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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. 4
1, 12, 4608, 1728, 1260, 509607936, 2985984, 144, 56358560858112, 5159780352, 302400, 6232805962420322304, 207360000, 887040, 201226394483583074212773888, 15407021574586368, 248832, 2286144000, 26623333280885243904, 522547200, 8430527379596857675529996470321152 (list; graph; refs; listen; history; text; internal format)
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^((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
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
Sequence in context: A288967 A361106 A009094 * A236067 A134821 A229669
KEYWORD
nonn
AUTHOR
Labos Elemer, Jun 18 2001
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
STATUS
approved

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Last modified June 25 18:08 EDT 2024. Contains 373707 sequences. (Running on oeis4.)