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A109429 Permutation of the A050376 numbers; such 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 (list; graph; refs; listen; history; internal format)
OFFSET

1,1

COMMENTS

A073904(2^n) is the product of the first n members of this sequence. Problem Generalite A073904(p^n) analogous: permutation of the numbers of the form q^(p^k), such that a(p^j)=p^(p^j); A073904(p^n)=(produkt of the first n members)^(p-1). - David WASSERMAN & Tom Ordowski

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. Yes? For n>4 all right? Yes!

CROSSREFS

Cf. A050376.

Sequence in context: A079308 A189825 A115399 * A114894 A183169 A053124

Adjacent sequences:  A109426 A109427 A109428 * A109430 A109431 A109432

KEYWORD

nonn

AUTHOR

Tomasz Ordowski (ordot(AT)poczta.onet.pl), Aug 26 2005

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Last modified February 16 06:46 EST 2012. Contains 205867 sequences.