cp's OEIS Frontend

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).