A355916 Variant of Inventory Sequence A342585 where indices are also counted (long version).

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

Offset

1,3

Comments

Similar to A342585, except that when we take inventory, we write down what we are counting as a subscript on the count. So if we have found k copies of m so far, we write down k_m, and include both the k and m values when we next take inventory.

More than the usual number of terms are shown, in order to match A355917.

Links

Rémy Sigrist, Table of n, a(n) for n = 1..10020

Rémy Sigrist, PARI program

N. J. A. Sloane, The first eight inventories, with better alignment.

Example

Initially we have no 0's, so the first inventory is 0_0. Just as in A342585, when we reach a count of zero, we take a new inventory.
Now we see two 0's, so we write down 2_0, followed by 0_1, since there are no 1's.
So the first two inventories are
  0_0,
  2_0, 0_1.
Now we see four 0's, so the next inventory starts 4_0, then 1_1, 1_2, and 0_3:
  4_0, 1_1, 1_2, 0_3.
The first eight inventories are:
  0_0,
  2_0, 0_1,
  4_0, 1_1, 1_2, 0_3,
  6_0, 4_1, 2_2, 1_3, 2_4, 0_5,
  8_0, 6_1, 5_2, 2_3, 3_4, 2_5, 2_6, 0_7,
  10_0, 7_1, 9_2, 4_3, 5_4, 4_5, 3_6, 2_7, 1_8, 1_9, 1_10, 0_11,
  12_0, 11_1, 11_2, 6_3, 7_4, 5_5, 5_6, 4_7, 2_8, 2_9, 2_10, 3_11, 1_12, 0_13,
  14_0, 13_1, 15_2, 8_3, 9_4, 8_5, 6_6, 5_7, 5_8, 4_9, 3_10, 4_11, 2_12, 2_13, 1_14, 1_15, 0_16,
...
The sequence is obtained by reading the inventories, with each count followed by its index: 0, 0, 2, 0, 0, 1, 4, 0, 1, 1, 1, 2, 0, 3, ...
If the indices are omitted, we get the short version, A355917. A355918 lists the highest index in each inventory.

Mathematica

nn = 9; c[_] = 0; a[1] = a[2] = 0; c[0] = 2; i = 3; 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, {n, 2, nn}]; Array[a, i - 1] (* Michael De Vlieger, Sep 25 2022 *)

Prog

(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, 3+terms//2):
        c = [inventory[num], num]
        num = 0 if c[0] == 0 else num + 1
        alst.extend(c)
        inventory.update(c)
    return alst[:terms]
print(aupton(128)) # Michael S. Branicky, Sep 25 2022

Crossrefs

Cf. A342585, A355917, A355918.

Sequence in context: A168511 A111146 A109077 * A309023 A353429 A137585

Adjacent sequences: A355913 A355914 A355915 * A355917 A355918 A355919

Keyword

nonn

Author

N. J. A. Sloane, Sep 24 2022

Status

approved