|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
The squares of the primes == 1 (mod 6), squares of A002476, are a subsequence: 49, 169, 361,... - R. J. Mathar, May 19 2020
|
|
LINKS
|
Eric Weisstein's World of Mathematics, Divisor
|
|
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
|
|
|
STATUS
|
approved
|
|
|
|