A361287 A variant of the inventory sequence A342585: now a row ends when the number of occurrences of the largest term in the sequence thus far has been recorded.
0, 1, 1, 1, 3, 0, 1, 2, 4, 1, 1, 1, 2, 7, 2, 1, 1, 0, 0, 1, 4, 10, 3, 2, 2, 0, 0, 1, 0, 0, 1, 8, 12, 5, 2, 2, 1, 0, 1, 1, 0, 1, 0, 1, 11, 17, 7, 2, 2, 1, 0, 2, 1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 17, 23, 10, 2, 2, 1, 0, 2, 1, 0, 2, 1, 1, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0
Offset: 0
Examples
The triangle begins: 0; 1, 1; 1, 3, 0, 1; 2, 4, 1, 1, 1; 2, 7, 2, 1, 1, 0, 0, 1; 4, 10, 3, 2, 2, 0, 0, 1, 0, 0, 1; 8, 12, 5, 2, 2, 1, 0, 1, 1, 0, 1, 0, 1; 11, 17, 7, 2, 2, 1, 0, 2, 1, 0, 1, 1, 1, 0, 0, 0, 0, 1; 17, 23, 10, 2, 2, 1, 0, 2, 1, 0, 2, 1, 1, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 1; ... a(0)=0 because we begin each row by recording the number of zeros in the sequence. As yet, there are none. 0 is still the largest term in the sequence, so we begin the next row. a(1)=1 because there is one 0. a(2)=1 because there is one 1. 1 is the largest term thus far, so we begin the next row. a(3)=1 (there is still one zero); a(4)=3 (there are three 1s); a(5)=0 (no 2s); the row ends with a(6)=1 because there is one 3 and three is the largest term thus far.
Links
- Samuel Harkness, Table of n, a(n) for n = 0..12000
Programs
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Mathematica
len = 84; K = {0}; While[Length@K <= len, i = 0; While[i <= Max@K, AppendTo[K, Count[K, i]]; i++]]; Print[Take[K, len + 1]] (* Samuel Harkness, Mar 12 2023 *) -
Python
from collections import Counter from itertools import count, islice def agen(): # generator of terms an, b, m, inventory = 0, 0, 0, Counter([0]) for n in count(1): yield an an = inventory[b] m = max(m, an) b = b + 1 if m > b else 0; inventory.update([an]) print(list(islice(agen(), 85))) # Michael S. Branicky, Mar 07 2023
Formula
a(n) is the number of all a(k) == b(n) 0 <= k < n and b(0) = 0 and b(n) = b(n-1)+1 if an a(k) > b(n) exists for 0 <= k < n, and 0 otherwise.
Extensions
More terms from Alois P. Heinz, Mar 07 2023