How do I define a function in mathemathica that uses a complicated recurrence relation?

Question

I am trying to write a little script to compute an integer sequence. The function I'm trying to write in code is the one int the blackboard, a(n). The problem is that I was expecting the function h(n) I defined in the script to give a number as a result, but it is giving something else: for h(2) it gives ArgMax[{p, HarmonicNumber[p] <= 1}, p, Integers] How can I correct that? (You have to understand that I am by no means a programmer, nor know much about mathematica. Thanks in advnace. The script I wrote is this one:

    h[n_] := (ArgMax[{p, 
      Sum[1/s, {s, 1 + Sum[h[k], {k, 1, (n - 1)}], p}] <= 1}, p, 
     Integers]) - Sum[h[k], {k, 1, (n - 1)}]; h[1] = 1;

a(n)=(maximum p such that the sum from s equals r to p is less or equal than one)-r+1, where r=1+the sum from k=1 to (n-1) of a(k), and a(1)=1

PD: Those letters that look like v's are r's. Sorry.

Answer 1

a[1] = 1;

a[n_] := Module[{sum = 0},
  r = 1 + Sum[a[k], {k, n - 1}];
  x = r;
  While[sum <= 1, sum += 1/x++];
  p = x - 2;
  p - r + 1]

Table[a[n], {n, 6}]

{1, 2, 6, 16, 43, 117}

The result for a[4] is 16 not 14.

To illustrate, when n = 4

r = 1 + Sum[a[k], {k, 4 - 1}]

  = 1 + a[1] + a[2] + a[3]       (* refer to established results for a[n] *)

  = 1 +  1  +  2  +  6  =  10

sum = 0;
x = r;
While[sum <= 1, sum += 1/x++];
p = x - 2;
p - r + 1

16

Or in another form

Total[Table[1/s, {s, 10, 25}]] <= 1   (* True *)

p - r + 1 = 25 - 10 + 1 = 16

Using memoisation, as mentioned by ogerard

Clear[a]

a[1] = 1;

a[n_] := a[n] = Module[{sum = 0},
   r = 1 + Sum[a[k], {k, n - 1}];
   x = r;
   While[sum <= 1, sum += 1/x++];
   p = x - 2;
   p - r + 1]

only reduces the following run time by 9 seconds

Timing[Table[a[n], {n, 14}]]

{40.8906, {1, 2, 6, 16, 43, 117, 318, 865, 2351, 6391, 17372, 47222, 128363, 348927}}