A347317 An alternative version of the inventory sequence A342585: now a row only ends when a 0 is reached that would not be followed by any further terms.

0, 1, 1, 0, 2, 2, 2, 0, 3, 2, 4, 1, 1, 0, 4, 4, 4, 1, 4, 0, 5, 5, 4, 1, 6, 2, 1, 0, 6, 7, 5, 1, 6, 3, 3, 1, 0, 7, 9, 5, 3, 6, 4, 4, 2, 0, 1, 0, 9, 10, 6, 4, 9, 4, 5, 2, 0, 3, 1, 0, 11, 11, 7, 5, 10, 6, 6, 3, 0, 3, 2, 2, 0

Offset: 0

Author

N. J. A. Sloane, Sep 09 2021

Keywords

• nonn
• tabf

Comments

This sequence has offset 0, which seems more appropriate than the offset 1 that A342585 has.

In both A342585 and the present sequence, a row records the numbers of 0's, 1's, 2's, etc., in the sequence so far.

The difference is that in A342585 a row ends when the first 0 is reached. The row beginning 7, for example, is 7, 9, 5, 3, 6, 4, 4, 2, 0, and ends, because there is no 8 in the sequence up to this point. However, there is a 9, so we can say that the row ended prematurely.

In the present sequence we continue the row until we reach a 0 which is 1 more than the highest term in the sequence up to that point, and then the row ends.

So the row beginning 7 is now 7, 9, 5, 3, 6, 4, 4, 2, 0, 1, 0.

From this point on the two sequences differ.

Unfortunately, this version has the drawback that most of the entries are zero!

Examples

			The triangle begins:
   0;
   1,  1,  0;
   2,  2,  2,  0;
   3,  2,  4,  1,  1,  0;
   4,  4,  4,  1,  4,  0;
   5,  5,  4,  1,  6,  2,  1,  0;
   6,  7,  5,  1,  6,  3,  3,  1,  0;
   7,  9,  5,  3,  6,  4,  4,  2,  0,  1,  0;
   9, 10,  6,  4,  9,  4,  5,  2,  0,  3,  1,  0;
  11, 11,  7,  5, 10,  6,  6,  3,  0,  3,  2,  2,  0;
...
		

Links

• Michael S. Branicky, Table of n, a(n) for n = 0..25000

Crossrefs

Cf. A342585. See A347318 for row lengths.

Programs

Mathematica

Block[{c, k, m, r = 0}, c[0] = 1; {0}~Join~Reap[Do[k = 0; While[k <= r, If[IntegerQ@ c[k], Set[m, c[k]], Set[c[k], 0]; Set[m, 0]]; If[m > r, r = m]; Sow[m]; If[IntegerQ@ c[m], c[m]++, c[m] = 1]; k++]; Sow[0]; c[0]++, 9]][[-1, -1]]] (* Michael De Vlieger, Oct 12 2021 *)

Python

from collections import Counter
def aupton(terms):
    num, alst, inventory = 0, [0], Counter([0])
    for n in range(2, terms+1):
        c = inventory[num]
        if c == 0 and num > max(inventory):
            num = 0
        else:
            num += 1
        alst.append(c); inventory.update([c])
    return alst
print(aupton(73)) # Michael S. Branicky, Sep 09 2021