A338226 - OEIS

%I #38 Dec 21 2020 17:43:20

%S 0,9,198,3087,41976,530865,6419754,75308643,864197532,9753086421,

%T 108641975310,1197530864199,13086419753088,141975308641977,

%U 1530864197530866,16419753086419755,175308641975308644,1864197530864197533,19753086419753086422,208641975308641975311,2197530864197530864200

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

%C 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.

%C a(n) = A019566(n) for n <= 9. This is an alternate generalisation of A019566 beyond n=9.

%C 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).

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (22,-141,220,-100).

%F a(n) = A052245(n) - A014824(n).

%F a(n+1) - a(n) = A033713(n+1).

%F a(n) = ((9*n - 11)*10^n + (9*n + 11))/81. - _Andrew Howroyd_, Oct 26 2020

%F From _Colin Barker_, Oct 26 2020: (Start)

%F G.f.: 9*x^2 / ((1 - x)^2*(1 - 10*x)^2).

%F a(n) = 22*a(n-1) - 141*a(n-2) + 220*a(n-3) - 100*a(n-4) for n>4.

%F (End)

%F E.g.f.: exp(x)*(11 + 9*x + exp(9*x)*(90*x - 11))/81. - _Stefano Spezia_, Oct 27 2020

%t LinearRecurrence[{22, -141, 220, -100}, {0, 9, 198, 3087}, 21] (* _Amiram Eldar_, Oct 26 2020 *)

%o (PARI) concat(0, Vec(9*x^2 / ((1 - x)^2*(1 - 10*x)^2) + O(x^20))) \\ _Colin Barker_, Oct 27 2020

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

%K nonn,base,easy

%O 1,2

%A _Abhinav S. Sharma_, Oct 17 2020