

A122842


a(n) = k such that A038547[2n1] = k^2.


2



1, 3, 9, 27, 15, 243, 729, 45, 6561, 19683, 135, 177147, 225, 105, 4782969, 14348907, 1215, 675, 387420489, 3645, 3486784401, 10460353203, 315, 94143178827, 3375, 32805, 2541865828329, 6075, 98415, 68630377364883
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OFFSET

1,2


COMMENTS

A038547[n] is least number with exactly n odd divisors. It appears that for n>1 a(n) is divisible by 3^m where m(n) = {0,1,2,3,1,5,6,2,8,9,3,11,2,1,14,15,5,...}. a(n) = 3^(n1) for n = (p+1)/2, where p is odd prime.
a(3k+2) = 5*3^k for k = {1,2,3,5,6,8,9,11,14,15,...} = A005097[n] (Odd primes  1)/2. Thus a(3(p1)/2 + 2) = 5*3^((p1)/2). a(9k+5) = 3^k*5*7 for k ={1,2,3,5,...}. 3 divides a(2), 5 divides a(5), 7 divides a(14), 11 divides a(41), 13 divides a(122), etc. The first occurrence of the odd prime divisor Prime[n] among divisors of a(n) occurs at n = {2,5,14,41,122,...} = A007051[n1] = (3^(n1) + 1)/2 for n>1. Thus a(A007051[n1]) = A070826[n] or a(3^(n1) + 1)/2) = p(n)#/2 = A002110[n]/2 for n>1.


LINKS

R. J. Mathar, Table of n, a(n) for n = 1..1000


FORMULA

a(n) = Sqrt[ A038547[2n1] ]. a(n) = Sqrt[ A119265[2n1,2n1] ].


EXAMPLE

A038547[n] begins {1,3,9,15,81,45,729,105,225,405,59049,315,...}.
a(1) = 1 because A038547(1) = 1.
a(2) = 3 because A038547(3) = 9.
a(5) = 15 because A038547(9) = 225.


CROSSREFS

Cf. A038547, A121858, A001227, A005179, A119265.
Cf. A007051, A070826, A002110.
Sequence in context: A070359 A271015 A131997 * A070358 A321542 A321540
Adjacent sequences: A122839 A122840 A122841 * A122843 A122844 A122845


KEYWORD

nonn


AUTHOR

Alexander Adamchuk, Sep 13 2006, Sep 25 2006


EXTENSIONS

More terms from R. J. Mathar, Sep 20 2006


STATUS

approved



