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A277793
Numbers k such that the arithmetic and geometric means of the divisors of k are both integers.
3
1, 49, 169, 361, 961, 1369, 1849, 3721, 4489, 5329, 6241, 8281, 9409, 10609, 11881, 14641, 16129, 17689, 19321, 22801, 24649, 26569, 32761, 37249, 39601, 44521, 47089, 49729, 52441, 58081, 61009, 67081, 73441, 76729, 80089, 87616, 90601, 94249, 97969, 109561, 113569, 121801, 134689
OFFSET
1,2
COMMENTS
Intersection of A000290 and A003601.
Union of squares of A107924 and squares of A107925.
The squares of the primes == 1 (mod 6), squares of A002476, are a subsequence: 49, 169, 361,... - R. J. Mathar, May 19 2020
LINKS
EXAMPLE
a(2) = 49 because 49 has 3 divisors {1,7,49} therefore (1 + 7 + 49)/3 = 19 and (1*7*49)^(1/3) = 7 are both integers.
MATHEMATICA
Select[Range[140000], Divisible[DivisorSigma[1, #1], DivisorSigma[0, #1]] && Mod[DivisorSigma[0, #1], 2] == 1 & ]
Select[Range[150000], AllTrue[{Mean[Divisors[#]], GeometricMean[ Divisors[ #]]}, IntegerQ]&] (* The program uses the AllTrue function from Mathematica version 10 *) (* Harvey P. Dale, Nov 21 2018 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Ilya Gutkovskiy, Oct 31 2016
STATUS
approved