1,2

It appears that A007952(n) is the index of the row where n first appears.

It appears that A007952(n)-1 is the index of the row where the last nonzero term of the n-th column is seen. - Michel Marcus, Aug 02 2025

Michel Marcus, Table of n, a(n) for n = 1..500

Triangle whose columns are summed.

m/n| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

----------------------------------------------------------------

1 | 1

2 | 1

3 | 2 1

4 | 2 1

5 | 3 2 1

6 | 3 2 1

7 | 3 2 1

8 | 3 2 1

9 | 4 3 2 1

10 | 4 3 2 1

11 | 5 4 3 2 1

12 | 5 4 3 2 1

13 | 5 4 3 2 1

14 | 5 4 3 2 1

15 | 5 4 3 2 1

16 | 5 4 3 2 1

17 | 6 5 4 3 2 1

18 | 6 5 4 3 2 1

19 | 6 5 4 3 2 1

20 | 6 5 4 3 2 1

...

The completed column for n=5 is definitely fully visible here because in column 6 for n=6 the divisor k=6 already appeared. That means that column 5 cannot have more divisors in it under the last k=5 in row 17 because in that row only k=7 may follow k=6 in theory, but 7 does not divide 5. So, all similarly proven, definitely fully visible completed columns in this sample array are readily summable by sight. E.g. column 5: a(5) = 1 + 5 + 5 + 5 + 5 + 5 + 5 = 31.

(PARI) \\ uses row(n) from A386755

a(n) = my(ok=1, k=1, last=-1, s=0, r); while(ok, r=row(k); if (#r >= n, s+=r[n]); k++; if (#r>=n, if ((last==n) && (r[n]==0), ok = 0, last = r[n]))); s; \\ Michel Marcus, Aug 02 2025

(PARI) \\ uses row(n) from A386755

lista(nn) = my(ok=1, k=1, vlast=vector(nn, i, -1), vs=vector(nn)); while(ok, my(r=row(k)); for (i=1, nn, if (#r>=i, vs[i]+=r[i])); k++; my(nbok=0); for (i=1, nn, if (#r>=i, if ((vlast[i]==i) && (r[i]==0), nbok++, vlast[i] = r[i]))); if (nbok == nn, ok = 0); ); vs; \\ Michel Marcus, Aug 02 2025

Cf. A386755.

Cf. A007952 (row number where k=n first appears).

Sequence in context: A232610 A235337 A151994 * A321992 A231806 A183782

Adjacent sequences: A386517 A386518 A386519 * A386521 A386522 A386523

Tamas Sandor Nagy, Jul 24 2025

approved