a(0) = 1, a(n) = sum of digits of all previous terms

Open in the 3-D viewerA004207 on the OEIS
| Terms | 100,000 (n = 1 … 100,000) |
|---|---|
| Decomposable (a > 2d) | 99,996 |
| Level class, k > L | 14,194 · 14.19 % |
| Weight class, k ≤ L | 85,802 · 85.81 % |
| Ties, k = L | 1 |
| On the level line L = 1 | 3 |
| Forced level, l ≤ d² | 12 |
| Range of a(n) | 1 … 2,609,882 |
| Range of the jump d | 1 … 50 |
| Largest weight k, level L | 289,937, 300,015 |
a(n) = a(n-1) + digitsum(a(n-1)), so 9 divides l at every term. Above a = 800 every level term lies on a line L in 9Z (a level term off 9Z needs l <= 3d^2, impossible past a = 10^4), and Lemma 3 (L <= d) caps the lines at the gap: with d <= 50 in this range only L = 9, 18, 27, 36 occur. The 12 level terms off 9Z are all below 800.
The decomposition
Every term of a strictly increasing sequence is written a(n) = k(n)·L(n) + d(n): the jump d = a(n+1) − a(n), the weight k the least divisor of a − d greater than d, the level L = (a − d)/k. A term decomposes when a > 2d; it is in the level class when k > L and in the weight class when k ≤ L. See decompwlj.com and arXiv:0711.0865, or how it works, with worked examples and a live one.