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A109429
Rearrange terms of A050376 so that a(2^j)=2^(2^j) for j>=0.
1
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
OFFSET
1,1
COMMENTS
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
FORMULA
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.
EXAMPLE
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.
CROSSREFS
Cf. A050376.
Sequence in context: A259476 A271363 A115399 * A358820 A366027 A114894
KEYWORD
nonn
AUTHOR
Thomas Ordowski, Aug 26 2005
EXTENSIONS
Definition edited by N. J. A. Sloane, Oct 27 2014
More terms from Thomas Ordowski, Jun 05 2015
STATUS
approved