1,2

a(11), if it exists, is greater than 10^12. - Ryan Propper (rpropper(AT)stanford.edu), Oct 10 2005

Comments from Lambert Klasen (lambert.klasen(AT)gmx.net), Oct 25 2005: "Sequence is infinite. For a prime p, a(p) has p^p as a factor. Factoring the a(n) gives the pattern for the exponents:

[2, 1]

[2, 2]

[2, 1; 3, 3]

[2, 5; 3, 1]

[2, 2; 3, 1; 5, 5]

[2, 2; 3, 1; 5, 1]

[2, 2; 3, 1; 5, 1; 7, 7]

[2, 10; 3, 1; 5, 1; 7, 1]

[2, 3; 3, 10; 5, 1; 7, 1]

[2, 3; 3, 2; 5, 1; 7, 1]

[2, 3; 3, 2; 5, 1; 7, 1; 11, 11]

[2, 3; 3, 2; 5, 1; 7, 1; 11, 1]

[2, 3; 3, 2; 5, 1; 7, 1; 11, 1; 13, 13]

[2, 3; 3, 2; 5, 1; 7, 1; 11, 1; 13, 1]

[2, 3; 3, 2; 5, 1; 7, 1; 11, 1; 13, 1]

[2, 19; 3, 2; 5, 1; 7, 1; 11, 1; 13, 1]

[2, 4; 3, 2; 5, 1; 7, 1; 11, 1; 13, 1; 17, 17]

[2, 4; 3, 2; 5, 1; 7, 1; 11, 1; 13, 1; 17, 1]

[2, 4; 3, 2; 5, 1; 7, 1; 11, 1; 13, 1; 17, 1; 19, 19]

[2, 4; 3, 2; 5, 1; 7, 1; 11, 1; 13, 1; 17, 1; 19, 1]

[2, 4; 3, 2; 5, 1; 7, 1; 11, 1; 13, 1; 17, 1; 19, 1]

[2, 4; 3, 2; 5, 1; 7, 1; 11, 1; 13, 1; 17, 1; 19, 1]

[2, 4; 3, 2; 5, 1; 7, 1; 11, 1; 13, 1; 17, 1; 19, 1; 23, 23]."

(1*4*54*96)^(1/4)= (20736)^(1/4) = 12.

a(5)=37500=2^2*3*5^5

a(11)=718985409939720=2^3*3^2*5*7*11^11

p = 1; Do[k = 1; While[ !IntegerQ[(p*k*n)^(1/n)], k++ ]; Print[k*n]; p *= (k*n), {n, 1, 10}] (Propper)

Cf. A095209, A095211.

Sequence in context: A111034 A109801 A099340 this_sequence A001545 A073863 A089205

Adjacent sequences: A095207 A095208 A095209 this_sequence A095211 A095212 A095213

nonn

Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Jun 08 2004

More terms from Ryan Propper (rpropper(AT)stanford.edu), Oct 10 2005

a(11) onwards from Lambert Klasen (lambert.klasen(AT)gmx.net), Oct 25 2005