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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A277834 Number of '4' digits in the set of all numbers from 0 to A014824(n) = Sum_{i=1..n} i*10^(n-i) = (0, 1, 12, 123, 1234, 12345, ...).

Original entry on oeis.org

0, 0, 1, 22, 344, 4671, 59053, 713985, 8374417, 96089849, 1084355281, 12078120713, 133126886145, 1454725651577, 15781824417015, 170163923182508, 1825096021948551, 19485528120720094, 207200960219546637, 2195466392318923180, 23189231824423799723
Offset: 0

Views

Author

M. F. Hasler, Nov 01 2016

Keywords

Examples

			For n=2 there is only one digit '4' in the sequence 0, 1, 2, ..., 12.
For n=3 there are 11 + 10 = 21 more digits '4' in { 14, 24, 34, 40, ..., 49, 54, ..., 114 }, where 44 accounts for two '4's.

Links

Crossrefs

Cf. A277830 - A277838, A277849, A277635, A272525, A083449, A014824.

Programs

  • PARI
    print1(c=N=0);for(n=1,8,print1(","c+=sum(k=N+1,N=N*10+n,#select(d->d==4,digits(k)))))
  • PARI
    A277834(n,m=4)=if(n>m,A277833(n,m+1)+(m+2)*10^(n-m-1),A277830(n)-(m>n))

Formula

a(n) = A277849(n) = A083449(n) = A277830(n) - 1 for n < 4,
a(n) = A277833(n) - 5*10^(n-4) for n >= 4, a(n) = A277835(n) + 6*10^(n-5) for n >= 5.
More generally, for m = 0, ..., 9, let a[m] denote A277830, ..., A277838 and A277849, respectively. Then a[0](n) = a[n](n) = a[m](n) + 1 for all m > n >= 0, and a[m-1](n) = a[m](n) + (m+1)*10^(n-m) for all n >= m > 1.

Extensions

More terms from Lars Blomberg, Nov 05 2016
Removed incorrect b-file. - David A. Corneth, Dec 31 2020

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
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