A033714 -id:A033714 - cp's OEIS frontend

A053541 a(n) = n*10^(n-1).

1, 20, 300, 4000, 50000, 600000, 7000000, 80000000, 900000000, 10000000000, 110000000000, 1200000000000, 13000000000000, 140000000000000, 1500000000000000, 16000000000000000, 170000000000000000

Offset: 1

Barry E. Williams, Jan 15 2000

This sequence gives the number of 1's (or any other nonzero digit) required to write all integers from 0 up to 10^n-1. - Jason D. W. Taff (jtaff(AT)jburroughs.org), Dec 05 2004 (improved by Bernard Schott, Nov 17 2022)

The corresponding number of 0's required to write all these integers from 0 up to 10^n-1 is A033714(n). - Bernard Schott, Nov 17 2022

Albert H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

Vincenzo Librandi, Table of n, a(n) for n = 1..100
Frank Ellermann, Illustration of binomial transforms.
Index entries for linear recurrences with constant coefficients, signature (20,-100).

Cf. A001787, A033714, A038303, A053464, A053469, A081045, A094798.

List([1..20], n-> n*10^(n-1)) # G. C. Greubel, May 16 2019

[n*10^(n-1): n in [1..30]]; // Vincenzo Librandi, Jun 06 2011

seq(n*10^(n-1), n = 1 .. 40); # Bernard Schott, Nov 17 2022

f[n_]:=n*10^(n-1);f[Range[40]] (* Vladimir Joseph Stephan Orlovsky, Feb 09 2011*)
LinearRecurrence[{20,-100},{1,20},20] (* Harvey P. Dale, Aug 08 2023 *)

a(n)=n*10^(n-1) \\ Charles R Greathouse IV, Dec 05 2011

[n*10^(n-1) for n in (1..20)] # G. C. Greubel, May 16 2019

a(n) = 20*a(n-1) - 100*a(n-2), with a(0)=0, a(1)=1, a(2)=20.
From Jason D. W. Taff (jtaff(AT)jburroughs.org), Dec 05 2004: (Start)
a(n) = 10*a(n-1) + 10*(n-1).
a(n) = Sum_{k=1..n} k*binomial(n,k)*9^(n-k).
a(n) = A094798(10^n - 1). (End)
From G. C. Greubel, May 16 2019: (Start)
G.f.: x/(1-10*x)^2.
E.g.f.: x*exp(10*x). (End)
From Amiram Eldar, Oct 28 2020: (Start)
Sum_{n>=1} 1/a(n) = 10*log(10/9).
Sum_{n>=1} (-1)^(n+1)/a(n) = 10*log(11/10). (End)
a(n) = Sum_{k=1..n} A081045(k-1). - Bernard Schott, Nov 17 2022

Offset changed from 0 to 1 by Vincenzo Librandi, Jun 06 2011

A212704 a(n) = 9n10^(n-1).

Original entry on oeis.org

9, 180, 2700, 36000, 450000, 5400000, 63000000, 720000000, 8100000000, 90000000000, 990000000000, 10800000000000, 117000000000000, 1260000000000000, 13500000000000000, 144000000000000000, 1530000000000000000, 16200000000000000000, 171000000000000000000, 1800000000000000000000

Offset: 1

Stanislav Sykora, May 25 2012

Main transitions in systems of n particles with spin 9/2.
Please, refer to the general explanation in A212697.
This particular sequence is obtained for base b=10, corresponding to spin S = (b-1)/2 = 9/2.
Number of 0 needed to write all numbers of n+1 digits. - Bruno Berselli, Jun 30 2014
Essentially the same as A113119. - Bernard Schott, Nov 15 2022
From Bernard Schott, Nov 22 2022: (Start)
Number of nonzero digits needed to write all integers from 1 up to 10^n - 1.
a(n) is a square iff n in { A016754 union A033583\{0} } (see formulas). (End)

Stanislav Sykora, Table of n, a(n) for n = 1..100
Stanislav Sýkora, Magnetic Resonance on OEIS, Stan's NMR Blog (Dec 31, 2014), Retrieved Nov 12, 2019.
Index entries for linear recurrences with constant coefficients, signature (20,-100).

Cf. A008591, A011557, A016754, A033583, A033713, A033714, A053541, A081045, A113119.

Cf. A001787, A212697, A212698, A212699, A212700, A212701, A212702, A212703 (for b=2..9).

Rest@ CoefficientList[Series[9 x/(10 x - 1)^2, {x, 0, 18}], x] (* or *)
Array[9 # 10^(# - 1) &, 18] (* Michael De Vlieger, Nov 18 2019 *)

mtrans(n, b) = n*(b-1)*b^(n-1);
a(n) = mtrans(n, 10);

def a(n): return 9*n*10**(n-1)
print([a(n) for n in range(1, 21)]) # Michael S. Branicky, Nov 14 2022

a(n) = n*(b-1)*b^(n-1) with b=10.
From R. J. Mathar, Oct 15 2013: (Start)
G.f.: 9*x/(10*x-1)^2.
a(n) = 9*A053541(n). (End)
From Bernard Schott, Nov 14 2022: (Start)
a(n+1) - a(n) = 9*A081045(n).
a(n) = A113119(n) for n > 1.
a(n) = A033713(n+1) - A033713(n) = A033714(n+1) - A033714(n).
a(A016754(n)) = (3 * (2n+1) * 10^(2*n*(n+1)))^2.
a(A033583(n)) = (3 * n * 10^(5*n^2))^2. (End)
From Elmo R. Oliveira, May 13 2025: (Start)
E.g.f.: 9*x*exp(10*x).
a(n) = A008591(n)*A011557(n-1).
a(n) = 20*a(n-1) - 100*a(n-2) for n > 2. (End)

A033713 Number of zeros in numbers 1 to 999..9 (n digits).

