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.

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A099673 Partial sums of repdigits of A002280.

Original entry on oeis.org

6, 72, 738, 7404, 74070, 740736, 7407402, 74074068, 740740734, 7407407400, 74074074066, 740740740732, 7407407407398, 74074074074064, 740740740740730, 7407407407407396, 74074074074074062, 740740740740740728, 7407407407407407394, 74074074074074074060, 740740740740740740726
Offset: 1

Views

Author

Labos Elemer, Nov 17 2004

Keywords

Examples

			6 + 66 + 666 + 6666 + 66666 = a(5) = 74070.

Links

Crossrefs

Cf. A002275-A002283, A014824, A057932, A099669-A099675.

Programs

  • Mathematica
    <Robert G. Wilson v, Nov 20 2004 *)

Formula

a(n) = (2/27)*(10^(n+1) - 9*n - 10). - R. Piyo (nagoya314(AT)yahoo.com), Dec 10 2004
From Elmo R. Oliveira, Apr 02 2025: (Start)
G.f.: 6*x/((1 - x)^2*(1 - 10*x)).
a(n) = 6*A014824(n).
E.g.f.: 2*exp(x)*(10*exp(9*x) - 9*x - 10)/27.
a(n) = 12*a(n-1) - 21*a(n-2) + 10*a(n-3) for n > 3. (End)

Extensions

More terms from Elmo R. Oliveira, Apr 02 2025

A338226 a(n) = Sum_{i=0..n-1} i*10^i - Sum_{i=0..n-1} (n-1-i)*10^i.

Original entry on oeis.org

0, 9, 198, 3087, 41976, 530865, 6419754, 75308643, 864197532, 9753086421, 108641975310, 1197530864199, 13086419753088, 141975308641977, 1530864197530866, 16419753086419755, 175308641975308644, 1864197530864197533, 19753086419753086422, 208641975308641975311, 2197530864197530864200
Offset: 1

Views

Author

Abhinav S. Sharma, Oct 17 2020

Keywords

Comments

Note that adding a constant k does not change the result: a(n) = (Sum_{i=0..n-1} (k+i) * 10^i) - (Sum_{i=0..n-1} (k+n-1-i) * 10^i). This means any set of consecutive numbers may be used to generate the terms.
a(n) = A019566(n) for n <= 9. This is an alternate generalization of A019566 beyond n=9.
For two numbers A = Sum_{i=0..n-1} (x_i) * b^i and A' = Sum_{i=0..n-1} (x'i) * b^i, A-A' is divisible by b-1 if Sum{i=0..n-1} (x_i) = Sum_{i=0..n-1} (x'_i). x_i and x'_i are sets of integers. This is because b^i == 1 (mod b-1). In this specific case b=10, hence all terms are divisible by 9 and are given by a(n) = 9*A272525(n-1).

Links

Crossrefs

Cf. A033713 (first differences), A019566 ("unique" numbers).

Programs

  • Mathematica
    LinearRecurrence[{22, -141, 220, -100}, {0, 9, 198, 3087}, 21] (* Amiram Eldar, Oct 26 2020 *)
  • PARI
    concat(0, Vec(9*x^2 / ((1 - x)^2*(1 - 10*x)^2) + O(x^20))) \\ Colin Barker, Oct 27 2020

Formula

a(n) = A052245(n) - A014824(n).
a(n+1) - a(n) = A033713(n+1).
a(n) = ((9*n - 11)*10^n + (9*n + 11))/81. - Andrew Howroyd, Oct 26 2020
From Colin Barker, Oct 26 2020: (Start)
G.f.: 9*x^2 / ((1 - x)^2*(1 - 10*x)^2).
a(n) = 22*a(n-1) - 141*a(n-2) + 220*a(n-3) - 100*a(n-4) for n>4.
(End)
E.g.f.: exp(x)*(11 + 9*x + exp(9*x)*(90*x - 11))/81. - Stefano Spezia, Oct 27 2020
Previous Showing 41-42 of 42 results.

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