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

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

Offset

1,2

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 generalisation 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

Table of n, a(n) for n=1..21.

Index entries for linear recurrences with constant coefficients, signature (22,-141,220,-100).

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

Mathematica

LinearRecurrence[{22, -141, 220, -100}, {0, 9, 198, 3087}, 21] (* Amiram Eldar, Oct 26 2020 *)

Prog

(PARI) concat(0, Vec(9*x^2 / ((1 - x)^2*(1 - 10*x)^2) + O(x^20))) \\ Colin Barker, Oct 27 2020

Crossrefs

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

Sequence in context: A180778 A110807 A019566 * A157563 A003026 A157594

Adjacent sequences: A338223 A338224 A338225 * A338227 A338228 A338229

Keyword

nonn,base,easy

Author

Abhinav S. Sharma, Oct 17 2020

Status

approved