cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

A212704 a(n) = 9*n*10^(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

Views

Author

Stanislav Sykora, May 25 2012

Keywords

Comments

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)

Links

Crossrefs

Cf. A008591, A011557, A016754, A033583, A033713, A033714, A053541, A081045, A113119.
Cf. A001787, A212697, A212698, A212699, A212700, A212701, A212702, A212703 (for b=2..9).

Programs

  • Mathematica
    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 *)
  • PARI
    mtrans(n, b) = n*(b-1)*b^(n-1);
a(n) = mtrans(n, 10);
  • Python
    def a(n): return 9*n*10**(n-1)
print([a(n) for n in range(1, 21)]) # Michael S. Branicky, Nov 14 2022

Formula

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)

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

Views

Author

Bernard Schott, Nov 16 2022

Keywords

Comments

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.

Examples

			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.

Links

Crossrefs

Cf. A064532, A081045, A212704.
Cf. A113119, A196563, A358854, A359271.

Programs

  • Maple
    seq((5*(9*n-1))*10^(n-2), n = 1 .. 30);
  • Mathematica
    a[n_] := 5*(9*n - 1)*10^(n - 2); Array[a, 22] (* Amiram Eldar, Nov 16 2022 *)

Formula

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

Views

Author

Bernard Schott, Nov 23 2022

Keywords

Examples

			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.

Links

Crossrefs

Cf. A081045, A113119, A212704.

Programs

  • Maple
    seq((9*(9*n+1))*10^(n-2), n = 1 .. 20);
  • Mathematica
    a[n_] := 9*(9*n + 1)*10^(n - 2); Array[a, 20] (* Amiram Eldar, Nov 23 2022 *)
  • PARI
    a(n)=(81*n+9)*10^(n-2) \\ Charles R Greathouse IV, Nov 29 2022
  • Python
    def A358620(n): return 9 if n == 1 else 9*(9*n+1)*10**(n-2) # Chai Wah Wu, Nov 29 2022

Formula

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.

A359271 Number of odd digits necessary to write all nonnegative n-digit integers.

Original entry on oeis.org

5, 95, 1400, 18500, 230000, 2750000, 32000000, 365000000, 4100000000, 45500000000, 500000000000, 5450000000000, 59000000000000, 635000000000000, 6800000000000000, 72500000000000000, 770000000000000000, 8150000000000000000
Offset: 1

Views

Author

Bernard Schott, Dec 23 2022

Keywords

Examples

			To write the integers from 10 up to 99, each of the digits 1, 3, 5, 7 and 9, must be used 19 times, hence a(2) = 19*5 = 95.

Links

Crossrefs

Cf. A113119, A196564, A279766, A358439.

Programs

  • Maple
    seq(5 * (9*n+1) * 10^(n-2), n=1..18);
  • Mathematica
    a[n_] := 5*(9*n + 1)*10^(n - 2); Array[a, 20] (* Amiram Eldar, Dec 23 2022 *)

Formula

a(n) = 5 * (9*n+1) * 10^(n-2).
a(n) = A279766(10^n-1) - A279766(10^(n-1)-1).
a(n) = A113119(n) - A358439(n).
From Stefano Spezia, Dec 24 2022: (Start)
O.g.f.: 5*x*(1 - x)/(1 - 10*x)^2.
E.g.f.: (exp(10*x)*(1 + 90*x) - 1)/20. (End)

A173906 Total number of digits of the squares of all n-digit numbers.

Original entry on oeis.org

16, 338, 5183, 69837, 878377, 10583772, 123837722, 1418377223, 15983772233, 177837722339, 1958377223398, 21383772233983, 231837722339831, 2498377223398316, 26783772233983162, 285837722339831620, 3038377223398316206, 32183772233983162066, 339837722339831620668, 3578377223398316206680
Offset: 1

Views

Author

Robert G. Wilson v, Nov 26 2010

Keywords

Comments

Sequence inspired by Claudio Meller's email to [seqfan] 07:01 a.m., Nov 26 2010, subject: About a problem of Bernardo Recamán Santos found in http://www.mathpuzzle.com/

Crossrefs

Cf. A000290, A062940, A113119.

Programs

  • Mathematica
    f[n_] := Block[{br = 1+ Floor[10^(n-1)*Sqrt@10]},(2n-1)(br-10^(n-1)) + 2n(10^n-br)]; Array[f, 20]

Formula

Limit_{n->oo} a(n)/(n*10^n) = 9/5. [corrected by Jason Yuen, Feb 07 2025]
From Jason Yuen, Feb 07 2025: (Start)
a(n) = (2*n-1)*A062940(2*n-1) + (2*n)*A062940(2*n).
a(n) = (9/5)*n*10^n - ceiling((sqrt(10)-1)*10^(n-1)) for n > 1. (End)
Showing 1-5 of 5 results.

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