.--------. |Notation| .--------. The sequence "a" denotes A374663, the lexicographically earliest sequence of positive integers such that for any n > 0, Sum_{k = 1..n} 1 / (k*a(k)) < 1. 1 For any n > 0, let r_n = 1 - Sum_{k = 1..n-1} ------. k*a(k) Note that, by construction, r_n > 0. .-----. |Lemma| .-----. ( 1 ) For any n > 0, a(n) = floor(-----) + 1. (n*r_n) Indeed: 1 r_n - ------ = r_{n+1} > 0 with a(n) a positive integer as small as possible. n*a(n) 1 1 ( 1 ) So r_n > ------ => a(n) > ----- => a(n) = floor(-----) + 1. n*a(n) n*r_n (n*r_n) QED .-------. |Theorem| .-------. We will now prove the following theorem: For any n > 1: P[n]: r_n is the inverse of a positive integer, say d_n, Q[n]: d_{n+1} is divisible by d_n, R[n]: d_n is divisible by all positive integers < n. Base case --------- The theorem is satified for small n: a(1) = a(2) = a(3) = 2, n r_n d_n - ---- --- 1 1 1 2 1/2 2 3 1/4 4 4 1/12 12 P[n] is satisfied for n = 1..4, Q[n] is satisfied for n = 1..3, R[n] is satisfied for n = 1..4. Induction step -------------- We assume that P[n], Q[n-1] and R[n] are satisfied for some n > 1. Let d_n = k_n * n + e_n with 0 < e_n < n. ( 1 ) (d_n) (k_n * n + e_n) So a(n) = floor(-----) + 1 = floor(---) + 1 = floor(-------------) + 1 = k_n + 1. (n*r_n) ( n ) ( n ) We consider the two following cases: 1) d_n is divisible by n, 2) d_n is not divisible by n. Case 1 ...... We have d_n = k_n * n. 1 1 1 n * (k_n + 1) - d_n r_{n+1} = r_n - ------------- = --- - ------------- = ------------------- n * (k_n + 1) d_n n * (k_n + 1) d_n * n * (k_n + 1) n*k_n + n - n*k_n n 1 r_{n+1} = ------------------- = ------------------- = --------------- d_n * n * (k_n + 1) d_n * n * (k_n + 1) d_n * (k_n + 1) So: P[n+1]: r_{n+1} is the inverse of the positive integer d_{n+1} = d_n * (k_n + 1), Q[n]: d_{n+1} is divisible by d_n (and by all positive integers < n), R[n+1]: as d_n is a multiple of n, d_{n+1} is also divisible by n, and d_{n+1} is divisible by all positive integers < n+1. Case 2 ...... We have d_n = k_n * n + e_n. n * (k_n + 1) - d_n n * k_n + n - k_n * n - e_n n - e_n r_{n+1} = ------------------- = --------------------------- = ------------------- d_n * n * (k_n + 1) d_n * n * (k_n + 1) d_n * n * (k_n + 1) n - e_n 1 r_{n+1} = ------- * ------------- d_n n * (k_n + 1) n - e_n is a positive integer < n, so it divides d_n, d_n and r_{n+1} is the inverse of the positive integer d_{n+1} = ------- * n * (k_n + 1), and d_{n+1} is divisible by n. n - e_n n * (k_n + 1) = n * k_n + n = n * k_n + e_n - e_n + n = d_n + n - e_n is divisible by n - e_n, n * (k_n + 1) so d_{n+1} = d_n * ------------- is also divisible by d_n (and by all positive integers < n). n - e_n Hence: P[n+1]: r_{n+1} is the inverse of a positive integer d_{n+1}, Q[n]: d_{n+1} is divisible by d_n, R[n+1]: d_{n+1} is divisible by all positive integers < n+1. QED .---------. |Corollary| .---------. For any n >= 0, 1 1 d_{n+1} - 1 A374983(n) Sum_{k = 1..n} -------- = 1 - r_{n+1} = 1 - ------- = ----------- = ----------. k * a(k) d_{n+1} d_{n+1} A375516(n) So A375516(n) = A374983(n) + 1.