A395531 Lay down bricks of length 1, 2, 3, ... on the top-right quadrant of the x-y plane. Place a brick as close to x=0 as possible, and on the highest row with bricks directly underneath where the new brick does not overhang. a(n) is the length of the leftmost brick on each row.

1, 3, 7, 11, 16, 25, 32, 40, 49, 59, 73, 85, 98, 112, 127, 147, 164, 182, 201, 221, 242, 272, 295, 319, 344, 370, 397, 425, 454, 484, 515, 553, 586, 620, 655, 691, 728, 766, 805, 852, 893, 935, 978, 1022, 1067, 1113, 1160, 1208, 1265, 1315, 1366, 1418, 1471, 1525, 1580, 1636, 1693

Offset

1,2

Comments

Positions of 0's in A233380. - Rémy Sigrist, Apr 30 2026

Conjecture: when n = a(m) for some m, we have a(n) = n(n+3)/2 - m. Verified experimentally up to m=1172, n=688491. - Aleksei Udovenko, May 07 2026

Links

Aleksei Udovenko, Table of n, a(n) for n = 1..10000

Aleksei Udovenko, Optimized C++ program

Example

After 11 bricks have been placed, the wall looks like this (with A for block 10 and B for block 11):
  BBBBBBBBBBB---------
  7777777AAAAAAAAAA---
  333666666999999999--
  12244445555588888888
The next brick, 12 units long, has to go on the bottom row because on any other row it would have to overhang the row below.

Prog

(Python)
from itertools import count
def a(): # Generator function for the sequence
    rows = []
    height = 0
    for n in count(1):
        if height==0 or n <= rows[height-1]:
            yield n
            height += 1
            rows.append(n)
        else:
            for i in range(height-1, -1, -1):
                if i==0 or rows[i] + n <= rows[i-1]:
                    rows[i] += n
                    break

Crossrefs

Cf. A233380.

Sequence in context: A131024 A053982 A118000 * A121640 A319258 A112786

Adjacent sequences: A395527 A395529 A395530 * A395533 A395534 A395535

Keyword

easy,nonn

Author

Christian Perfect, Apr 27 2026

Status

approved