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A361287
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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.
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1
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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
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OFFSET
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0,5
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LINKS
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FORMULA
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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.
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EXAMPLE
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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.
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MATHEMATICA
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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 *)
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PROG
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(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])
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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