A355917 Variant of Inventory Sequence A342585 where indices are also counted (short version).

0, 2, 0, 4, 1, 1, 0, 6, 4, 2, 1, 2, 0, 8, 6, 5, 2, 3, 2, 2, 0, 10, 7, 9, 4, 5, 4, 3, 2, 1, 1, 1, 0, 12, 11, 11, 6, 7, 5, 5, 4, 2, 2, 2, 3, 1, 0, 14, 13, 15, 8, 9, 8, 6, 5, 5, 4, 3, 4, 2, 2, 1, 1, 0, 16, 16, 18, 10, 12, 11, 8, 6, 7, 5, 5, 6, 4, 3, 2, 2, 3, 0

Offset: 1

Author

N. J. A. Sloane, Sep 24 2022

Keywords

• nonn

Comments

See A355916 for further information.

Examples

			The first eight inventories are:
  0,
  2, 0,
  4, 1, 1, 0,
  6, 4, 2, 1, 2, 0,
  8, 6, 5, 2, 3, 2, 2, 0,
  10, 7, 9, 4, 5, 4, 3, 2, 1, 1, 1, 0,
  12, 11, 11, 6, 7, 5, 5, 4, 2, 2, 2, 3, 1, 0,
  14, 13, 15, 8, 9, 8, 6, 5, 5, 4, 3, 4, 2, 2, 1, 1, 0,
  ...
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 1..10030
• Rémy Sigrist, PARI program

Crossrefs

Cf. A342585, A355916, A355918.

Programs

Mathematica

nn = 9; c[] = 0; i = 1; Do[k = 0; While[c[k] > 0, Set[{a[i], a[i + 1]}, {c[k], k}]; c[a[i]]++; c[a[i + 1]]++; i += 2; k++]; Set[{a[i], a[i + 1]}, {c[k], k}]; c[a[i]]++; c[a[i + 1]]++; i += 2, nn]; Array[a[2 # - 1] &, (i - 1)/2] (* _Michael De Vlieger, Sep 25 2022 *)

PARI

See Links section.

Python

from collections import Counter
def aupton(terms):
    num, alst, inventory = 0, [0, 0], Counter([0, 0])
    for n in range(3, 2*terms):
        c = [inventory[num], num]
        num = 0 if c[0] == 0 else num + 1
        alst.extend(c)
        inventory.update(c)
    return alst[:2*terms:2]
print(aupton(82)) # Michael S. Branicky, Sep 25 2022

Extensions

More terms from Rémy Sigrist, Sep 25 2022