.--------.
|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.