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A367555
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Number of zeros (or ones) in each row of the iterates of the Christmas tree pattern map (A367508).
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4
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1, 1, 3, 3, 3, 6, 2, 6, 2, 6, 6, 10, 5, 5, 10, 5, 5, 10, 5, 10, 10, 15, 3, 9, 3, 9, 9, 15, 3, 9, 3, 9, 9, 15, 3, 9, 9, 15, 9, 15, 15, 21, 7, 7, 14, 7, 7, 14, 7, 14, 14, 21, 7, 7, 14, 7, 7, 14, 7, 14, 14, 21, 7, 7, 14, 7, 14, 14, 21, 7, 14, 14, 21, 14, 21, 21, 28
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listen;
history;
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internal format)
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OFFSET
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1,3
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COMMENTS
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See A367508 for the description of the Christmas tree patterns, references and links.
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LINKS
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EXAMPLE
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The following diagram shows the first 4 tree pattern orders, along with the corresponding number of zeros = number of ones.
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Order 1: |
0 1 | 1
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Order 2: |
10 | 1
00 01 11 | 3
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Order 3: |
100 101 | 3
010 110 | 3
000 001 011 111 | 6
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Order 4: |
1010 | 2
1000 1001 1011 | 6
1100 | 2
0100 0101 1101 | 6
0010 0110 1110 | 6
0000 0001 0011 0111 1111 | 10
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MATHEMATICA
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With[{imax=9}, Map[Total, NestList[Map[Delete[{If[Length[#]>1, Rest[#], Nothing], Join[{First[#]}, #+1]}, 0]&], {{0, 1}}, imax-1], {2}]] (* Generates terms up to order 9 *)
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PROG
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(Python)
from itertools import islice
from functools import reduce
def uniq(r): return reduce(lambda u, e: u if e in u else u+[e], r, [])
def agen(): # generator of terms
R = [["0", "1"]]
while R:
r = R.pop(0)
yield sum(e.count("1") for e in r)
if len(r) > 1: R.append(uniq([r[k]+"0" for k in range(1, len(r))]))
R.append(uniq([r[0]+"0", r[0]+"1"] + [r[k]+"1" for k in range(1, len(r))]))
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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