Consider the recursion b(1,n) = 1, b(k+1,n) = b(k,n) + (b(k,n) reduced mod(k+n)); then there is a number y such that b(k,n)-b(k-1,n) is a constant (= A074482(n)) for k > y. Sequence gives values of y.

A074483

Consider the recursion b(1,n) = 1, b(k+1,n) = b(k,n) + (b(k,n) reduced mod(k+n)); then there is a number y such that b(k,n)-b(k-1,n) is a constant (= A074482(n)) for k > y. Sequence gives values of y.