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

%I A355916 #36 Sep 25 2022 21:35:03
%S A355916 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,
%T A355916 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,
%U A355916 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
%N A355916 Variant of Inventory Sequence A342585 where indices are also counted (long version).
%C A355916 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.
%C A355916 More than the usual number of terms are shown, in order to match A355917.
%H A355916 Rémy Sigrist, <a href="/A355916/b355916.txt">Table of n, a(n) for n = 1..10020</a>
%H A355916 Rémy Sigrist, <a href="/A355916/a355916.gp.txt">PARI program</a>
%H A355916 N. J. A. Sloane, <a href="/A355916/a355916.png">The first eight inventories</a>, with better alignment.
%e A355916 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.
%e A355916 Now we see two 0's, so we write down 2_0, followed by 0_1, since there are no 1's.
%e A355916 So the first two inventories are
%e A355916   0_0,
%e A355916   2_0, 0_1.
%e A355916 Now we see four 0's, so the next inventory starts 4_0, then 1_1, 1_2, and 0_3:
%e A355916   4_0, 1_1, 1_2, 0_3.
%e A355916 The first eight inventories are:
%e A355916   0_0,
%e A355916   2_0, 0_1,
%e A355916   4_0, 1_1, 1_2, 0_3,
%e A355916   6_0, 4_1, 2_2, 1_3, 2_4, 0_5,
%e A355916   8_0, 6_1, 5_2, 2_3, 3_4, 2_5, 2_6, 0_7,
%e A355916   10_0, 7_1, 9_2, 4_3, 5_4, 4_5, 3_6, 2_7, 1_8, 1_9, 1_10, 0_11,
%e A355916   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,
%e A355916   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,
%e A355916 ...
%e A355916 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, ...
%e A355916 If the indices are omitted, we get the short version, A355917. A355918 lists the highest index in each inventory.
%t A355916 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 *)
%o A355916 (PARI) See Links section.
%o A355916 (Python)
%o A355916 from collections import Counter
%o A355916 def aupton(terms):
%o A355916     num, alst, inventory = 0, [0, 0], Counter([0, 0])
%o A355916     for n in range(3, 3+terms//2):
%o A355916         c = [inventory[num], num]
%o A355916         num = 0 if c[0] == 0 else num + 1
%o A355916         alst.extend(c)
%o A355916         inventory.update(c)
%o A355916     return alst[:terms]
%o A355916 print(aupton(128)) # _Michael S. Branicky_, Sep 25 2022
%Y A355916 Cf. A342585, A355917, A355918.
%K A355916 nonn
%O A355916 1,3
%A A355916 _N. J. A. Sloane_, Sep 24 2022