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A356784
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Inventory of positions as an irregular table; row 0 contains 0, subsequent rows contain the 0-based positions of 0's, followed by the position of 1's, of 2's, etc. in prior rows flattened.
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12
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0, 0, 0, 1, 0, 1, 2, 3, 0, 1, 2, 4, 3, 5, 6, 7, 0, 1, 2, 4, 8, 3, 5, 9, 6, 10, 7, 12, 11, 13, 14, 15, 0, 1, 2, 4, 8, 16, 3, 5, 9, 17, 6, 10, 18, 7, 12, 21, 11, 19, 13, 22, 14, 24, 15, 26, 20, 23, 25, 28, 27, 29, 30, 31, 0, 1, 2, 4, 8, 16, 32, 3, 5, 9, 17, 33
(list;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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0,7
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COMMENTS
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The n-th row contains A011782(n) terms, and is a permutation of 0..A011782(n)-1.
The leading term of each row is 0, and is followed by powers of 2, and then by positive nonpowers of 2.
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LINKS
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FORMULA
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a(n) = 1 iff n belongs to A000051 \ {2}.
a(n) = 2 iff n belongs to A052548 \ {3, 4}.
a(n) = 3 iff n belongs to A005126 \ {2, 4}.
T(n, 0) = 0.
T(n, k) = 2^(k-1) for k = 1..n-1.
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EXAMPLE
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Table begins:
0,
0,
0, 1,
0, 1, 2, 3,
0, 1, 2, 4, 3, 5, 6, 7,
0, 1, 2, 4, 8, 3, 5, 9, 6, 10, 7, 12, 11, 13, 14, 15,
...
For n = 5:
- the terms in rows 0..4 are: 0, 0, 0, 1, 0, 1, 2, 3, 0, 1, 2, 4, 3, 5, 6, 7,
- we have 0's at positions 0, 1, 2, 4, 8,
- we have 1's at positions 3, 5, 9,
- we have 2's at positions 6, 10,
- we have 3's at positions 7, 12,
- we have one 4 at position 11,
- we have one 5 at position 13,
- we have one 6 at position 14,
- we have one 7 at position 15,
- so row 5 is: 0, 1, 2, 4, 8, 3, 5, 9, 6, 10, 7, 12, 11, 13, 14, 15.
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PROG
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(C++) See Links section.
(Python)
terms = [0, ]
for i in range(1, 10):
new_terms = []
for j in range(max(terms)+1):
for k in range(len(terms)):
if terms[k] == j: new_terms.append(k)
terms.extend(new_terms)
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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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STATUS
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approved
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