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A109429
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Rearrange terms of A050376 so that a(2^j)=2^(2^j) for j>=0.
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1
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2, 4, 3, 16, 5, 7, 9, 256, 11, 13, 17, 19, 23, 25, 29, 65536, 31, 37, 41, 43, 47, 49, 53, 59, 61, 67, 71, 73, 79, 81, 83, 4294967296, 89, 97
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OFFSET
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1,1
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COMMENTS
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A073904(2^n) is the product of the first n members of this sequence. Generalization: for any prime p, we may consider the analogous permutation of numbers of the form q^(p^k) such that a(p^j)=p^(p^j); then A073904(p^n)=(product of the first n members)^(p-1). - David Wasserman and Thomas Ordowski. Corrected by Thomas Ordowski, Jun 06 2015
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LINKS
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FORMULA
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a(2^j)=2^(2^j). So a(1)=2 for j=0; a(2)=4 for j=1; a(4)=16 for j=2.
A073904(2^n)=2*4*3*...*a(n) for every n.
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EXAMPLE
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Numbers: 2, 3, 2^2, 5, 7, 3^2, 11, 13, 2^(2^2), 17, ..., 2^(2^3), ...
Permutation: 2, 2^2, 3, 2^(2^2), 5, 7, 3^2, 2^(2^3), 11, 13, 17, ...
If n=4 then A073904(16)=2*4*3*16=384.
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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