A081045 -id:A081045 - 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)

A113119 Total number of digits in all n-digit nonnegative integers.

Original entry on oeis.org

10, 180, 2700, 36000, 450000, 5400000, 63000000, 720000000, 8100000000, 90000000000, 990000000000, 10800000000000, 117000000000000, 1260000000000000, 13500000000000000, 144000000000000000, 1530000000000000000, 16200000000000000000, 171000000000000000000

Offset: 1

Alexandre Wajnberg, Jan 03 2006

			a(1)=10 because there are ten one-digit numbers (including the 0).
a(2)=180 because there are 100-10=90 two-digit numbers, for a total of 90*2=180 digits.

Colin Barker, Table of n, a(n) for n = 1..950
The Math Forum, Ask Dr. Math: Really Counting to One Billion, 2001.
Index entries for linear recurrences with constant coefficients, signature (20,-100).

Cf. A053541, A081045.

Essentially the same as A212704.

LinearRecurrence[{20,-100},{10,180,2700},20] (* Harvey P. Dale, Dec 09 2021 *)

Vec(10*x*(1-2*x+10*x^2)/(1-10*x)^2 + O(x^20)) \\ Colin Barker, Aug 05 2016

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

For n > 1, a(n) = 9*n*10^(n-1).
From Colin Barker, Aug 05 2016: (Start)
a(n) = 20*a(n-1) - 100*a(n-2) for n > 3.
G.f.: 10*x*(1 - 2*x + 10*x^2) / (1-10*x)^2.
(End)
From Bernard Schott, Nov 14 2022: (Start)
a(n) = A212704(n) for n > 1.
a(n) = 9 * A053541(n) for n > 1.
a(n) = 9 * A081045(n-1) + A212704(n-1), for n > 1 (means a(n) = number of nonzero digits + number of zero digits). (End)
E.g.f.: x*(1 + 9*exp(10*x)). - Stefano Spezia, Dec 24 2022

More terms from Joshua Zucker, May 08 2006

a(17) corrected by Colin Barker, Aug 05 2016

A081044 9th binomial transform of (1,8,0,0,0,0,0,0,.....).

Original entry on oeis.org

1, 17, 225, 2673, 29889, 321489, 3365793, 34543665, 349156737, 3486784401, 34480423521, 338218086897, 3295011258945, 31914537622353, 307565765227809, 2951106226689969, 28207085096966913, 268687927383516945

Offset: 0

Paul Barry, Mar 04 2003

Also number of (n+1)-digit numbers with exactly one '9' in their decimal expansion. Nine can be replaced by any nonzero digit 1..9. - Zak Seidov, Jul 11 2016

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (18,-81).

Cf. A081043, A081045.

Table[(8n+9)9^(n-1),{n,0,30}] (*or*) LinearRecurrence[{18, -81}, {1, 17}, 40] (* Vincenzo Librandi, Feb 23 2012 *)

a(n) = (8*n+9)*9^(n-1); \\ Altug Alkan, Jul 18 2016

a(n) = 18*a(n-1)-81*a(n-2), a(0)=0, a(1)=17.
a(n) = (8n+9)*9^(n-1).
a(n) = Sum_{k=0..n} (k+1)*8^k*binomial(n, k).
G.f.: (1-x)/(1-9x)^2.
E.g.f.: (1 + 8*x)*exp(9*x). - Ilya Gutkovskiy, Jul 18 2016

A358439 Number of even digits necessary to write all positive n-digit integers.

Original entry on oeis.org

4, 85, 1300, 17500, 220000, 2650000, 31000000, 355000000, 4000000000, 44500000000, 490000000000, 5350000000000, 58000000000000, 625000000000000, 6700000000000000, 71500000000000000, 760000000000000000, 8050000000000000000, 85000000000000000000, 895000000000000000000

Offset: 1

Bernard Schott, Nov 16 2022

If nonnegative n-digit integers were considered, then a(1) would be 5.
Also, a(n) is the total number of holes in all positive n-digit integers, assuming 4 has no hole. Digits 0, 6 and 9 have 1 hole, digit 8 has 2 holes, and other digits have no holes or (circular) loops (as in A064532).
Proof of the first formula: For n>=2, to write all positive n-digit integers, digits 6, 8, 9 occur A081045(n-1) = (9n+1)*10^(n-2) times each, and digit 0 occurs A212704(n-1) = 9*(n-1)*10^(n-2) times; so a(n) = 4*A081045(n-1) + A212704(n-1).
For a(1), if 0 were included then there would be 5 holes in the 1-digit numbers 0..9.

			To write the integers from 10 up to 99, each of the digits 2, 4, 6 and 8 must be used 19 times, and digit 0 must be used 9 times hence a(2) = 4*19 + 9 = 85.

Index entries for linear recurrences with constant coefficients, signature (20,-100).

Cf. A064532, A081045, A212704.

Cf. A113119, A196563, A358854, A359271.

