1,1

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 ORDO

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.

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!

Cf. A050376.

Sequence in context: A091861 A079308 A115399 this_sequence A114894 A053124 A071970

Adjacent sequences: A109426 A109427 A109428 this_sequence A109430 A109431 A109432

nonn

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