Original entry on oeis.org

0, 9, 189, 2889, 38889, 488889, 5888889, 68888889, 788888889, 8888888889, 98888888889, 1088888888889, 11888888888889, 128888888888889, 1388888888888889, 14888888888888889, 158888888888888889, 1688888888888888889, 17888888888888888889, 188888888888888888889, 1988888888888888888889

Offset: 1

Olivier Gorin (gorin(AT)roazhon.inra.fr)

Also the first n places of 1, ..., n-digit numbers in the almost-natural numbers (A007376). - Erich Friedman.

a(n+1) is also the total number of digits in numbers from 1 through 999..9 (n digits). - Jianing Song, Apr 17 2022

M. Kraitchik, Mathematical Recreations. Dover, NY, 2nd ed., 1953, p. 49.

F. Calogero, Cool irrational numbers and their rather cool rational approximations, Math. Intell. 25 (4), 72-76 (2003).
Index entries for linear recurrences with constant coefficients, signature (21,-120,100).

Cf. A033714, A058183.

Table[ Sum[9i*10^(i - 1), {i, 1, n}], {n, 0, 16}]
LinearRecurrence[{21,-120,100},{0,9,189},30] (* Harvey P. Dale, Jan 24 2012 *)

a(n)=((n-1)*(10^n)-n*10^(n-1)+1)/9 \\ Charles R Greathouse IV, Feb 19 2017

From Stephen G Penrice, Oct 01 2000: (Start)
a(n) = (1/9)*((n-1)*(10^n)-n*10^(n-1)+1).
G.f.: (9*x^2)/((1-x)(1-10x)^2). (End)
a(n) = Sum_{i=1..n} 9*i*10^(i-1).
a(1)=0, a(2)=9, a(3)=189, a(n)=21*a(n-1)-120*a(n-2)+100*a(n-3). - Harvey P. Dale, Jan 24 2012
a(n+1) = A058183(10^n-1) for n >= 1. - Jianing Song, Apr 17 2022
E.g.f.: exp(x)*(1 + exp(9*x)*(9*x - 1))/9. - Stefano Spezia, Sep 13 2023

More terms from Erich Friedman.

a(18)-a(21) from Stefano Spezia, Sep 13 2023

A268839 a(n) = Sum_{j=1..10^n-1} 2^f(j) where f(j) is the number of zero digits in the decimal representation of j.

Original entry on oeis.org

9, 108, 1197, 13176, 144945, 1594404, 17538453, 192922992, 2122152921, 23343682140, 256780503549, 2824585539048, 31070440929537, 341774850224916, 3759523352474085, 41354756877214944, 454902325649364393, 5003925582143008332, 55043181403573091661

Offset: 1

Michel Lagneau, Feb 14 2016

We calculate the number of integers between 1 and 10^n - 1 having k zeros in their decimal representation. To form a such number consisting of m digits (k < m), place k zeros in m-1 possible positions, then we must choose m-k digits different from zero. Thus, the number of integers between 1 and 10^n - 1 having k zeros in their decimal representation is: Sum_{m=k+1..n} binomial(m-1, k)*9^(m-k).

Hence the sum: Sum_{m=1..n} Sum_{k=0..m-1} binomial(m-1,k)*9^(m-k)*2^k = Sum_{m=1..n} 9^m*(11/9)*(m-1) = (9/10)*(11^n - 1).

			a(1) = 9 because 2^f(1) + 2^f(2) + ... + 2^f(9) = 2^0 + 2^0 + ... + 2^0 = 9;
a(2) = 108 because 2^f(1) + 2^f(2) + ... + 2^f(99) = 9*10 + 2*9 = 108, where f(10) = f(20) = ... = f(90) = 1 and f(i) = 0 otherwise.

Colin Barker, Table of n, a(n) for n = 1..950
Index entries for linear recurrences with constant coefficients, signature (12,-11).

Cf. A016123, A033713, A033714, A055641, A119291.

[(9/10)*(11^n-1): n in [1..20]]; // Vincenzo Librandi, Feb 15 2016

for n from 1 to 100 do: x:=(9/10)*(11^n-1):printf(‘%d, ‘,x):od:

Table[Table[(9/10) (11^n - 1), {n, 1, 20}]] (* Bruno Berselli, Feb 15 2016 *)
CoefficientList[Series[9/((1 - 11 x) (1 - x)), {x, 0, 33}], x] (* Vincenzo Librandi, Feb 15 2016 *)

Vec(9*x/((1-11*x)*(1-x)) + O(x^30)) \\ Colin Barker, Feb 22 2016

a(n) = (9/10)*(11^n-1) = 9*A016123(n-1).
From Vincenzo Librandi, Feb 15 2016: (Start)
G.f.: (9*x)/((1-11*x)*(1-x)).
a(n) = 11*a(n-1) + 9. (End)
E.g.f.: 9*exp(x)*(exp(10*x) - 1)/10. - Stefano Spezia, Sep 13 2023

Name edited by Jon E. Schoenfield, Sep 13 2017

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A053541 a(n) = n*10^(n-1).

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A212704 a(n) = 9*n*10^(n-1).

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A033713 Number of zeros in numbers 1 to 999..9 (n digits).

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A268839 a(n) = Sum_{j=1..10^n-1} 2^f(j) where f(j) is the number of zero digits in the decimal representation of j.

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A212704 a(n) = 9n10^(n-1).