Maple
```
seq((5*(9*n-1))*10^(n-2), n = 1 .. 30);
```

a[n_] := 5*(9*n - 1)*10^(n - 2); Array[a, 22] (* Amiram Eldar, Nov 16 2022 *)

a(n) = 5*(9n-1)*10^(n-2).
Formulas coming from the name with even digits:
a(n) = A358854(10^n-1) - A358854(10^(n-1)-1).
a(n) = A113119(n) - A359271(n) for n >= 2.
Formula coming from the comment with holes:
a(n) = Sum_{k=10^(n-1)..10^n-1} A064532(k).

A358620 Number of nonzero digits needed to write all nonnegative n-digit integers.

Original entry on oeis.org

9, 171, 2520, 33300, 414000, 4950000, 57600000, 657000000, 7380000000, 81900000000, 900000000000, 9810000000000, 106200000000000, 1143000000000000, 12240000000000000, 130500000000000000, 1386000000000000000, 14670000000000000000, 154800000000000000000

Offset: 1

Bernard Schott, Nov 23 2022

			a(1) = 9 because there are 9 one-digit numbers that are > 0.
a(2) = 171 because there are 90 two-digit numbers, so 90*2 = 180 digits are needed to write these integers, nine of these integers end with 0, and 180-9 = 171.

Index entries for linear recurrences with constant coefficients, signature (20,-100).

Cf. A081045, A113119, A212704.

Maple
```
seq((9*(9*n+1))*10^(n-2), n = 1 .. 20);
```

a[n_] := 9*(9*n + 1)*10^(n - 2); Array[a, 20] (* Amiram Eldar, Nov 23 2022 *)

a(n)=(81*n+9)*10^(n-2) \\ Charles R Greathouse IV, Nov 29 2022

def A358620(n): return 9 if n == 1 else 9*(9*n+1)*10**(n-2) # Chai Wah Wu, Nov 29 2022

a(n) = 9 * (9*n+1) * 10^(n-2).
a(n) = 20*a(n-1) - 100*a(n-2); a(1)=9, a(2)=171.
a(n) = 9 * A081045(n-1).
a(n) = A113119(n) - A212704(n-1), for n >= 2.

A380747 Array read by ascending antidiagonals: A(n,k) = [x^n] (1 - x)/(1 - k*x)^2.

Original entry on oeis.org

1, -1, 1, 0, 1, 1, 0, 1, 3, 1, 0, 1, 8, 5, 1, 0, 1, 20, 21, 7, 1, 0, 1, 48, 81, 40, 9, 1, 0, 1, 112, 297, 208, 65, 11, 1, 0, 1, 256, 1053, 1024, 425, 96, 13, 1, 0, 1, 576, 3645, 4864, 2625, 756, 133, 15, 1, 0, 1, 1280, 12393, 22528, 15625, 5616, 1225, 176, 17, 1

Offset: 0

Stefano Spezia, Jan 31 2025

			The array begins as:
   1, 1,   1,    1,     1,     1, ...
  -1, 1,   3,    5,     7,     9, ...
   0, 1,   8,   21,    40,    65, ...
   0, 1,  20,   81,   208,   425, ...
   0, 1,  48,  297,  1024,  2625, ...
   0, 1, 112, 1053,  4864, 15625, ...
   0, 1, 256, 3645, 22528, 90625, ...
   ...

Cf. A000012 (k=1 or n=0), A000567 (n=2), A001792 (k=2), A007778, A060747 (n=1), A081038 (k=3), A081039 (k=4), A081040 (k=5), A081041 (k=6), A081042 (k=7), A081043 (k=8), A081044 (k=9), A081045 (k=10), A103532, A154955, A380748 (antidiagonal sums).

A[0,0]:=1; A[1,0]:=-1; A[n_,k_]:=((k-1)*n+k)k^(n-1); Table[A[n-k,k],{n,0,10},{k,0,n}]//Flatten (* or *)
A[n_,k_]:=SeriesCoefficient[(1-x)/(1-k*x)^2,{x,0,n}]; Table[A[n-k,k],{n,0,10},{k,0,n}]//Flatten (* or *)
A[n_,k_]:=n!SeriesCoefficient[Exp[k*x](1+(k-1)*x),{x,0,n}]; Table[A[n-k,k],{n,0,10},{k,0,n}]//Flatten

A(n,k) = ((k - 1)*n + k)*k^(n-1) with A(0,0) = 1.
A(n,k) = n! * [x^n] exp(k*x)*(1 + (k - 1)*x).
A(n,0) = A154955(n+1).
A(3,n) = A103532(n-1) for n > 0.
A(n,n) = A007778(n) for n > 0.

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

GAP

Magma

Maple

Mathematica

PARI

Sage

Formula

Extensions

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Python

Formula

A113119 Total number of digits in all n-digit nonnegative integers.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Python

Formula

Extensions

A081044 9th binomial transform of (1,8,0,0,0,0,0,0,.....).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A358439 Number of even digits necessary to write all positive n-digit integers.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A358620 Number of nonzero digits needed to write all nonnegative n-digit integers.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Python

Formula

A380747 Array read by ascending antidiagonals: A(n,k) = [x^n] (1 - x)/(1 - k*x)^2.

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