotients:

%e 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,...].

%e 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,...].

%e 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,...].

%e 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,...].

%e 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,...].

%e (End)

%t Table[10^n - 1, {n, 0, 22}] (* _Michael De Vlieger_, Sep 27 2015 *)

%o (Magma) [(10^n-1): n in [0..20]]; // _Vincenzo Librandi_, Apr 26 2011

%o (PARI) a(n)=10^n-1; \\ _Charles R Greathouse IV_, Jan 30 2012

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

%o makelist(A002283(n),n,0,20); /* _Martin Ettl_, Nov 08 2012 */

%o (Haskell)

%o a002283 = subtract 1 . (10 ^) -- _Reinhard Zumkeller_, Feb 21 2014

%o (Python) def a(n): return 10**n-1 # _Michael S. Branicky_, Feb 25 2023

%Y Cf. A000533, A003020, A007138, A007953, A008591, A010888, A048379, A066138, A073668, A168624, A276352.

%Y Cf. A002275, A002276, A002277, A002278, A002279, A002280, A002281, A002282.

%Y Cf. A075412, A075415, A178631, A178632, A178633, A178634, A178635.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_

%E More terms from _Michael De Vlieger_, Sep 27 2015