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.
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
Index entries for linear recurrences with constant coefficients, signature (22,-141,220,-100).
FORMULA
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
KEYWORD
nonn,base,easy
AUTHOR
Abhinav S. Sharma, Oct 17 2020
STATUS
approved