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.

1, 12, 4608, 1728, 1260, 509607936, 2985984, 144, 56358560858112, 5159780352, 302400, 6232805962420322304, 207360000, 887040, 201226394483583074212773888, 15407021574586368, 248832, 2286144000, 26623333280885243904, 522547200, 8430527379596857675529996470321152

Offset: 0

Author

Labos Elemer, Jun 18 2001

Keywords

• nonn

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^((k-1)/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 (8k-4) d(m) - (2k+1) d(m^2) = k, so the GCD is k. - Dean Hickerson, Jun 23 2001

All the terms are in A025487 because A061680(m) = gcd(d(m^2), d(m)) depends only on the prime signature of m. - Amiram Eldar, Nov 26 2023

Examples

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

Links

• Amiram Eldar, Table of n, a(n) for n = 0..28

Crossrefs

Cf. A000005, A000290, A025487, A048691, A061680.

Formula

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

Extensions

More terms from David Wasserman, Jun 20 2002

a(12)-a(13) corrected and a(17)-a(20) added by Amiram Eldar, Nov 26 2023