

A061701


Smallest number m such that GCD of d(m^2) and d(m) is 2n+1 where d(m) is the number of divisors of m.


3



1, 12, 4608, 1728, 1260, 509607936, 2985984, 144, 56358560858112, 5159780352, 302400, 6232805962420322304, 1587600, 900900, 201226394483583074212773888, 15407021574586368, 248832
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OFFSET

0,2


COMMENTS

a(n) exists for every n. In other words, every positive odd integer k is equal to the GCD of d(m^2) and d(m) for some m. To see this, let m = 2^(k^2  1) * 3^((k1)/2). Then d(m) = k^2 * (k+1)/2 and d(m^2) = (2 k^2  1) * k. Both of these are divisible by k and (8k4) d(m)  (2k+1) d(m^2) = k, so the GCD is k.  Dean Hickerson, Jun 23 2001


LINKS



FORMULA

a(n) = Min[m : GCD[d(m^2), d(m)] = 2n+1].


EXAMPLE

For n = 7, GCD[d(20736),d(144)] = GCD[45,15] = 15 = 2*7+1.


CROSSREFS



KEYWORD

nonn,more


AUTHOR



EXTENSIONS



STATUS

approved



